Cha pt e r:7Poi nt e r s 7. 1 7. 2
7 . 3
7. 4 7. 5 7. 6 7. 7 7. 8 7. 9 7. 10 7. 11
7. 12 7. 13
7. 14 7 . 1 5 7. 16
I nt r oduc t i on Dec l ar at i onofPoi nt er s 7. 2. 1 Ref er enceOper at or( &) 7 . 2 . 2 P oi nt e rv a r i a bl e s 7 . 2 . 3 S i z eo fpo i nt e r s 7 . 2 . 4 De r e f e r e n c eOpe r a t o r( * ) I ni t i a l i z a t i o no fP oi nt e r s 7. 3. 1 Nul lPoi nt er 7. 3. 2 Voi dPoi nt er Poi nt e rAr i t hm me e t i c Poi nt e r sa ndAr r a ys Poi nt e r sandSt r i ngs Mul t i di mens i onalar r a ysa ndpoi nt er s Poi nt er st os t r uc t ur es Poi nt ert oobj ec t s Poi nt er sandc ons t ant s Poi nt e r sa ndFunc t i onar gume nt 7 . 1 1 . 1Ca l lb yv a l ue 7 . 1 1 . 2Ca l lb yr e f e r e nc e 7 . 1 1 . 3Ca l lb ypoi nt e r s 7 . 1 1 . 4Func t i onr e t ur ni ngPoi nt e r s Poi nt ert oPoi nt er s Dynam mi i cme mem mo or yal l oca t i on/deal l oca t i onoper at or s 7. 13. 1ne w oper at or 7 . 1 3 . 2de l e t eope r a t or 7 . 1 3 . 3Poi nt e r sa ndobj e c t s 7. 13 . 4Me mor yLea ks Thi sP oi nt e r S e l fRe f e r e nt i a ls t r uc t ur e s Ref er enc evar i abl es
7 . 1I nt r oduc t i on C++ i saver yr i c hl anguagewi t hs omany f eat ur es. Poi nt eri soneoft hepowe wer f ul f e a t ur eo fC++. P oi nt e r sa l l o ws wsdi r e c tde a l i ngwi t hme mo mo r ya ddr e s s e sa ndhe l psi n c r e a t i nga ndma ni pul a t i ngdyna mi mi cda t as t r uc t ur e sl i kel i nke dl i s t, que ue sa nds t a c ks . De fini ngpo i nt e r sa n dus i ngt he mt oa c c e s st hev a r i a b l epo i nt e db yt he mi sa ni mpo r t a nt c o nc e pt ,t hor oughl yc o ve r e di nt hi sc ha pt e r .Theba s i cpoi nt e rc o nc e pt s ,poi nt e r swi wi t h ot he rf e at ur e sofC++l i kea r r a ys ,s t r i ngs ,f unc t i ons ,s t r uc t ur e sa ndobj e c tva r i a bl e sar e a l s oe xpl a i ne di nt hi sc ha pt e r . Adv a nt a g e so fpo i nt e r s
1.P .Poi nt eri mpr ovet heeffic ffic i encyoft hece r t ai nr out i nesbys har i ngt hem bet wee n numbe rofe n t i t i e s .I ta l l o ws wsdi ffe r e nts e c t i onsoft hec o det os h ar ei nf or ma t i on e a s i l y 2 . Poi nt e r se na bl ec ompl e x" l i nke d"da t as t r uc t ur e sl i kel i nke dl i s t sa ndbi na r yt r e e s &gr a phs 3. .I I ti sa wayt hr ough whi c h mem mo or yl oca t i on oft he var i abl esc an bedi r ec t l y mani pul at edasandwhenr equi r ed
7 . 2 De c l a r a t i onofPoi nt e r s I ne a r l i e rc h apt e r s, Weha v ea l r e a dydi s c us s e dho wv a r i a bl e sc a nbede fine di nC++. Wec a nde finei nt e ge rv ar i a bl e,floa t, c ha r a c t e r, s t r uc t ur e, orc l a s sv ar i a bl e.Ava r i a bl e i napr ogr am i ss ome met hi ngwi wi t haname me& t heval ueofwhi c hcanvar y. c o ns i de rt hef ol l o wi wi ngs t a t e me me nt: i nt x ; Thi ss t a t e me me ntde c l a r e sava r i a bl exwh whi c hc a nt a kei nt e ge rva l ue s. Thi ss t a t e me me nta l s o i mpl i e st ha tt hes pa c eof2byt e si sr e s e r v edi nt heme mo mor yf ort heva r i a bl e‘ x ’ . x=57 ; t hi ss t a t e m me e nta s s i gnt heva l ue57i nt heva r i a bl e‘ x ’ . Al lt hevar i abl esdefinedar es t or edi nt hec omp mput er ’ sme memo mor y.Theme memo mor yof c o mp mput e rc a nbei ma gi ne da sas uc c e s s i o no fme mo mo r yc el l s , e a c ho neo ft hemi mi ni ma ls i z e
o fo neb yt e . The s es i ngl e b yt eme me mo mo r yc e l l sa r enumbe r e di nac o ns e c u t i v ewa wa y ,s oa s , wi t hi na n yb l o c ko fme mo mo r y ,e v e r yc e l lha st hes a me menumb e ra st hepr e v i o uso ne, pl us one.
Ev e r yv a r i a b l ede fine dha su ni quea ddr e s si nt heme me mo mo r y.l i k ec o ns i de rt hi sd i a gr a m:
Thev a r i a b l exha st heme me mo mo r ya ddr e s s1 0 0 3a n dha st hev a l ue5 7 .Thea ddr e s so ft he v a r i a bl ec a nb eobt a i ne db ya na ddr e s so pe r a t or‘ &’( o ra l s okno wn wna sr e f e r e nc e o pe r a t or ). Thi so pe r a t orwhe na ppl i e dt ov a r i a bl e sg i v e st heme me mo mor ya ddr e s so ft ha t v a r i a bl e .s ot hef o l l o wi wi ngpr o gr a ms e gme nt x=57; c o ut<<” Va l ueo fxi s“ <
7 . 2 . 1
Re f e r e nc eope r a t or( &)
Whe nwe wede c l a r ev a r i a b l e , t hea mo mo unto fme mo mo r yi sde fine df o rt ha tv a r i a b l e de fine da ndt heuni quea ddr e s si sa s s i gne dt ot ha tv a r i a bl eb yt hec o mp mpi l e rdur i ng r unt i meo rc o mp mpi l et i me. S ome t i me s, wewa ntt oope r a t eo nt hea ddr e s soft he s ev ar i a bl e si nor de rt oope r a t e wi t hr e l at i vepos i t i onst oi t .
Thea ddr e s st ha tl oc a t e sav ar i a bl ewi t hi nme mor yi sc a l l e da sr e f e r e nc et ot ha t va r i a bl e .Thi sr e f e r e nc et oava r i a bl ec a nbeobt a i ne db ypr e c e di ngt hei de nt i fie ra n a mpe r s a nds i gn( &) , kno wna sr e f e r e nc eo pe r a t o ro ra ddr e s so pe r a t o r ,a ndi tc a nb e l i t e r a l l yt r a ns l a t e da s" a ddr e s sof " . I ti sauna r yope r a t orwhi c hgi ve st headdr e s sof i t sope r a nd. Thea ddr e s s&ope r a t ora ppl i e st oobj e c t si nme mor yt ha ti sv a r i a bl e s, a r r a y e l e me nt s, s t r uc t ur ea ndc l a s sobj e c t s. I tc a nnotbea ppl i e dt oe xpr e s s i ons , c ons t a nt s , orr e gi s t e rvar i abl e s . i nty=5 ; c out<<&y; wi l ldi s pl a yt hea ddr e s so ft heva r i a bl ey .
7 . 2. 2
Poi nt erva r i a bl e s
As s umet ha tt he r ei sa n ot he rv ar i a bl ept rwhi c hho l dst hea ddr e s so fxme a ns: i nt * pt r ; pt r=&x; t hi ss t a t e me nta s s i gnst hea ddr e s so fxt ot heva r i a bl ept ra nd pt ri ss ai dt o‘ poi ntt o’x.
Pt ri sc a l l e da spoi nt e r. I tdoe snots t or es i mpl eva l ue s. i ns t e adoft hi si ts t or er e f e r e nc e ora ddr e s st oa no t he rv a r i a bl e.A poi nt e rhol dst hea ddr e s soft heve r yfir s tb yt eoft he me mor yl oc a t i onwhe r ei ti spoi nt i ngt o.t hea ddr e s soft hefir s tb yt ei sa l s okno wna s ba s ea ddr e s s .
De f Apoi nt e ri sav a r i a bl ewhi c hc o nt a i nst hea ddr e s sofa not he rv a r i a bl e .
pt ri sapo i nt e rv ar i a bl ewhi c hc o nt a i nst hea ddr e s sofs omeo t he rv a r i a bl e(o ft ype i nt e g e ri nt hi sc a s e ).The nob vi ous l yapoi nt e rv ar i a bl ei sno tt hes a met ypea si nt e g e r v ar i a bl e.t he nho wt he ya r ede c l a r e dl i ke
whe r ept ri snota ni nt e g e rva r i a bl ebutpt rc a nhol dt headdr e s so ft hei nt e ge rv a r i a bl e poi nt e rt oa ni nt e ge rv a r i a bl e ’ i . e .i ti sa‘ , butt hede c l a r a t i onofpoi nt e rv ar i a bl edoe s no tma k et he m po i ntt oa n yl o c a t i o n.
i nti =5 7
i
57
a s s i gni ngaddr e s so fit opt r,pt r=& i ; wi l lr e s ul ti n
Thi si st hei ni t i a l i z at i on oft hepoi nt e rva r i a bl e .s i nc et hepoi nt e rc an poi ntt oa ny por t i onoft heme mor y, i ti sago odpr ogr a mmi ngpr a c t i c et oi ni t i a l i z et hepoi nt e r. ( Ar r o wss h owni nt hefigur e sa r ej us tt h el o gi c a li l l us t r a t i o no ft h ec o nc e p to t he r wi s epo i nt e r c o nt a i nst hea ddr e s so ft hev ar i a bl e, whi c hi sa s s i gne dt oi t . ) wec a na l s ode c l a r et hepoi nt e rva r i a bl e so fa nyo t he rda t at ype s. F ore x ampl e : do ubl e*dpt r ; c ha r*c h; floa t*f pt r ; The s ea r et hr e ede c l a r a t i onso fpoi nt e r s .Ea c honei si nt e nde dt opoi ntt oadi ffe r e ntda t a t ype , buti nf a c ta l lo ft he ma r epo i nt e r sa nda l loft he m wi l lo c c upyt hes a mea mountof s pac ei nmemor y( t hes i zei nmemor yofapoi nt erdependsont hepl at f or m wher et he c o dei sg oi ngt or un) .Ne v e r t he l e s s ,t heda t at owhi c ht he ypoi ntt odono toc c upyt he s a mea mo unto fs pa c en ora r eoft hes a met y pe :t hefir s to nepo i nt st oa ndo ub l e,t he s e c o ndonet oac h ara n dt hel a s to net oaflo at .The r e f o r e , a l t hought he s et hr e ee x a mpl e v a r i a bl e sa r ea l lo ft he m po i nt e r swhi c ho c c upyt hes a mes i z ei nme mor y ,t he ya r es a i d t oha v edi ffe r e ntt ype s :i nt * ,c ha r *a ndfloa t *r e s pe c t i v e l y ,de pe ndi ngont het ypet he y poi ntt o.
7 . 2 . 3
Si z eofpoi nt e r s
Thes i z eofapo i nt e ri sde pe nde ntupont hea r c h i t e c t ur eoft hec o mput e r— a3 2 b i t c o mput e rus e s3 2 b i tme mo r ya ddr e s s e s— c o ns e q ue n t l y , apo i nt e ro na1 6 b i tma c h i ne i s1 6bi t s( 2byt e s ) ., ona32bi tma c hi ne,apoi nt e rwoul dbe32bi t si . e .4byt e s , Ona64 bi tmac hi ne ,apoi nt e rwoul dbe64bi t s( 8byt e s ) . Ex ampl e: 7 . 1Showi ngt hes i z eo fpoi nt e r s # i nc l ude
# i nc l ude
i nt* i pt r ; c har* c pt r ; floa t* f pt r ; voi dmai n( ) { c l r s c r ( ) ; c out <<" s i z eofi nt e ge rpoi nt e ri s: " <
No t et ha tt hes i z eofpoi nt e rdoe sno tde pe ndupont heda t at ypeo ft hev al uei ti s poi nt i ngt o. I tc a nbenot i c e df r om t hi se x a mpl et ha t, t hes i z eoft hepoi nt e ri sa l wa yst hes a me .Thi s i sb e c a us eapo i nt e ri sj us tame mo r ya ddr e s s ,a ndt henumb e ro fbi t sn e e de dt oa c c e s sa memor yaddr es sonagi venmac hi nei sal waysc ons t ant . Poi nt er sar ea ver y power f ulf ea t ur eoft he C++ l anguaget hathasmany us esi n advanc edpr ogr ammi ng.Fur t herahead,wewi l ls eehow t hi st ypeofvar i abl ei sus ed a ndde c l a r e d.
7 . 2. 4
De r e f er enc eope r a t or( * )
Apoi nt e rva r i a bl ec a nbeus e ddi r e c t l yt oa c c e s st hev al ues t or e di nt hev ar i a bl ewhi c h i tpoi nt st o.Fort hi s , wes i mpl yha vet opr e c e det hepoi nt e r ' si de nt i fie rwi t ha na s t e r i s k ( * ) ,whi c ha c t sa si ndi r e c t i onorde r e f e r e nc eope r a t ora ndt ha tc a nbel i t e r a l l yt r a ns l a t e d t o" v al uepoi nt e db y" . De f Theuna r yope r a t or( * )i st hede r e f e r e nc i ngpoi nt e rori ndi r e c t i onpoi nt e r, whe n a ppl i e dt ot hepoi nt e rc a na c c e s st heva l uet hepoi nt e rpoi nt st o.
The r e f or e ,f ol l o wi ngwi t ht heva l ue soft hepr e vi ouse x a mpl e ,i fwewr i t e : c o ut<<*pt r
Thi swi l ldi s pl a yt heva l ueoft heva r i a bl eia s5 7. whi c hi ss a mea s c o ut <
c o ut<
Vva l ueo fii s: 1 5 Addr e s so fi i s: 1 00 3 Va l ueofi :1 5 Addr e s sofii s:1 00 3 Addr e s sofyi s: 2 7 12 i no t he rwo r dswhe npt ri sa s s i gne dwi t ht hea ddr e s so fit he n: Pt ri st hes a mea s&i * pt ri st hes a mea si
Now haveal ookatt hi sc ode: Ex ampl e:7 . 2
Showi ngt heus eofp oi nt e r s .
# i nc l ude i nt mai n( ) { i nt fir s t ,s e c o nd; i nt *mypo i nt e r ; mypoi nt er=&fir s t ; * mypoi nt er=10; mypoi nt er=&s ec ond; * mypoi nt er=20; c out<<" fir s tva l uei s" <<fir s t<
o u t p u t
fir s tva l uei s1 0 s e c ondv al uei s2 0 De r e f e r e nc e dpoi nt e rma yal s obeus e do nt hel e f ts i deoft hea s s i gnme ntope r a t orl i ke
* pt r=1 0 whi c hmakest heval ueofyas10 .
ptr 10 10 03 03
i
10
I ti sa l s ous e dt or e adi nt heva l ueofif r o mt heus e rl i ke c i n>>* pt r ; t heva l uee nt e r e db yt heus e ri ss t or e di nt heva r i a bl ei .
Re me mbe r a ddr e s sope r a t or( &)a ndde r e f e r e nc i ngope r a t or( * )a r euna r yope r a t orwhi c har e e va l ua t e dr i ghtt ol e f ti nt hee xpr e s s i on.
MYNOTES ntk ; Av a r i a b l ei sde c l a r e db ygi vi ngi tat ypea ndana me( e . g.i ) Apo i nt e rv a r i a b l ei sde c l a r e db ygi vi ngi tat ypea ndana me( e . g.i )wh e r e nt* p t r t hea s t e r i s kt e l l st hec o mpi l e rt ha tt h ev a r i a bl ena me dt sapo i nt e rv a r i a bl ea nd p ri t het ypet e l l st hec o mpi l e rwha tt ypet hep o i nt e ri st opo i ntt o( i nt e g e ri nt hi s c a s e ) . Onc eav a r i a b l ei sd e c l a r e d, wec a ng e ti t sa d dr e s sb ypr e c e di ngi t sna mewi t ht he u na r y& o pe r a t o r ,a si n&k. • •
•
• • •
•
Thes i z eo fapo i nt e ri sd e pe nde ntupo nt h ea r c h i t e c t ur eo ft hec o mput e r. Thes i z eo ft hep o i nt e ri sa l wa yst hes a mei r r e s p e c t i v eo ft hed a t at ypei ti sp o i nt i ngt o .
Wec a n"de r e f e r e nc e "apo i nt e r ,i . e . r e f e rt ot hev a l ueo ft ha twhi c hi tp o i nt st o ,b y us i ngt heuna r y' * 'o pe r a t o ra si n* . p t r An"l v a l ue "o fav a r i a b l ei st hev a l ueo fi t sa ddr e s s , i . e . whe r ei ti ss t o r e di n
me mo r y .The"r v a l ue "o fav a r i a bl ei st hev al ues t o r e di nt ha tv a r i a b l e( a tt ha t a ddr e s s ) . &i st h er e f e r e nc eo p e r a t o ra ndc a nb er e a da s"a ddr e s so f " *i st hed e r e f e r e nc eo pe r a t o ra ndc a nb er e a das" va l uep o i nt e dby" Av a r i a b l er e f e r e nc e dwi t h& c a nb ede r e f e r e nc e dwi t h* . • • •
7 . 3I ni t i a l i z a t i onofpoi nt e r s consider the following snippet : int x,y,*ptr; x=57; y=*ptr;
here ptr is used without being assigned a proper address
yi sa s s i gne dava l uewha t e v erpt ri spoi nt i ngt o. butpt ri snotpoi nt i ngt oa nywhe r e, i t ma yc ont ai nt headdr es sofanypor t i onoft hememor ywhi c hi sunknownor uni ni t i a l i ze ds o, t he r ei sapos s i bi l i t yyi sa s s i gne daj unkval ue . Ensure that pointers are assigned and initialized with valid addresses before their usage.
P oi nt e rc a nb ei ni t i a l i z e dwi t ha0o rNULLo rava l i da ddr e s s Whe ny oude c l a r et hepoi nt e r si tne e dt oe xpl i c i t l yi ni t i a l i z e dwi t ht heva l i da ddr e s s i ntnum; i nt*pt r ; pt r =#
/ /poi nt e ri si ni t i a l i z e dwi t ht hea ddr e s soft hei nt e ge rv ar i a bl e
now t hepoi nt e ri ni t i a l i z at i ont a ke spl a c ea nda s s i gni ngt hepoi nt e rr e f e r e nc et ot he var i abl enum. t hec ompi l e ra l s oal l o wst hes pe c i a lc a s et ha twewa ntt oi ni t i a l i z et hec ont e nta twhi c h po i nt e rpoi nt st owi t hc o ns t a nt sa tt hes a memo me ntt hepoi nt e ri sde c l a r e d: c ha r* s t r=“ He l l o” ;
I nt hi sc a s e ,me mor ys pa c ei sr e s e r v edt oc ont a i n" he l l o"a ndt he napoi nt e rt ot hefir s t c ha r a c t e ro ft hi sme mor ybl oc ki sa s s i gne dt os t r .I fwei ma gi net ha t" he l l o"i ss t or e da t t heme mor yl oc a t i onst ha ts t a r ta ta ddr e s s e s1 0 0 3 ,wec a nr e pr e s e n tt hepr e vi ous de c l a r a t i ona s
7.3.1
The null pointer
S ome t i me si ti sus e f ult oma keourpoi nt e r si ni t i a l i z e dt onot hi ng. Thi si sc a l l e danul l p o i n t e r . Anul lpo i nt e ri st hepo i nt e rdoe sn otpoi ntt oa nyv a l i dr e f e r e nc eo rme mor y a ddr e s s. Wea s s i gnapoi nt e ranul lva l ueb ys e t t i ngi tt oaddr e s s0 : i nt* i pt r ; i pt r =0; or byus i ngt heke ywor dde finei nC++c ompi l e r i pt r=NULL: Thi ss t a t e me nta s s i gna ddr e s s0t oi Pt r. Thi ss t a t e me ntde s c r i beas pe c i a lpr e pr oc e s s or de finec a l l e dNULLt ha te v a l ua t e st o0. he nc eAnul lpoi nt e ri sapoi nt e rt ypewhi c hha s as pe c i a lva l uet ha ti ndi c a t ei ti snotpoi nt i ngt oa nyva l i dr e f e r e nc e. Eve nt hought hi si s no tt e c h ni c a l l ypa r to fC++, i t ’ sus a g ei sc ommone no ught ha ti twi l lwo r ki ne v e r yC++ c o mpi l e r .
7 . 3 . 2
Voi dPoi nt e r
Ev e r ypoi nt e rpoi nt st oas pe c i ficda t at ypee x c e pt‘apoi nt e rt ovoi d‘whi c hc a n poi nt t oa nyda t at ypef or m doubl et oc ha r a c t e r s buti tc a nno tbedi r e c t l yde r e f e r e nc e d. I t ha sagr e a tf a c i l i t yofpoi nt i ngt oa nyda t at ype .I tc a nonl ybede r e f e r e nc e db yc a s t i ng t hea ddr e s si nt hev oi dpoi nt e rt oac o nc r e t eda t at ypebe f or ede r e f e r e nc i ngi t.Thi s c o nv e r s i o no fpoi nt e rt os omeot he rv a l i dda t at ypei sa c h i e v e db yus i ngt hec o nc e p to f t ype c a s t i ng . I ti sg e ne r a l l yus et opa s sg ene r i ca r g ume nt st of unc t i ons . v oi d*vpt r ; i nti ; floa tf ; doubl ed; t hes t a t e me nt s vpt r=&i vpt r=&f v pt r=&d
/ / de c l a r a t i onofvoi dpoi nt e r
/ / ma ki ngt hepoi nt e rvpt rpoi ntt oi nt e ge rt ypeofv a r i a bl e / / ma ki ngt hepoi nt e rvpt rpoi ntt ofloa tt ypeofv a r i a bl e / / ma ki ngt hepo i nt e rvpt rpoi ntt odo ubl et y peo fv a r i a bl e
Al lt he s es t a t e me nt sa r eva l i d,a sv oi dpoi nt e rc a npoi ntt oobj e c t so fa nyda t at ype . Di r e c tde r e f e r e nc i ngofv oi dpoi nt e ri sno tpe r mi t t e d. Thevoi dpoi nt e rmus tfir s tbe e xpl i c i t l yc as tt oanot he rpoi nt e rt ypebe f or ei ti sde r e f e r e nc e d.Al s oi ti snotpos s i bl et o dopoi nt e ra r i t hme t i co navoi dpoi nt e r .
MYNOTES •
•
•
•
A pointer which has been declared as a variable can point to any portion of the memory therefore need to be initialized.
Po i nt e rc a nb ei ni t i a l i z e dwi t ha0o rNULLo rav a l i da d dr e s s Anul lpo i nt e ri st hepo i nt e rdo e sno tpo i ntt oa nyv a l i dr e f e r e nc eo rme mo r ya ddr e s s. We a s s i gnapo i nt e ranul lv a l ueb ys e t t i ngi tt oa ddr e s s0 : Ev e r ypo i nt e rpo i nt st oas pe c i ficd a t at ypee x c e p t‘ap oi nt e rt ov oi d‘whi c hc a n po i ntt oa nyda t a t ypef o r mdo ub l et oc h a r a c t e r s
Pl e a s eno t et ha ta s t e r i s ks i gn( * )t ha tweus ewhe nde c l a r i ngapoi nt e ronl yme a nst ha t t heva r i a bl ebe i ngde c l a r e di sapoi nt e r. I ti spa r tofi t st ypec ompounds pe c i fie r ,a ndi t
s h oul dnotbec o nf us e dwi t ht hede r e f e r e nc eope r a t ort ha tweha v es e e nabi te a r l i e r( * ) . The ya r es i mpl yt wodi ffe r e ntt hi ng sr e pr e s e n t e dwi t ht hes a mes i g n.
7 . 4Poi nt e ra r i t hme t i c Hopey oumus tha v eunde r s t o odt hede c l a r a t i ona ndi ni t i a l i z a t i o no fv a r i a bl e s.No w we wi l ls e eho w wec a nma ni pul a t et hepoi nt e rv a r i a bl e s .Poi nt e ra r i t hme t i cope r a t i onsa r e di ffe r e ntf r om no r ma la r i t hme t i cope r a t i o ns . The s eo pe r a t i o nsma ybea l l o we dt ope r f or mo nP oi nt e r s : •
•
•
Addors ubt r a c ti nt e ge r st o/f r om apoi nt e r .Ther e s ul ti sapoi nt e r . S ubt r a c tt wopoi nt e r st ot hes a met ype .Ther e s ul ti sa ni nt . Mul t i pl yi ng, a ddi ngt wopoi nt e r s , e t c . don' tma kes e ns e .
Onl yt woa r i t hme t i co pe r a t i o ns ,a ddi t i ona nds u bt r a c t i ona r emo s t l yus e d ope r a t i o n o npoi nt e r s .Bo t ha ddi t i o na nds u bt r a c t i onope r a t i o nde pe ndupo nt hes i z eofda t at ype poi nt e rva r i a bl epoi nt st o. c o ns i de rt hef ol l owi ngpoi nt e r s: i ntx; i nt* i pt r, *pt r ; c har* c pt r ; Addi ngors ubt r a c t i ngi nt e ge rt o/f r om poi nt e r
S i mpl ea ddi t i onors ubt r a c t i onso pe r a t i onsc a nbepe r f or me donpoi nt e rv ar i a bl e s I f* i Pt rpoi nt st oa ni nt e ge r ,* i pt r+1i st headdr e s so ft hene xti nt e ge ri nme mor ya f t e r * i Pt r .* i Pt r-1i st hea ddr e s soft hepr e vi ousi nt e ge rbe f or e* i Pt r . x=57 i pt r =&x;
i pt r =i pt r+1 or i pt r++; Thi si st hemo s tc o mmo na r i t hme t i co pe r a t i ondo neus i ngpo i nt e r si . e .i nc r e me nt i ngt he po i nt e rb y1 .Thepo i nt e ri ss hi f t e dt one x tme mo r yl o c a t i o nno tb yo neby t ebutb y2 byt e sasi pt rt ypei si nt e ge randi nt e ge rwi l lt akei nt e r nal l yt wobyt e sofs t or age .
I ti st onot et ha twhe npoi nt e r sa r ei nc r e me nt e db y1t hes i z eoft heva r i a bl et owhi c hi ti s po i nt i ng t o (1f orc ha r ,2f ori nt,4f o rflo at ,8f o rdoub l e ) i st a ke ni nt oa c c ount . I nc r e me nt i ngapoi nt e rwi l lr e s ul ti nt hef ol l o wi ngc o mput a t i on: ne wa ddr e s sofpoi nt e r=ol da ddr e s so ft hepoi nt e r+s c a l i ngf a c t or
whe r es c al i ngf a c t ori st hes i z eofvar i a bl ei ti spoi nt i ngt o.
s i mi l ar l y i pt r =i pt r-1or i pt r-; Thepoi nt e ri ss h i f t e dt opr e v i o usme mor yl oc a t i o n1 0 0 3t o1 0 01 no tb yo neb yt eb utb y 2byt e sa si pt rb as et ypei si nt e ge ra ndi nt e ge rwi l lt a kei nt e r na l l yt wobyt e sofs t or a ge .
Re me mbe r:Poi nt e ri sa l wa ysi nc r e me nt e dorde c r e me nt e da spe rt het ypeo fv al uei ti s poi nt i ngt o.
No t et ha ti Pt r +1doe sn otr e t ur nt hea d d r e s s af t e ri Pt r ,butt hene x to bj e c to ft het y pe t ha t i Pt rpoi nt st o.I fi Pt rpoi nt st oa ni nt e ge r( s i z ei s2byt e s ) ,i Pt r +3me ans3i nt e ge r sa f t e r i Pt r ,whi c hi s6a ddr e s s e sa f t e ri Pt r .I fc Pt rpoi nt st oac ha r ,whi c hi sal wa ys1b yt e , c Pt r +3me a ns3c ha r sa f t e rc pt r ,whi c hi s3a ddr e s s e sa f t e rc Pt r . Ex ampl e: 7 . 3 findt heout putoft hef ol l owi ngi ft headdr e s sofni s1 0 0.
floatn=7; i nt* pt r =&n; c out <
108 1 1 2 No t et ha tpoi nt e rpt ri si nc r e me nt e db y4byt e sa tat i measi ti spoi nt i ngt ofloa tt ypeof da t a . Ex ampl e: 7 . 4s howi ngpoi nt e rar i t hme t i c
char * myc har ; s h o r t * myshor t ; l o n g * myl ong; and t hatwe know t hatt he y poi ntt o me mor yl oc at i ons 1000,20 00 and 3000 r e s pe c t i ve l y,S oweWr i t e myc har ++; or myc har =myc har +1 mys hor t ++; or mys hor t=ms hor t+1 myl ong++; or myl ong=myl ong+1
Ex ampl e7 . 5 findt heout putoft hef ol l owi ng
# i nc l ude i ntone [ ] ={ 1, 2, 3} ; voi dmai n( ) { i nt* pt r ; p t r = o n e ; c o ut <<* pt r <, e ndl ;
p t r + + ; c o u t < < * p t r } out put 1 2
Bo t ht hei nc r e a s e( ++)a ndde c r e a s e( )ope r a t or sha v egr e a t e ro pe r a t orpr e c e de nc et ha n t hede r e f e r e nc eope r a t or( * ) , buti ts ho wsdi ffe r e ntbe ha vi our, c o ns i de rt hef ol l owi ng e x pr e s s i o nma yl e a dt oc o nf us i on: *p++ Be c a us e++ha sgr e a t e rpr e c e de nc et ha n* ,t hi se x pr e s s i oni se q ui va l e ntt o* ( p++) . The r e f or e ,wha ti tdoe si st oi nc r e a s et heva l ueofp,butbe c a us e++i sus e da spos t fix t hewhol ee x pr e s s i oni se va l ua t e da st heva l uepoi nt e db yt heor i gi na lr e f e r e nc e( t he a ddr e s st hepoi nt e rpoi nt e dt obe f or ebe i ngi nc r e a s e d) . No t i c et hedi ffe r e nc ewi t h: ( * p) ++ He r e , t hee x pr e s s i o nwo ul dha v eb e e ne v a l ua t e da st hev a l uepo i nt e db ypi nc r e a s e db y o ne . Thev a l ueo fp( t hepoi nt e ri t s e l f )wo ul dno tb emodi fie d( wha ti sb e i ngmo di fie di s wha ti ti sbe i ngpoi nt e dt ob yt hi spoi nt e r ) . p++=* q++; I fwewr i t e :*
Be c a us e++ha sah i ghe rpr e c e de nc et ha n* , bo t hpa ndqar ei nc r e a s e d,butbe c a us ebo t h i nc r e a s eo pe r a t or s( ++)a r eus e da spos t fixandno tpr e fix ,t heva l uea s s i gne dt o* pi s* q be f or ebot hpandqar ei nc r e as e d.Andt he nbot har ei nc r e as e d.I twoul dber oughl y e qui va l e ntt o: * p=* q; ++p; ++q; I ti sr e c o mme nde dt ous epa r e nt he s e s( )i no r de rt oa v oi dune x pe c t e dr e s ul t sa ndt ogi v e mor el e gi bi l i t yt ot hec ode .
s i gnmentoper at i on ( =)bet TheAs wee nt wopoi nt er smaket hem poi ntt ot hes ame t hi ng.Co ns i de rt hee xa mpl ebe l o w whe r et heva l ueoft hepoi nt e ri pt ri sa s s i gne dt o a not he rpoi nt e rv a r i a bl eoft hes a met ypepoi nt e rp t r. pt r=i pt r Af t e rt hi sa s s i gnme nts t a t e me ntb ot ht hepoi nt e r s pt ra ndi pt rp oi nt i ngt os a meva l ue t husma ki ngpt rp oi ntt owha t e v e ri pt rp oi nt e dt o.
x 1003
P oi nt e rwhi c hr e f e rt ot hes a mea ddr e s sl o c a t i o ni sc a l l e da s“s h a r i ng”.Us i ngt hi s f e a t ur eo fpo i nt e r s, as i ng l eme mo r ys t r uc t ur ei nt heme mo r yc a nb es ha r e db yt woo r mo r et ha nt woe nt i t i e swhi c hi sav e r yi mpor t a ntgoa l . No wo bs e r v et hef o l l o wi ngs t a t e me n t s: i ntx, y, *pt r 1 ,*pt r 2 ; x=45; y=90 ; pt r 1=&x; pt r 2=& y;
wha t e v e r pt r 2i spoi nt i ngt o( he r ey)i ss e te qua lt owha t e v e rpt r 1i spoi nt i ngt o( he r e x ), Thusyt a k e so nt hev a l ueo fxa ndb e c o me4 5 . Li t t l emor ec ompl i c a t e dma ni pul a t i onofpoi nt e r si sa l s opos s i bl el i ke: c o ns i de rt hef ol l o wi ng: pt r=&x; // Thi ss t at e me ntmake spt rpoi nt st ox y=*pt r+1 ; Thi ss t a t e me ntge t swha t e ve rp t rpoi nt sa t, a dds1t oi t, a nda s s i gnt hefina lva l uet oy; i ft heva l ueofxi s5 7t heywi l ldi s pl a yt heva l ue5 8. c o ns i de ra not he rs t a t e me ntl i ke * i p =0 Thi ss t a t e me ntwha t e v eri pi spoi nt i ngt owi l lbes e tt oz e r o. i . e .xwi l lbe c omez er o.
MYNOTES •
incrementing a pointer by 1 makes it point to memory location given by the formula new address = old address+ scaling factor.
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Decrementing a pointer by 1 makes it point to the memory location given by the formula : newaddress = old address scaling factor
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Bo t ha ddi t i o na nds ub t r a c t i o no pe r a t i o nde p e ndupo nt hes i z eo fda t at y pep o i nt e r
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v a r i a bl epo i nt st o . Bo t ht hei nc r e a s e( ++)a ndde c r e a s e( )o pe r a t o r sh a veg r e a t e ro p e r a t o rpr e c e d e nc et ha n
•
t hed e r e f e r e nc eo pe r a t o r( * ) , T h e s e t we e nt wop o i nt e r sma k et h e mp oi ntt ot h es a met h i ng A s i g nme ntope r at i on( =)b
7.6Pointer and arrays The r ei sas t r o ngr e l a t i o ns h i pb e t we e nt hepo i nt e ra ndt hea r r a y s .An yo pe r a t i onwhi c hc a nbe pe r f o r me dus i nga r r a ys u bs c r i pt i ngc a na l s obepe r f o r me dus i ngpo i nt e r s. Thepo i nt e rv e r s i on wi l lbege ne r a l l yf a s t e rt ha nt hea r r a yv e r s i on oft ha tpr ogr a m. c o ns i de rt hef ol l o wi ngde c l a r a t i on:
i ntl i s t [ 1 0] ; t hi sde fine sa na r r a yof1 0me mo r yl o c a t i o nswhi c hc a ns t o r e1 0i nt e g e rnumb e r s, whi c hc a nbe r e f e r r e da sl i s t [ 0 ]……………l i s t [ 9 ] . Li s ta r r a yc a nber e pr e s e nt e da s:
T oa c c e s st hee l e me nt soft hea r r a y ,s ubs c r i ptc a n beus e d.l i kea [ 8 ]r e f e rt ot he8th e l e me ntoft hear r a yi . e .3 4.The not a t i ona [ i ]r e f e r st ot heith e l e me ntoft hea r r a y .But ha v eyoue v e rt hi nkt ha t ,wha tt heva r i a bl el c o nt a i nswi t houtt hes ubs c r i pt? i s t
The‘ l i s t ’c o nt a i nst hea ddr e s soft hefir s te l e me nto ft hea r r a y.S i nc eapoi nt e ri sa va r i a bl ewhi c hc ont a i nst hea ddr e s sofa not he rva r i a bl e . ,i ti se vi de ntt ha tl i s ti sal s oa poi nt e rv ar i a bl ebe c a us ei tc o nt a i nt headdr e s so ft hefir s te l e me ntoft hear r a y . S owha ti ti spoi nt i ngt o?l i s ti spoi nt i ngt ot hefir s te l e me ntoft hear r a y .
no wi fy ouwi l lgi v et hes t a t e me nt c out <<* l i s t; i twi l ldi s pl a yt heva l ue6 Thusnow i ti sc l e a rt ha tna meoft hear r a yi sa c t ua l l yapoi nt e rt ha tpoi nt st ofir s t e l e me ntoft hear r a y . Re me mbe r
Thena meo ft hea r r a ya l wa ysc o nt a i nt hea ddr e s soft hefir s te l e me ntoft hea r r a y . Aspe rpoi nt e ra r i t hme t i ca ddi ng1t opoi nt e ri . e .l i s t+1wi l lpoi ntt ot hene xtobj e c tof t hea r r a y, he nc el i s t+1gi ve st hea ddr e s so ft hes e c o nde l e me nt, t hel i s t+2gi ve st he a ddr e s soft het hi r de l e me nt……. l i s t+igi v e st hea ddr e s soft he( i +1 )th e l e me nto ft he a r r a y . T odi s pl a yt heva l ue soft hefif t he l e me ntoft hea r r a ywec a nwr i t ea nyo ft he s e s t a t e me nt s: c out <
wec a nc ha ng eorma ni pul a t et hec o nt e ntoft hea r r a yus i ngpoi nt e rr e f e r e nc e sl i ke
* (list + 4) =15
Thepa r e n t he s e sa r ene e de di nt hea bo ves t a t e me ntbe c a us et heuna r y` ` c o nt e nt sof ' ' p e c e de nc e ope r a t or*ha shi ghe rr ( i . e . , bi ndsmor et i ght l yt ha n)t headdi t i onope r a t or .I f wewr o t e* l i s t+4 ,wi t houtt hepa r e n t he s e s ,we ' dbef e t c hi ngt heva l uepoi nt e dt ob y ‘ l i s t ’ ,a nda ddi ng4t ot ha tva l ue .Thee xpr e s s i on* ( i p+1 ) , ont heot he rha nd,a c c e s s e s t hefif t hva l ueoft hea r r a y
Reme mbe r
Re f e r e nc et ol i s t [ i ]c a na l s ober e f e r r e da s* ( l i s t+i )whe r el i s ti ft hena meoft hea r r a y . Bo t ht het wof or msa r ee qui v a l e nt. Al s o&l i s t [ i ]andl i s t +iar ei de nt i c al ,asl i s t+ii st headdr e s soft he( i +1)th el e me ntAnd t h &l i s t [ i ]i sa l s or e f e r st ot headdr e s soft he( i +1) e l e me ntoft hear r a y . The r ei so nl yo nedi ffe r e nc eb e t we e nana r r a yna mea ndt hepo i nt e r.Apo i nt e ri sa va r i a bl es o i nt*pt r ; pt r=l i s t ; v a r i a b l e
//a s s i gni nga ddr e s soft hefir s te l e me ntt oa not he rpoi nt i ng
pt r++i sava l i dope r a t i ona ndi sl e ga l . butanar r a ynamei sc ons t antpoi nt e rvar i abl ewhos eval uec annotbec hange d. he nc e t h eope r a t i onl i kel i s t ++i snotava l i dope r a t i on. remember : Array name is a constant pointer
Ex ampl e: 7 . 6Pr og r a mt oc a l c u l at et hes um ofal lt hea r r aye l e me n t su s i ngar r ay var i abl eandpoi nt e rvar i abl e: Ans Us i ngAr r aySubsc r i pt # i nc l ude voi dmai n( ) {
i nta[ 10] ; c o ut<<” Ent e rt hee l e me nt so ft hea r r a y: ” ; i nts um =0; f or( i nti =0; i <10; i ++) { c o ut <<” Ent e rt hee l e me nt: “ ; c i n>>a [ i ] ; s um +=a[ i ] ;
} c o ut <<“Sum o fa l lt hee l e me nt so ft hea r r a yi s: ” < voi dmai n( ) { i nta[ 10] ; c o ut<<” Ent e rt hee l e me nt so ft hea r r a y: ” ; i nts um =0; f or( i nt* p=a; p>a [ i ] ; s um +=* p; } c o ut <<“Sum o fa l lt hee l e me nt so ft hea r r a yi s: ” <
Ex ampl e: 7 . 7 Wr i t et heout putoft hef ol l owi ng: ( As s umet ha tt h eb a s ea dd r e s so ft h ea r r a ynumb e ri s5 02 0)
# i nc l ude i ntma i n( ) { i ntnumbe r [ ]={31 ,2 8,3 1, 30 ,3 1,3 0, 3 1, 3 1,3 0, 3 1, 3 0, 31} ; c o ut<<" Ani nt e g e roc c upi e s"<
r e t ur n0 ; } Thi swoul dpr oduc e: Out put: Ani nt e g e ro c c upi e s4b yt e s Number : 50 20 &number [ 0] :5020 Number +1:50 24 &Number : [ 1]50 24 Number +2:50 28 &Number : [ 2]50 28
Remember :,Byaddi ngnumber st ot henameoft hevar i abl e,wear eabl et ogett headdr es sofany memberoft hear r ay .
From the above declaration it can be derived that ointers do not have to point to single variables! "hey can also point at the cells of an array i!e! it can point to first element of the array that means the array can also be declared li#e this
int list!
""Array can be declared using pointers
$eading and writing in this type of array can be ta#e place using the following method : E#ample 7.7 $eading and writing of arrays when it is declared as pointers. %include&iostream!h' void main() int *list ; for (int i=+; i&=5;i) cout&&-.nter a number: -; cin''*(listi); "" reading of array elements / cout&&-"he elements of the array are :-&&endl;t
for (i=+; i&=5 ;i) cout&&*(listi)&&endl; /
00 display of array elements
output
.nter a number: +1 .nter a number: 23 .nter a number: 54 .nter a number: 7 .nter a number: 6 "he elements of the array are : +1 23 54 47 7
ointers are not restricted to integers only ! t can also be used with character arrays as explained e xplained in the next section of this chapter!
MYN!"# •
•
•
•
•
Th ena meo ft h ea r r a ya l wa ysc o nt a i nt h ea d dr e s so ft h efi r s te l e me nto ft h ea r r a y . Re f e r e nc et ol i s t [ i ]c a na l s ober e f e r r e da s* ( l i s t+i )whe r el i s ti ft h ena me meo ft h ea r r a y . Bo t ht h et wof o r msa r ee q ui v al e nt. t h Al s o&l i s t [ i ]a ndl i s t +ia r ei de nt i c a l ,a sl i s t+ii st h ea ddr e s so ft h e( i +1) e l e me ntAnd t h &l i s t [ i ]i sa l s or e f e r st ot h ea dd r e s so ft he( i +1) e l e me nto ft h ea r r a y .
Array name is a constant pointer
Bya d di ngnumb e r st ot h ena meo ft h ev a r i a b l e ,wea r ea b l et og e tt h ea d dr e s so fa ny me mb e ro ft h ea r r a y .
7.% 7.% &oin &ointe ters rs and and str strin ings gs S t r i nga r eo nedi me ns i ona lc ha r a c t e ra r r a ysc ha r a c t e ra r r a y si nC++whi c hi st e r mi na t e d byNULLc har ac t e ri . e .‘ /0’ . t ode fineas t r i ngwec a ngi vet hes t at e me ntl i ke
c ha rs t r [ 1 0] =” He He l l o” ; t hi ss t a t e m me e ntwi l lde fineas t r i ngof9c ha r a c t e r sa ndl a s tc ha r a c t e ri sr e s e r v edf ort he nul lc ha r a c t e r .I tc a nber e pr e s e nt e di nt heme me mo mor yl i ket hi s:
H 1 1 0 0 0 0
1 0 0
1 1 0 0 0 0
E 1 0 0
L 1 0 0
L 1 1 0 0 0 0
O 1 0 0
\0 1 1 0 0 0 0
1 1 0 0 0 0
1 0 1
1 1 0 0 1 1
Li ke ,t hei nt e ge ra r r a yst hena me meoft hes t r i nga r r a ya l s oc o nt a i nt hea ddr e s so ft hefir s t e l e me me ntoft hea r r a y. I nt hedi a gr a ma ddr e s soft hefir s te l e me me nti s1 0 03.He nc et hes t r t r c ont a i nst hi sva l ue .Si nc es c ont a i nt headdr e s s,i tc a nbec al l e da sapoi nt e rv ar i a bl e , buti ti sac ons t antpoi nt e rvar i abl ewhos eval uec annotbec hange d. s oas t r i ngc a na l s obede fine dl i ke c ha r* s t r=“ He He l l o” ; Butt he r ei sao nev e r yi mpor t a ntdi ffe r e nc ebe t we e nt het wode c l a r a t i o ns: c hars t r 1[ 10] =“HE HELLO”; c har*s t r 2=“HE HELLO”;
s t r 1i sa nc ha r a c t e ra r r a yt ohol dt hes e que nc eofc ha r a c t e r s. s t r 1c o nt a i nst hea ddr e s sof t hefir s te l e me me ntoft hea r r a y .s t r 1a l wa ysr e f e rt ot hes a me mes t or a gel oc a t i onwhi c hc a nnot bec hange d. be c aus es t r 1i sapoi nt e rc ons t ant . o nt heo t he rha nds t r 2i spo i nt e rt oac h a r a c t e rwho s ev al uec a nb emo mo di fie da sa n d whenr equi r ed.
Bo t hs t r 1a nds t r 2a r epoi ntt oH a ndt hel oc a t i on1 0 03 butv al ueo fs t r 1c a nnotbe mo di fie d.I twi l lr e ma ma i ns a me met hr o ugho utt hee x e c u t i o no ft hepr o gr a mb utt hev a l ueo f s t r 2c a nb ema ni pul a t e d.
Examp mpl e7. 8 Codet oc omp mpar et wost r i ngs
# i nc l ude # i nc l ude c har* ms g1,* ms g2; voi dmai n( ) { c o ut<<" Ent e raS t r i ng1 : " ; g e t s ( ms g 1 ) ; c o ut <<" Ent e rS t r i ng2 : " ; g e t s ( ms g 2 ) ; c har* pt r 1 , * pt r 2 ; f or( pt r 1 =ms g 1, pt r 2 =ms g 2; ( * pt r 1 ==* pt r 2&&* pt r 1 ! =' \0 ' && * pt r 2 ! =' \0 ' ) ; pt r 1 ++, pt r 2 ++) ; i f( * pt r 1 ==' \0 ' &&* pt r 2 ==' \0 ' ) c out <<" \n Twos t r i ngsa r ei de nt i c a l " ; e l s e c out <<" \nS t r i ngsa r enoti de nt i c a l " ; }
output Ent e ras t r i ng1:Ar c ha na Ent e ras t r i ng2:Ar c ha na T wos t r i ngsa r ei de nt i c a l
Ex ampl e: 7 . 9Toc o untl e ng t hoft hes t r i ng
# i nc l ude # i nc l ude voi dmai n( ) { i ntc o unt =0 ; c h ar* p; c o ut <<” Ent e rt hes t r i ng“ ; ge t s ( p) ; f or(; * p! =' \0' ; * p++) {c o unt ++;} c o ut <<” Thel e ngt hoft hes t r i ngi s: ” <
Ex ampl e: 7 . 1 0
Pr og r am t oc o pyones t r i ngi n t oanot he r
# i nc l ude # i nc l ude c ha r* my_ s t r c py( c ha r* de s t i na t i on, c ha r* s o ur c e ) { c ha r* p=de s t i na t i on; whi l e( * s our c e ! =' \0' ) { * p++=* s our c e++;
} * p=' \0' ; r e t ur nde s t i na t i on; } i ntma i n( v oi d) { c har*s t r A,* s t r B; c out <<" Ent e rs t r i ng: " ; ge t s ( s t r A) ; s t r B=my_ s t r c py( s t r B, s t r A) ; c o ut <<" \n Thec opi e ds t r i ngi s: " ; put s ( s t r B) ; } output Ent e rs t r i ng: pr o gr a mmi ng Thec o pi e ds t r i ngi s: pr o gr a mmi ng Ex a mp l e7 . 1 1
Pr og r a mt oc o n c a t e n at et wos t r i n g
# i nc l ude # i nc l ude voi dc onc a t (c ha r* , c ha r* , c ha r*) ; / *t hi sf unc t i onsc opi e st hes t r i ngsaa ndbt ot hede s t i na t i ons t r i ngc* / voi dc onc a t (c ha r* a ,c ha r* b, c ha r* c ) { whi l e (* a) { /*whi l e (* c ++=* a ++) ;* / * c=* a ;++a ;++c ; } whi l e (* b) { * c=* b; ++b;++c ; } * c=' \0' ; }
mai n( ) { c ha r* s t r i ng1 ; c ha r* s t r i ng2 ; c ha r* s t r i ng3 ; c out <<" Ent e rt hes t r i ng1 : " ; ge t s ( s t r i ng1) ; c o ut <<" \nEnt e rt hes t r i ng2: " ; ge t s ( s t r i ng2) ; c onc at (s t r i ng1, s t r i ng2, s t r i ng3) ; c o ut <
T oc o untt henumb e rofv owe l si nas t r i ng
# i nc l ude # i nc l ude c ons ti nts i z e=7 ; c ha rs t r [ s i z e ]=" Comput e rSc i e nc e " ; i ntvo we l s=0 ; f or( c ha r* Pt r=s t r ;Pt r
c a s e' U' : v owe l s ++;br e a k; } } c out<
"his program uses a pointer to step through each of the elements in an array! .ach element is dereferenced by the switch expression, and if the element is a vowel, vowels is incremented! "he for loop then uses the operator to advance the pointer to the next character in the array! "he for loop terminates when all characters have bee n examined!
MYNOTES •
• •
o ned i me ns i o na lc h a r a c t e ra r r a ysc a nb ede c l a r e dbyd e c l a r i ngapo i nt e ra ndi ni t i a l i z i ngi t. Thi si sb e c a us et hena meo fac ha r a c t e ra r r a yi sa c t ua l l yapo i nt e rt ot hefir s te l e me nto ft he a r r a y . Cha r a c t e ra r r a yc a nb ep r i nt e dus i ngt h ep o i nt e rna meo nl y . De c l a r a t i o no fac ha r a c t e rt ypepo i nt e rd o e sno ta s s i gna nyv a l uet oi t. The r e f o r ei tc a nno tb e us e dt oi nputs t r i ng s , unl e s si ni t i a l i z e d.
7. 7Mul t i di mens i onalAr r a ysa ndPoi nt er s c o ns i de rt hede fini t i on i nwhi c hwec r e a t et wodi me ns i o na la r r a y s i ntm[ 5 ] [ 6 ] ; i nt*b[ 5] ; Thi sde c l a r a t i o no f‘ b’c r e a t e sfiv epo i nt e r sa nd e a c h poi nt e rpoi nt st oa na r r a yof i nt e g e r s . Asp e rt hea bo ve de fini t i onst hel oc a t i o ns m[ 2 ] [ 3 ]a nd b [ 2 ] [ 3 ]b ot ha r e r e f e r r i ngt oas i ngl ei nt e g e r, butt hea r r a ym i sat r uet wodi me ns i ona la r r a yi nwhi c h5x 6=3 0i nt e ge rl oc a t i onsa r er e s e r ve df ort hee l e me nt soft hea r r a y .F orbt hede fini t i on o nl ya l l oc a t e s5 poi nt e r sa nd do e s no ti ni t i a l i z et he m ( i ni t i a l i z a t i on c a nb e done e x pl i c i t l ywi t hc o de) . Compi l e rwi l lr e s e r vet heme mor ys pa c ef ort hefiv epoi nt e r s .
Thebi g ge s ta dv a nt a g eoft hepo i nt e ra r r a y i st ha tr o wso ft hea r r a yma ybeofv ar i a bl e l engt h.Eac h el ementofB need notpoi ntt os i xel ement s .s omema y poi ntt ot wo e l e me nt s,s omepoi nt st os i xe l e me nt s,s o mepoi nt st ononeata l l,s i nc et hec o mpi l e r c annotpr edi ctandc nnotdec i deont henumberofmember sofeac har r ay.T oa c c e s st he fir s tpo i nt e r ,y ouwoul dt ype* ( b +0 ) ,whi c hi st hes a mea s* ( b)or* b .Thes e c o ndpoi nt e r c an be ac c es s ed wi t h* ( b+1) .Kee pi n mi nd t hat ,* bi sa poi nt erand nota val ue. The r e f or e ,t oa c c e s same mb e roft hea r r a y ,y oumus tfir s ts p e c i f yt hede s i r e dpoi nt e r . The n, us i ngi t si nde x ,y ouc a ng e tt hec o r r e s pondi ngv al ue .
Wec a na l s ode c l a r et wodi me ns i ona lc h ar a c t e ra r r a ya s: c h a r*na me [ 5 ] ; he r efiv epoi nt e r sa r ebe i ngde c l a r e d,e a c hofwhi c hi spoi nt i ngt oc ha r a c t e r. s i nc eo ne po i nt e rt oac h ar a c t e rc a nbeus e dt or e pr e s e ntas t r i ng,as e toffiv es t r i ngsc a nbe r e pr e s e nt e da sa r r a yoffivec ha r a c t e rpoi nt e r s . Af t e rde c l a r a t i on t he i ndi vi dua ls t r i ngsc a n be i ni t i a l i z e db ys e pa r a t ea s s i gnme nt s t a t e me nt sl i ke: name [ 0] =“Moni ka” ; na me [ 1 ] =” Ra c h i t ” name[ 2] =”Hi mans hi ”; name[ 3] =”Hr i s habh”; ori tc a na l s obei ni t i a l i z edasf ol l ows: c har* name={“Moni ka”, “ Ra c hi t ” , “Hi mans hi ”, “Hr i s habh” } ;
Adva nt a geofha vi nga na r r a yofpoi nt e r si st ha t 1.Aswehavedi s c us s edabovet he, t hateac hel ementofnameca npoi ntt ovar abl e l e ngt hs t r i ng 2 .Ma ni pul a t i o no fs t r i ng sb e c o mec o nv e ni e nt.F ore x a mpl ewec a nwr i t e name [ 0] =”Moni ka” whi c hi sno tpos s i b l ei nno r ma lc h ar a c t e ra r r a y .I nnor ma la r r a ywene e dt ous es t r c p y f unc t i ont oc opy s t r i ngl i t e r a li nt os t r i ngvar i abl el i ke c hars t r [ 10] ; s t r =” He l l o”; //I nva l i ds t a t e me nt s t r c py( s t r , ” he l l o” ) ; //c or r e c ts t a t e me nt
MY N!"# • • •
!wo dimensional arrays can also be declared using pointers. A set of strings can be declared as an array of pointers to characters. "t may be initialized by assigning values to one string at a time or all at one time at the time of declaration .
7.$Pointers to str%&t%res As we know, we can declare the form of a block of data containing dierent data types by means of a structure declaration . consider the example of employee struct which contains the following elements : struct employee { char ename [2!" int ecode"
#oat salary" $ %ust like another data types we can also declare the pointer &ariables which can store the address of the structure &ariable . employee e" employee (sptr"
''declaration of structure &ariable ''declaration of a pointer to a structure
similarly like arrays ,structure &ariable name contain the address of the )rst element of the array .
sptr = 'e
assin sptr as pointer to te str%&t%re
struct employee { char ename[*!" int ecode" #oat salary" $" see, normally structure members are accessed through structure &ariables using +. operator like .
Example 7.13 -include iostream.h/ -include string.h/ &oid main01 { employee e" strcpy0e.ename, 3r. Adish1" e.ecode 4 55" e.salary 4 *5.67" cout e.ename endl e.ecode endl e.salary endl" $ 8ow we will see how through structure pointers sptr , de)ned abo&e we will access the members of the structure. 9tructure members are access through arrow operator 01 . o refer to the structure members using operator the general form is
str%&t%re pointer lie sptr / ena,e
str%&t%re ,e,-er sptr/e&ode sptr/ salary
;et us see this program how we can access the structure elements in the program:
Example 7.14 -include iostream.h/ -include string.h/ struct employee { char ename[*!" int ecode" #oat salary" $" &oid main01 { employee e, (sptr" sptr 4
>ointers to structures are &ery useful when the whole structure is passed as the arguments . 8e #now the name of an array stands for the address of its 9eroth element the same concept applies for names of arrays of structures! uppose item is an array variable of struct type product! onsider the following declaration:
s t r uc tpr oduc t { c h arn ame [ 3 0 ] ; i ntma nuf a c ; floa tne t ; } ; pr o duc ti t e m[ 2 ] , * pt r ; t hi ss t a t e me ntde c l a r e si t e ma sa r r a yoft woe l e me nt s , e a c ht ypes t r uc tpr oduc t sa ndpt r a sapoi nt e rda t aobj e c t so ft ypes t r uc tpr oduc t s , t hea s s i gnme nt
pt r =i t e m; wo ul da s s i gnt hea ddr e s so fz e r o t he l e me ntt opr o duc t [ 0 ] .I t sme mbe r sc a nbea c c e s s e d byus i ngt hef ol l owi ngnot at i on. pt r ->na me ; pt r ->ma nuf a c ; pt r ->ne t ;
Thes ymb ol->i sc a l l e da r r o w po i nt e ra ndi sma deupo fmi nuss i gna ndgr e a t e rt ha n s i gn. Not et ha tpt r ->i ss i mpl ea not he rwa yofwr i t i ngpr oduc t [ 0 ] . Whe nt hepoi nt e ri si nc r e me nt e db yo nei ti sma det opo i ntt one x tr e c o r di ei t e m[ 1 ] .The f ol l o wi ngs t a t e me ntwi l lpr i ntt heva l ue so fme mbe r so fa l lt hee l e me nt so ft hepr oduc t a r r a y . f or( pt r =i t e m;pt r name<manuf ac <net ; Wec o ul da l s ous et heno t a t i o n ( * pt r ) . numbe r t oa c c e s st heme mb e rn umb e r .Thepa r e nt he s i sa r o undpt ra r ene c e s s a r yb e c a us et he me mbe ro pe r a t o r‘ . ’ha sahi ghe rpr e c e de nc et ha nt heope r a t or*
7 . 9Poi nt e rt oobj e c t s Weh a v ea l r e a dyl e a r ne dt ha t: Ac l a s s i sa ne x pa nde dc o nc e ptofada t as t r uc t ur e : i ns t e a dofhol di ngonl yda t a ,i tc a nhol dbo t hv ar i oust ypeo fda t aa ndf unc t i onswhi c h c a no pe r a t eont ha tda t a . b j e c t Ano i sa ni ns t a nc eofac l a s s . I nt e r mso fC++va r i a bl e s , ac l a s swoul dbet het ype , a nda nob j e c two ul db et hev a r i a bl e . Rul e sofde fini ngc l a s sobj e c tpoi nt e ri ss i mi l a rt o de fini ngpoi nt e rt os t r uc t ur eva r i a bl e a sweha v edi s c us s e da bo ve . C++a l l o wst oc r e a t epoi nt e r st ha tpo i ntt oc l a s s e s .Onc e
y oude fineac l a s si tb e c o me sav a l i dt y pe ,& wec a nus et hec l a s sna mea st het y pef o r b j e c t t he po i nt e r .The s e poi nt e r swhi c ha r e po i nt i ng t oc l a s s ob j e c t sa r ec a l l e do poi nt e r s . ;et us de)ne a class bank:
Example 7.15 c l a s sBa nk { i nta c c ount no ; c h art ype ; c ha rc us t _ na me [ 2 0 ] ; flo atba l a nc e ;
publ i c : v oi dr e a dda t a() / / F unc t i ont or e a dt hev a l ue sofda t ame mbe r s { c o ut <, ” Ent e ra c c ountnumbe r: ” ; c i n>>ac c ount no; c o ut <, ” Ent e rt ypeofa c c o unt( S /C): ” ; c i n>>t y pe ; c o ut <, ” Ent e rc u s t o me rna me : ” ; c i n>>c us t _name; c o ut <, ” Ent e rt heb a l a nc ea mo unt: ” ; c i n>>amount ; } v oi ds ho w( ) / / F unc t i o nt os ho wt hev a l ue so fda t ame mbe r s { c o ut<<“ Ac c ountde a t i a l s“ ; c out <<” Ac c ountnumber: <
} } ; T ode c l a r eac l a s spoi nt e ro f‘ ba nk’t ypewes ha l lwr i t e:
bank
*bpt r;
c l a s sna me poi nt e rt oc l a s s
whe r eba nki st hea l r e a dyde fine dc l a s s.Whi l ea c c e s s i ngt heme mb e r so ft hec l a s st he c e s st hepr i vat e ar r ow oper at or( ) i ns t e ad of( ‘ . ’ )ope r a t ori sus e d . Wecannotac memberoft hecl ass . A poi nt e rc a npoi ntt oonl ypubl i cme mbe r so ft hec l a s s. l e tuss e e how wec an ac c es st hepubl i cmemberoft hec l as st hr ough c l as spoi nt er. Us e() ope r a t ort oa c c e s st heda t ame mbe r so ft hec l a s s. v oi dma i n( ) { ba nk b,* bpt r ; bpt r=&b; bptr r e a dda t a ( ) ; b s how( ) ; $
/ /c a l l i ngofpubl i cme mbe r sus i ngobj e c tpoi nt e r s
Example : Show use of pointers to class objects
# i nc l ude # i nc l ude c l a s ss t ude nt { pr i va t e : c ha rc o ur s e [ 1 0] ; c h a rna me [ 3 0 ] ; publ i c : voi ds e t da t a ( i nt* x ,c ha r* c ) ; voi ddi s pl a y( v oi d) ; } ;
voi dmai n( ) { wor ke r* PGr a dua t e ,* gr a dua t e ;//De c l a r e daspoi nt e rt oc l a s sobj e c t pgr aduat e=new s t udent ; gr aduat e=new s t udent ; pgr a dua t e >s e t da t a ( “ MS c ( OR) ” ,"Mudi tJ a i n" ) ; gr a dua t e >s e t da t a ( “ BS c ( CS ) ,"S us h mi t aGe r a " ) ; pgr a dua t e >di s pl a y( ) ; gr a dua t e >di s pl a y( ) ; } voi ds t ude nt : : s e t dat a( i nt* i , c har* s ) { s t r c py( c our s e, i ) ; s t r c py( na me ,s ) ; } voi ds t ude nt : : di s pl a y( voi d) { c out<
Array of 'b(ects 8e can also declare the array of ob?; @rray name always contains the address of the first element in the array! t means the array name is a pointer as it is holding address ! f we ma#e an ob
bank b[ 10] , * bpt r ; t hi ss t a t e me ntde c l a r e s‘ b’a sa na r r a yoft e ne l e me nt s , e a c ht ypec l a s sba nk a ndbpt ra s apoi nt e rda t aobj e c t soft ypeb ank. Thea s s i gnme nt bpt r =b; wo ul da s s i g nt hea ddr e s so fz e r o t he l e me n tb [ 0 ]t obpt r .No wt hec l a s sme mb e r sc a nb e a c c e s s e db yus i ngt hef ol l o wi ngnot a t i on.
bpt r ->r e a dda t a ( ) ; bpt r >s h ow( ) ;
Thes ymb ol->i sc a l l e da r r o w po i nt e ra ndi sma deupo fmi nuss i gna ndgr e a t e rt ha n -> s i gn. Not et ha tbpt r i ss i mpl ea not he rwa yofwr i t i ngb[ 0 ] . Whe nt hepoi nt e ri si nc r e me nt e db yonei ti sma det opoi ntt one x tr e c o r di . e. b[ 1 ] .The f ol l o wi ngs t a t e me ntwi l lpr i ntt heva l ue so fme mbe r so fa l lt hee l e me nt so ft hepr oduc t a r r a y . f or( bpt r =b;pt r s ho w( ) ;
//di s pl a yi nga l lt heobj e c t soft hea r r a y
Pa s s i ngobj e c tpoi nt e r st of unc t i ons A whole class object may be returne or pass by pointers easily !with in a function for better e"ciency . Ac l a s so bj e c tpa s s e d/r e t ur ne db ypoi nt e r sa v oi dsc r e a t i onoft e mpor a r yobj e c t s . T e mpor a r yobj e c t sc a l lc o ns t r uc t or sa ndde s t r u c t or s( whi c hi si ne ffic i e nt ) . Pa s s i ng/r e t ur n i ngb ypoi nt e r sdoe s n' tc r e a t et e mpor a r yo bj e c t sbe c a us eapoi nt e r onl y s t or e st hea ddr e s st oanobj e c tt ha ta l r e a dye xi s t s . •
•
Gi v e n:
c l a s sBo ok { //. . . } ;
c l a s sBa c k pa c k { publ i c : Book*g e t Cont e nt s ( ) ; / / f unc t i onr e t ur ni ngpoi nt e rt oc l a s s / vo i d s e t Cont e nt s ( Bo ok*bk) ; / f unc t i onpa s s i ngobj e c tpoi nt e ra sa r gume nt s pr i va t e: Book myTe xt ; } ; Bo ok&Ba c k pa c k : : g e t Co nt e nt s ( ) {
r e t ur nmyT e x t ; } v oi dBa c k pa c k : : s e t Co nt e nt s ( Bo ok&b k) { myTe xt =bk; } Theme mbe rf unc t i o nsg e t Co nt e nt s ( )a nds e t Co nt e nt s ( )us er e f e r e nc e st oa v oi d i ne ffic i e ntt e mpor a r i e s : i ntma i n( ) { Bookfict i on; Books ome book; Bac kpac kmyPac k; myPa c k. s e t Cont e nt s ( fic t i on) ;/i / ns e t Cont e nt ss our c e , bki sar e f e r e nc e / /t ofic t i o n, no ta ni ne ffic i e ntt e mpo r a r y
s o me b oo k=my Pa c k . g e t Co nt e nt s ( ) ; /myTe / x ti sr e t ur ne dbyr e f e r e nc e r e t ur n0 ; }
7.10
Pointers and &onstants
o n s tqua You haveal r ea dyknow t hatc l i fie ri sus e dt ode c l a r es ymbol i cc ons t a nt s, t he val ueswhi c hs houl dnotbemodi fied.Thes amekeywor di sus edt odefinet hepoi nt er c o ns t a nt s. Ac o ns t a ntpo i nt e rme a nst ha tt hepoi nt e rwi l la l wa y spoi ntt ot hes a mea ddr e s s .I t s a ddr e s sc a nno tb emo di fie d.T ode fineac o ns tpo i nt e r,t y pet hec o ns tke ywor d ont he l e f ts i deoft heda t at ypeoft hepoi nt e rva r i a bl e.
i nt*c ons tpt r=&x ; whe r ept ri sac o ns t a ntpoi nt e rpoi nt i ng t oa i nt e g e r.c o ns t a ntpo i nt e rs houl d be i ni t i a l i z eda tt het i meofde c l a r a t i ononl y .
c ons ti nt*pt r ; his statement de)ne a pointer which is pointing to a constant &alue, means the statements for changing the &alue ptr is pointing to will not be executed like statements (ptr 47 " ''in&alid statement : try to modify the &alue of x
c ons ti nt*c ons tpt r ; his statement de)ne a constant pointer which is pointing to a constant &alue, means the statements for changing the &alue of ptr or the &alue ptr is pointing to will not be executed like statements (ptr 47 " ''in&alid statement : try to modify the &alue of x ptr??" '' in&alid statement : try to modify constant pointer
Example 7.15
# i nc l ude # i nc l ude i ntx =1 0 ;
i nt*c o ns tc pt r=&x ; voi dmai n( ) { c l r s c r ( ) ; c out <<" va l ueofpt r:" <<* c pt r ; * c pt r =5 ; c o ut <<" ne w va l ueofpt ri s: " <<* c pt r ; ge t c h( ) ; }
his statement will de)ne a constant pointer &ariable cptr , which is pointing to the same address
@onstant pointers can be dereferenced and &alues they are
MYN!"# • • •
Ano nc o ns tpo i nt e rc a nb er e di r e c t e dt opo i ntt oo t he ra ddr e s s e s . Ac o ns tpo i nt e ra l wa y spo i nt st ot h es a mea ddr e s s , a ndt hi sa ddr e s sc a nno tb ec ha ng e d. Apo i nt e rt oano nc o ns tv a l uec a nc ha ng et hev a l uei ti spo i nt i ngt o .
•
Apo i nt e rt oac o ns tv a l uet r e a t st hev a l uea sc o ns t( e v e ni fi ti sno t ) ,a ndt husc a nno tc ha ng e t hev a l uei ti spo i nt i ngt o .
7 . 1 2
Poi nt e r sa ndFunc t i onar gume nt
@ pointer can also be pass as an argument to a function t is used to pass the single element as well as the whole array ! Aere we modified the concept of passing arguments to functions : n B class we have learned that there are two ways to pass arguments to function! +! all by value : n this method value of each actual argument is copied in the calling function is copied onto the corresponding formal arguments of the called function !
7.1).1
*all by value
# i nc l ude v oi ds o me f unc ( flo a tf v a r ) { f va r =9 9. 9 ; } i ntma i n( ) { flo a tfl=4 . 3 7 ; somef unc ( fl) ; c out<<fl<
I nt hef unc t i ont hes t a t e me ntf va r =9 . 9 9wi l lc ha nget heva l ueoff va rt o9 . 9 9
Af t e rt hef unc t i one x e c ut i oni so ve rt hec ont r olr e t ur nt ot hema i nf unc t i on& di s pl a yt heva l ueoff 1whi c hi sunc ha nge di . e .i twi l ldi s pl a y4 . 3 7.
7 . 1 2 . 2Ca l lbyr e f e r e nc e: I nt hes i t ua t i onswhe r ewewoul dl i ket oc ha nget heva l ue so ft heva r i a bl e si nt hec a l l i ng pr o gr a mt he nwepa s st hepa r a me t e ri nt hef unc t i o nb er e f e r e n c e.Whe npa r a me t e r sa r e pa s s e db yr e f e r e nc et he nt hef or ma lpa r a me t e r si nt hec a l l e df unc t i onbe c o mer e f e r e nc e s ( ora l i a s e s )t ot heac t ua lpa r a me t e r si nt hec a l l i ngf unc t i on.Thi sme anst ha tt hef unc t i on doe sno tc r e a t ei t so wnc o pyofor i gi na lv al ue s, buti twor kso nt heor i gi na lda t ahe nc e t hec ha ng e sdo nei nt hef o r ma lv a r i a bl e sa r er e fle c t e dba c ki nt hea c t ua la r gume nt so f t hec a l l i ngpr ogr a m. # i nc l ude
voi ds omef unc ( float&f var ) { f va r =9 9. 9 ; } i ntma i n( ) { flo a tfl=4 . 3 7 ; somef unc ( fl) ; c out<<fl<
s oanyc ha ngei nt hef va ri sr e fle c t e dba c ki nt hef 1.Ac t ua l l ybot ht heva r i a bl e sa r e r e pr e s e nt i ngt hes a meme mo r yl o c a t i o n. Thusi nt hi sme t ho df unc t i o nwo r kswi t ht he or i gi na la r gume nt s, a ndt heva l ue sa r enotdupl i c a t e d. Thec a l lb yr e f e r e nc eme t hodof s e ndi nga r gume nt st of unc t i o ni smor ee ffic i e nta si tne e dt opa s sonl yt hea ddr e s so ft he obj e c tnott hee nt i r eobj e c t. Thi sme t hoda v oi dsc r e a t e a t i onoft e mpor a r yobj e c t s be c aus er e f e r e nc ei snot hi ngbutal i ast oa nobj e c tt hatal r e adye xi s t s .
7. 12 . 3 Cal lby Poi nt er s . Li keot he rv a r i a bl e s ,apoi nt e rc a nbepa s s e dt oaf unc t i on.P a s s i ngapo i nt e ra sa na r gume nt
t oaf unc t i o ni ss i mi l a rt ot heme t hodo fp as s i nga r gume ntb yr e f e r e nc e.The ybo t h pe r mi tt heva r i a bl et obemodi fie di nt hec a l l i ngpr ogr a m wi t houtc r e a t i ngi t sdupl i c a t e c o pi e soft hear gume nt s. I nc a l lb yr e f e r e nc eme t hodar e f e r e nc eora l i a soft heor i gi na l va r i a bl ei sc r e at e d, whi l ei nc a l lbypoi nt e rme t hod poi nt e rt ot heor i gi na lva r i a bl ei s c r e a t e d.
Poi nt e ra r gume nte na bl e saf unc t i on e na bl eaf unc t i ont oac c e s sa ndc ha ng eobj e c t si n t hef unc t i ont ha tc a l l e di t. Ex a mp l e: 7 . 1 6T wove r s i o nsofapr o gr am t os wa pv al ue so ft wov ar i a bl e sus i n g f unc t i ons wap( )ar es hownhe r e. not et hec hang ei nt hes ynt axofbot hme t hods .
1.pas sar gument sbyr ef er ence 2.pas sar gument sbypoi nt er s Bypas s i ngr e f e r e nc e
Bypas s i ngPoi nt e r s
# i nc l ude
# i nc l ude
voi dmai n( )
voi dmai n( )
{
{
v oi ds wa p( i nt&, i nt&) ;
voi ds wa p( i nt* x, i nt* y) ;
c o ut<<” Ent e rt hev a l ue so faa ndb: ” ;
c o ut<<” Ent e rt hev a l ue so faa ndb: ” ;
c i n>>a>>b;
c i n>>a>>b;
c o ut <<” \nOr i gi na lv a l ue sa r e: ” ;
c o ut <<” \nOr i gi na lv al ue sa r e: ” ;
c out <<”\na=“<
c out <<”\na=“<
c out <<”\nb=“<
c out <<”\nb=“<
s wa p( a ,b) ;
s wa p( a ,b) ;
c out <<”\nSwapped val uesar e: ”;
c out <<”\nSwapped val uesar e: ”;
c out <<”\na=“<
c out <<”\na=“<
c out <<”\nb=“<
c out <<”\nb=“<
}
}
v oi ds wa p( i nt&x ,i nt&y )
v oi ds wa p( i nt*x , i nt*y )
{
{
i ntt e mp;
i ntt e mp;
t e mp=x;
t emp=* x;
x=y;
* x=* y ;
y=t e mp;
* y=t emp;
}
}
Out put
Out put
Ent e rt hev a l ue so faa ndb: 1 2
Ent e rt heva l ue so faa ndb: 1 2
4 5
45
Or i gi na lv a l ue sa r e:
Or i gi na lva l ue sa r e:
a=12
a=12
b=4 5
b=45
Swappedval uesar e:
Swappedval uesar e:
a=45
a=45
b=1 2
b=12
Pa s s i nga r r a yst ot hef unc t i on Thena meoft hea r r a ya l l o wst hec o mpi l e rt opa s st hewhol ea r r a yt ot hef unc t i o n. I ns t e a do fp a s s i nga na r r a ya sa r gume n t ,i ns t e a do ft hi swec a nus eapo i nt e r. Onc ea po i nt e rha sbe e ni ni t i a l i z e da sho l di ngt hea ddr e s sofa na r r a y ,t hena meo ft hea r r a y a ndt hena meo ft hepo i nt e rpo i nt i ng t ot hes a mea ddr e s s .Thi sme a ns, t hena meo ft he poi nt e rc a nbeus e d, whe nc a l l i ngs uc haf unc t i on. exampl ebel ow wi l ls how how t hewhol ear r aycanbeas s edus i ngpoi nt er s: Ex a mp l e: 7 . 1 7Sh owi n gpa s s i n ga r r a ysa spoi n t e r s
# i nc l ude i nt* l i s t i ntAr r ays um( , i nts i ze ) { i nts um =0; f or ( i nti=0; i
S i mi l a r l yt wodi me ns i ona la r r a ysc a na l s obepa s s e dt ot hef unc t i one x c e ptava r i a bl et ha tpoi nt s t oat wodi me ns i ona la r r a y , pr o vi det het ypeofv ar i a bl ef ol l o we db ya na s t e r i s kt ha ti ndi c a t e s t ha tt hea r gume nti sapoi nt e r ,a ndf ol l o we db yapa i rofs qua r ebr a c ke t s . l i kef orc a l c ul a t i ngt he s um ofnumber sofdoubl edi mens i onalar r aywec anwr i t e # i nc l ude us i ngnames pac es t d; voi d Di s pl a y( i nt* Li s t [ ] ) ; i ntma i n( ) { i ntnumbe r [ 2 ] [ 6 ]={{3 1, 2 8,3 1, 3 0,3 1,3 0} , {31, 31,30, 31,30,3 1}} ; i nt* pt r [ 2] ; * pt r=number[ 0] ; ( * p t r ) [ 0 ] =numbe r [ 0 ] [ 0 ] ; ( * p t r ) [ 1 ] =numbe r [ 0 ] [ 1 ] ; ( * p t r ) [ 2 ] =numbe r [ 0 ] [ 2 ] ; ( * p t r ) [ 3 ] =numbe r [ 0 ] [ 3 ] ; ( * p t r ) [ 4 ] =numbe r [ 0 ] [ 4 ] ; ( * p t r ) [ 5 ] =numbe r [ 0 ] [ 5 ] ; * ( pt r +1 )=numb e r [ 1 ] ; ( * ( pt r +1 ) ) [ 0 ]=numbe r [ 1 ] [ 0 ] ; ( * ( pt r +1 ) ) [ 1 ]=numbe r [ 1 ] [ 1 ] ; ( * ( pt r +1 ) ) [ 2 ]=numbe r [ 1 ] [ 2 ] ; ( * ( pt r +1 ) ) [ 3 ]=numbe r [ 1 ] [ 3 ] ; ( * ( pt r +1 ) ) [ 4 ]=numbe r [ 1 ] [ 4 ] ; ( * ( pt r +1 ) ) [ 5 ]=numbe r [ 1 ] [ 5 ] ; cout<<" Li s tofNumber s " ; Di s pl a y( pt r ) ; r e t ur n0 ; } voi d Di s pl a y( i nt* l i s t [ ] ) { c out<<" \n( * pt r ) [ 0 ] ="<<( * l i s t ) [ 0 ] ; c out<<" \n( * pt r ) [ 1 ] ="<<( * l i s t ) [ 1 ] ; c out<<" \n( * pt r ) [ 2 ] ="<<( * l i s t ) [ 2 ] ; c out<<" \n( * pt r ) [ 3 ] ="<<( * l i s t ) [ 3 ] ; c out<<" \n( * pt r ) [ 4 ] ="<<( * l i s t ) [ 4 ] ;
c out<<" \n( * p t r ) [ 5 ] ="<<( * l i s t ) [ 5 ]<>x; c out <<" \nEnt ers ec ondnumber: " ; c i n>>y; m=max ( x, y) ; c out <<" \nMaxi s: " <<* m; ge t c h( ) ;
//de c l a r eapo i nt e r‘ m’poi nt i ngt oi nt e g e r
} i nt* ma x ( i nt&a ,i nt&b) { i f ( a>b) r et ur n&a; e l s e r et ur n&b; } Out put Ent e rfir s tnumb e r:3 4 Ent ers ec ondnumber: 12 Ma xi s: 3 4
Ano t he re x ampl eoff unc t i onr e t ur ni ngpoi nt e rt oac ha r a c t e r.
Ex a mpl e:7 . 1 9pr ogr a mi sus e dt os e a r c hs t r i ng1i nt ot hea no t he rs t r i ng2.Thef unc t i on ge t s ubs t rr e t ur nt hepoi nt e rt ot hebe gi nni ngoft hes ubs t r i ngi ns t r i ng2, i ff ound ot her wi s er et ur nNULL. # i nc l ude # i nc l ude c har* g et _s ubs t r ( c har* s ub,c ha r* s t r ) ; i ntma i n( ) { c har* s ubs t r , * s t r ; c out <<" Ent e rt hes t r i ng: " ; ge t s ( s t r ) ; c o ut <<" \nEnt e rt hes ubs t r i ng: " ; ge t s ( s ubs t r ) ; s ubs t r=ge t _ s ubs t r ( s ubs t r , s t r ) ; i f( s ubs t r ! =' \0' ) c o ut<<" s ubs t r i ngf ound:"<
f or ( t =0;s t r [ t ] ! =' \0' ;t ++) { p=&s t r [ t ] ; s t a r t=p; p2=s ub; / /c he c kf ors ubs t r i ng whi l e ( * p2 ! =' \0 ' &&* p2 ==* p) { p++; p2++; } /*I fa te ndofp2( i . e . ,s ubs t r i ng) ,t he nama t c hha sb e enf ound.* /
i f ( ! * p2) r e t ur ns t ar t ; //r e t ur npoi nt e rt obe gi nni ngofs ubs t r i ng } r e t ur n0 ; } } Out put Ent e rt hes t r i ng: I nt r oduc t i ont oC++ Ent e rt hes ubs t r i ng: t o s ubs t r i ngf o und: t oC++ Ent e rt hes t r i ng: Comput e rS c i e nc e Ent e rt hes u bs t r i ng: S CI ENCE s ubs t r i ngno tf ound Ent e rt hes t r i ng: Ha ppyBi r t hda y Ent e rt hes ubs t r i ng: Bi r t hda y s ubs t r i ngf ound: Bi r t hda y
7. 13 Poi nt er st opoi nt er s C++a l l owst heus eofpoi nt e r st ha tpoi ntt opoi nt e r s ,t ha tt he s e ,i ni t st ur n, poi ntt oda t a ( ore ve nt oot he rpoi nt e r s ) .I nor de rt odot ha t ,weonl yne e dt oa dda na s t e r i s k( * )f or e ac hl e ve lofr e f e r e nc ei nt he i rde c l a r a t i ons : c h a rc h; c ha r* b; c ha r** c ; c h=‘ a ’ ; b=& c h; c =&b;
Thi s , s uppo s i ngt her a ndo ml yc h os e nme mo r yl o c a t i o nsf o re a c hv a r i a b l ec , ba ndc h c o ul db er e pr e s e nt e da s : 2000
3000
&
-
1000
&
3000
1000
+
Theva l ueofe ac hva r i a bl ei swr i t t e ni ns i deea c hc e l l ;unde rt hec e l l sa r et he i rr e s pe c t i ve addr es s esi nmemor y. Thene wt hi ngi nt hi se x a mpl ei sv a r i a bl ec ,whi c hc a nbeus e di nt hr e edi ffe r e ntl e v e l s o fi ndi r e c t i o n, e a c ho neoft he m woul dc o r r e s po ndt oadi ffe r e ntv a l ue : •
•
•
c ha st ypec h a r * *a ndav al ueo f3 0 0 0 * c ha st ypec h ar *a ndav a l ueof1 0 0 0 * * c ha st ypec ha ra ndava l ueof' a '
Example 7.#$
# i nc l ude voi dmai n( ) { i nta[ ]={ 3, 4, 5, 6} i nt* p, * * q, * s,* t, * * s s ; p=a; s=p+1; q=&s ; t=(* q+1 ) ; s s=&t ; r=&s s ; c out<<* p<<“\t“<<* * q<<“\t ”<<* * * r<
5
Dyna mi ci ni t i a l i z at i onoft heva r i a bl e s Dynami cmemor y al l oca t i on i st he abi l i t y oft heC++ pr ogr am t o ac qui r ebl ockof memor yf r om dynami cmemor ypoolc al l edas f r ees t or eorheap, dur i ngr unt i meof t hepr ogr am . s omet i mest hepr ogr am doesnotknow t heamountofdat at hatnee dst o bes t or e dwhe nt hepr ogr am i se xe c ut i ng.i nt he s es i t uat i onspr ogr am c anr e que s tt he me mor yt obeal l oc at e df or mt hedynami cme mor ypool.Now l e tuss e ehow wec an al l oca t ememor ybl ocksdur i ngr unt i me. Ther ear et wot ypesofmemor ymanagement o pe r a t or si nC++: •
•
ne w de l e t e
Thes et womemor ymanagementoper at or sa r eus edf ora l l oca t i ngandf r ee i ngmemor y bl oc ksi ne ffic i e ntandc onve ni e ntwa ys . Weha v es e e nt hee x a mpl e si nwhi c ht hea r r a yi sde c l a r e dl i k e i ntl i s t[ 5 0] ; r e s e r v eme mor yf or5 0i nt e ge r s .De c l a r i nga r r a yl i ket hi sha sones e r i ousdi s a dv ant a ge s i z eo ft hea r r a yne e dt obepr e de fine d. I nnumb e ro fs i t ua t i o nswedo e sn otk no w ho w muc hmemor ywenee dunt i lr unt i me.s owedefinet hebi ggers i zear r ayt ohol dt he l ar ges tnumberofel ement swhi c hi st hewas t ageofmemor y.Thes epr obl emsc anbe a v oi de db ydyna mi c a l l ya l l oc a t i nga na r r a yoft her i ghts i z e ,orr e a l l oc a t i nga na r r a y whe ni tne e dst oe x pa nd.Bo t ho ft he s ea r edo neb yde c l a r i nga na r r a ya sapo i nt e ra nd us i ngt hene wo pe r a t o rt oa l l o c a t eme mo r y ,a ndde l e t et of r e eme mo r yt ha ti sn ol o ng e r ne e de d. Us i ngpoi nt e r sa r r a yc a nbede c l a r e dl i ke: i nt* l i s t ; Declaring the array li#e this has two main advantages +! i9e of the array need not be defined! so the problem of overutili9ation of memory (declaring very large arrays) C underutili9ation (declaring smaller si9e arrays than reEuired ) may be avoided! 1! "his is variable si9e arrays which can be expanded as per reEuirement by using new operator!
7 . 1 4Thene w Ope r a t or I nC++ pr ovi dest hewayt oc r eat edynami cvar i abl es.Dynami cvar i abl esar et hose v a r i a bl e sf o rwhi c ht heme mor ys p ac ec a nbea l l o c a t e do rde a l l o c a t e da tt her unt i mea s pert heneed oft he pr ogr am.Dynami cvar i abl esdoesnotha ve namel i ke s t at i c va r i a bl e s, t he ya r eonl yr e f e r r e db ypoi nt e r s . Whe ne v e rwede c l a r ea nypo i nt e rv ar i a bl e sl i k e i nt*pt r ; pt r
Thi si sada ngl i ngpoi nt e rpoi nt i ngt ono whe r e. wec a nc o nne c tt hi sda ngl i ngpoi nt e rt o s t a t i cva r i a bl eorb ydyna mi cv ar i a bl e . weha v ea l r e a dydi s c us s e dho wt oc o nne c tada ngl i ngpoi nt e rwi t has t a t i cv a r i a bl el i k e i ntx; pt r =&x; no wl e tuss e eh ow i tc a nbec o nne c t e dwi t hady na mi cv a r i a b l e s . Theme mo r yc a nbea l l o c a t e ddur i ngr unt i meus i ngao pe r a t o rc a l l e dne w t hege ne r a lf or m ofus i ngt hi so pe r a t ori s:
poi nt e rv a r i a bl e = ne w Pointer o type
!e
da t at ype !ype o dyna,i& aria-le re%ired 8ny alid 9++
Thi so pe r a t ora l l oc a t e sme mor yofs i z e, e qua lt ot hes i z eofs pe c i fie dda t at ypea nd r e t ur napoi nt e rt ot hene wl ya l l oc a t e da r e a . f o re x a mpl e:
pt r=new i nt ;
//the new operator allocates sufficient memory to hold the object of datatype int and returns a pointer to its starting point
Thi ss t a t e me ntr e t ur nsaa ddr e s so ft hea l l o c a t e dl oc a t i o no fme mo r y whi c hc a nho l d onei nt e ge r .s opt ri sapoi nt e rpoi nt i ngt ot hea l l oc a t e dme mor ys pa c e .
pt r
:yna,i& aria-le
S i nc edyna mi cv a r i a bl edo e sno tha v ea nyna mei ti so nl yr e f e r r e dt hr oughapoi nt e r, poi nt i ngt oi t.Dyna mi cv ar i a bl e sa r ene v e ri ni t i a l i z e db yt hec ompi l e r .The r e f or e ,t he pr o gr a mme rs ho ul dma k ei tapr a c t i c et ofir s ta s s i gnt he m av a l ue * pt r=4 0 ;
pt r
40
The assignment can be made in either of the two ways:
i nt* a=ne wi nt ; * a=4 0; or i nt* a=ne wi nt ( 4 0 ) ;
S i mi l a r l yt hepoi nt e r st oot he rda t at ype sc a na l s obede fine dl i ke: c har*c pt r=ne wc har; //de c l ar at i ona nda s s i gnme ntc anbe c ombi ned flo a t*f pt r=ne w flo a t; now wec anwr i t e *c pt r=‘ a ’ ; *f pt r=1 8. 9 0; Reme mbe r Nul lpoi nt eri sr et ur nedbyt henew oper at orwhent her ei si ns uffic i entmemor y a va i l a bl ef ora l l oc a t i on.
Ex ampl e: 7 . 2 1Wr i t eap r o gr am t or e a dt hei n t e g e ri ndynami cv ar i ab l e, c a l c u l at e sand di s pl ayi t ss q uar e.
#i nc l ude mai n( ) { i nt* p; p=new i nt ; c o ut <<” \nEnt e ra ni nt e g e r: “ ; c i n>>* p; c out <<“\nSquar eoft henumberi s:“<<* p**p; } Out put Ent e ra ni nt e ge r: 5 S qua r eo ft henumb e ri s: 2 5
T ode c l a r eav ar i a bl et ha twi l lpoi ntt oadyna mi c a l l ya l l oc a t e da r r a yde c l a r ei ta sa poi nt e rt ot hee l e me ntt ype . Forexa mpl e i nt*a=NULL //poi nt e rt oa ni nt , i ni t i a l l yt onot hi ng t hi sde c l a r a t i onc r e a t e sapoi nt e rb utdoe snota l l oc a t et heme mor yt oi t. Thes y nt a xofdyna mi c a l l yc r e a t i ngana r r a yofpoi nt e r si s :
Dat aType A r ayName aType Di me ns i ons *r =ne w Dat [ ] ;
t aType Tos t a r t ,t ypet heki ndofa r r a yyouwa ntt oc r e at e :t hi si st heDa . Th e r i st hena meoft hea r r a y.S i nc et hi si sapoi nt e r ,t hea r r a yna memus t A r ayName bepr e c e de dbyanas t e r i s kope r at or .
e w As s i g nt hen o pe r a t ort ot hea r r a yna me . Then a t o ri sf o l l o we db yt hes a meki nd o fda t at y peoft hefir s tpa r a me t e r , e w oper Da t a T y p e .
Di me ns i on i si nc l ude dbe t we e ns q ua r eb r a c ke t s , i st het o t a lnumbe rofe l e me nt st obe a l l oc a t e di nt hear r a y . / Ar r a yof1 2e l e me nt soft ypedoubl e do ub l e* Di s t a nc e=ne w do ub l e [ 1 2 ] ; /
i nt* pRa nge s=ne wi nt [ 120] ;
Ar r a yof1 2 0e l e me nt so ft ypei nt e g e r //
float* Pr i c e s=ne w float [ 44] ;
r a yof4 4e l e me nt so ft y peflo a t //Ar
Af t e rdyna mi c a l l yc r e a t i nga na r r a y ,t hec o mpi l e ra l l oc a t e st hene c e s s a r yme mor ys pa c e f o ra l lt heme mb e r so ft hea r r a y ,b a s e do nt heda t at ypea nda c c o mmo da t i nge a c h.J us t l i k ea nyv ar i a bl e ,t heme mor ya l l oc a t e df ore a c hme mbe roft hea r r a yc o nt a i nsg a r ba g e . I ti st hepr ogr a mme r sr e s pons i bi l i t yt ofil li tupf ora ppr opr i a t eva l ue s . Thi sc a nbet a ke n c a r eo fb ya s s i gni ngava l uet oe ac hme mbe ro ft hea r r a y. Re me mbe r :Poi nt e r sma ybes ubs c r i pt e dj us ta sa r r a y sa r e . Ex ampl e: 7 . 2 2T or e a di nanumb e randal l oc a t e st hats i z ea r r ay.
i nt*a=NULL; i ntn; i nt *a=NULL; //Poi nt e rt oi nt ,i ni t i a l i z et onot hi ng. i ntn; //Si z ene e de df ora r r a y c o ut <<” Ent e rt hes i z eoft hea r r a y“ ; c i n>>n; //Re a di nt hes i z e ; //Al l oc a t eni nt sa nds a v ept ri na . a=new i nt [ n] f or( i nti =0;i
S i mi l a r l ywec a na l l oc a t et heme mor yf ort hes t r uc t ur eva r i a bl ea l s ous i ngne w ope r a t or . wec a ndi r e c t l yr e a di nt heva l ue si nt hes t r uc t ur eva r i a bl e . Ex a mp l e: S howi n gt h eus eo fn e w op e r a t o rwi t hs t r u c t u r ev ar i a bl e
Example 7.#3 # i nc l ude # i nc l ude s t r uc tc l i e nt { c h a rna me [ 3 0 ] ; i nta c c t num; doubl ea mount ; } ; voi dmai n( ) { c l i e nt* c l ; c l= new c l i e nt ; / /Al l oc at i ngt heme mor yf ort hes t r uc t ur ec l i ent s t r c py( c l >na me ," Mr .Adi s h" ) ; c out <<” Thec l i e ntde t a i l sa r e: \n” ; c l >ac c t num =10012; c l >a mo unt=5 6 7 0 0 . 5 6 ; c out<<“\nCl i entnamei s: ”<name<ac c t num <amount<
7 . 1 4 . 2 De l e t eoper a t or T oe n s ur es a f ea nde ffic i e nt ,us eo fme mo r yi ti sr e q ui r e dt ha tt heme mo r yr e s e r v e db y ne w ope r a t ormus tber e l e a s e dwhe ne v e ri ti sno tr e q ui r e di nt hepr ogr a m. Thi si sdo ne t hr oughde l e t eope r a t o r .Dyna mi cv a r i a bl ec a nber e t ur ne dt ot hes y s t e mb yus i ngt hi s a nope r a t or. Ge ne r a ls y nt a xofus i ngt hi so pe r a t ori s
de l e t e poi nt e rva r i a bl e; whe r ede l e t ei st heke ywor dt ode l e t eapoi nt e rv a r i a bl e
po i nt e rv a r i a bl ei st hena meo ft hepo i nt e rwho s ea l l o c a t e dme mo r yne e dt ob ede l e t e d f o re x a mpl e: de l e t ept r
Thi sp oi nt e rwi l lf r e et heme mo r yr e s e r v e db yt hepoi nt e rpt r .de l e t i ngt heme mor y doe snotde l e t et hepoi nt e rb uta f t e rde l e t es t a t e me ntwel e f twi t hda ngl i ngpoi nt e r s whi c hpoi nt snowher e.
pt r
Ex a mp l e: 7 . 2 4T os h o wt h ec o nc e pto fn e wa ndde l e t eo pe r a t o r
# i nc l ude voi dmai n( ) { / Al l oc a t e sme mor yus i ngne w ope r at o rf ors t or i ngai nt e g e rdat at ype i nt* a =ne wa ;/ * a = 1 0 0 ; c out<<"TheOut puti s=" <
TheOut puti s: 1 0 0 Use
[] when
deleting arrays
Youmus ts pe c i f y" [ ] "whe nde l e t i nga na r r a y ,butnotf oras i ngl eva l ue .I ti s n' tpos s i bl e t ode l e t eonl ypa r tofa na r r a y . de l e t e[]l i s t
/ / whe r el i s ti st hedyna mi c a l l yc r e at e da r r a ypoi nt e r .
Commoner r or i nt* l i s t [ ] ; …… de l e t el i s t
delete instead of delete? when deleting a dynamically allocated array can cause data corruption or other problems!
remember a#e a habit of assigning your pointers to > both when they are declared (unless assigned to another address), and after they are deleted! t will save you a lot of grief!
7. 14 . 3 Poi nt er sandobj ec t s No w wewi l ll e a r nho wt ous epoi nt e r st oa c c e s st hec l a s so bj e c t sa ndt he i rme mbe r s. a poi nt e rc a npoi ntt oa nobj e c tc r e a t e db yt hec l a s s. Thede c l a r a t i onf or ma ti ss a mea s o t he rda t at ype s: c l a s ss t ude nt { i ntr no; c h a rna me [ 2 0 ] ; floatmar ks; publ i c : v oi dge t da t a ( ) { c o ut <<” Ent e rt her o l l no: “ ; cin>>rno; c out <<”\nEnt ert hename: “; c i n>>na me ; c o ut <<” Ent e rt hema r ks: “ ; c i n>>mar ks ; } voi ds howdat a( ) { c out <<”\nRol lNo:“<
} } ; l e tusde c l a r eans t ude ntt ypeva r i a bl e‘ s ’ a ndapoi nt e rs pt rt o‘ s ’ a sf ol l ows : s t ude nt*s pt r ; s pt r=ne ws t udent No wt hepub l i cme mb e r so ft hedy na mi c a l l ya l l o c a t e do bj e c tc a nbea c c e s s e db yt he obj e c tpoi nt e r‘ s p t r ’wi t ht hehe l pofa r r o w ope r a t or(>)a ss ho wnbe l o w: T opr oc e s st hea bo vec l a s swec a nwr i t et hema i n( )f unc t i ona sf ol l o ws : voi dmai n( ) { s t ude nt* s pt r ; s pt r =news t udent; s pt r >ge t da t a ( ) ; s p t r >s h owda t a ( ) ; } I fac l a s sha sac o ns t r uc t orwi t ha r gume nt sa nddo e sno ti nc l udea ne mpt yc o ns t r uc t o r t he nwemus ts uppl yt hea r g ume nt swhe nt heo bj e c ti sc r e a t e d. Wec a na l s oc r e a t ea na r r a yofc l a s so bj e c t sus i ngpoi nt e r s.f ore xa mpl e: s t ude nt* s p t r=ne ws t ude nt[ 1 0 ] ; Thi ss t a t e me ntwi l lc r e at eana r r a yof1 0obj e c t sofc l a s ss t ude nt.
Ex ampl e: 7 . 2 5wor ki ngo nar r ayofa c l as sob j e c twi t hpoi nt e r .
# i nc l ude # i nc l ude # i nc l ude # i nc l ude # i nc l ude c l a s sb ook { pr i va t e : floa tpr i c e ; c har* t i t l e ; publ i c : voi da s s i gn( ) ; voi ddi s pl a y( v oi d) ; } ; voi dmai n( )
{ book*bpt r [ 5] ; i ntn=1 ; i ntc hoi c e =0 ; c l r s c r ( ) ; do { bpt r [ n] =ne w book; bpt r [ n] >as s i gn( ) ; n++; c o ut <<" \nDoyouwa ntt oe nt e rmo r eb oo ks ( 1f ory e s, 0f orno ): " ; c i n>>c h oi c e ; }whi l e( c hoi c e ) ; cout<<"\n\n"; c o ut <<" \nBo okDe t a i l s" ; c out <<" \n" ; f or( i nti =1;i di s pl a y( ) ; } voi dbook: : a s s i gn( ) { c out<<" \nEnt ert hebookname: " ; ge t s ( t i t l e ) ; c o ut <<" \nEnt e rt hepr i c e: " ; c i n>>pr i c e ; } v oi dbook: : di s pl a y( v oi d) { c out<<" \n" <
7.1+.+
,emory lea-s
Dyna mi c a l l ya l l oc a t e dme mor yus i ngne w ope r a t or s t a y sa l l oc a t e dunt i li ti se x pl i c i t l y de a l l oc a t e dorunt i lt hepr o gr a me nds De f A me mor yl e aki sat ypeofpr ogr ammi ngbuge at supt heme mor y, i toc c ur swhena pr ogr am al l ocat esmor ememor yt hani tf r ee s .Thes ec hunksofal l ocat edmemor ywhi c h mor yl e a k orme mor y i sno tr e l e a s e d, c a nno tber e a l l oc a t e d orr e us ea r ec a l l e d me mor yl e aksr e ducet hef r e eme mor y whi l et hepr ogr am i sr unni ng, b l e e d i n g . Me ma ki ngl e s sme mor ya va i l a bl eno to nl yt ot hi spr o gr a m,b utt oot he rpr ogr a msa swe l l . Thi sway ,anappl i c at i onc anr unoutofmemor yandc aus et hes yst em t or uns l owl yor c r as h.Topr eve ntmemor yl eaks ,younee dt oknow whent heyocc urmos tf r equent l y and bec ar ef ulwi t hyourus eoft he" new"and " del et e"C++ oper at or s .ut mos tc ar e s ho ul dbet a ke nwhi l ema ni pul a t i ngt hedyna mi cva r i a bl e s.I ti sa dvi s a b l et or e t ur na dyna mi cv a r i a bl ewhe ne v e ri ti sn otr e q ui r e di nt hepr o gr a m.F ol l o wi ngs t e pss ho ul dbe t a k e nt oa v o i dt hel e a k a geo ft heme mo r y .
1 .TheC++o pe r a t o r" ne w"a l l oc a t e she a pme mor y . The" de l e t e "o pe r a t o rf r e e she a p me mo r y . F ore v e r y" ne w, "y ous ho ul dus ea" de l e t e "s ot ha ty ouf r e et hes a me memor yyoual l oca t ed: c h a r *s t r=ne wc ha r[ 3 0 ] ; / /Al l o c a t e3 0by t e st os t o r eas t r i ng . e a rt h os e3 0b yt e sa ndma kes t rp oi n t e rn owh e r e . de l e t e[ ]s t r ; / /Cl
2 .Be f or er e a l l oc a t i ng me mo r y, t hepr e v i ousme mo r ys h oul db er e l e a s e s. I nt heg i v e n c o debe l o w, s t ra c qui r e sane wa ddr e s swi t ht hes e c onda l l oc a t i on. Thefir s ta ddr e s s i sl os ta nds oar et he1 0b yt e st ha ti tpoi nt e dt o.No wt he y' r ei mpos s i bl et of r e eup t hose10byt es, andyouhaveamemor yl eak:
3 . Ev e r ydyna mi cv a r i a b l e( a l l o c a t e dme mo r yont hehe a p)ne e dst ob ea s s o c i a t e dwi t h apoi nt e r .Whe nady na mi cva r i a bl eb e c ome sdi s a s s o c i a t e df r om i t spoi nt e r ( s ) ,i t be c ome s i mpos s i bl e t o e r as e . Agai n, t hi s r e s ul t s i n a me mor y l e ak:
i nt*a pt r,*bpt r
//de c l a r epoi nt e r s
pt r 1=ne wi nt ;
//Al l oc at edyna mi cme mor y
pt r 2=ne wi nt ;
//Al l oc at edyna mi cme mor y
* pt r 1=2 0 ; * pt r 2=6 0 ; pt r 1=pt r 2;
//ol daddr e s sl os t, me mor yl e ak
4 .Cons i de rt hef ol l o wi ngf unc t i on:
voi ds omef unc ( ) { i nt*p=ne wi nt ; } Thi sf unc t i ona l l oc a t e sme mor ybutne v e rf r e e si tus i ngde l e t e .Apoi nt e ry oude c l a r e i naf unc t i oni sa l l oc a t e dont hes t a c k,butt hedyna mi cv ar i a bl ei tpoi nt st oi s a l l oc a t e dont hehe ap.I fyoudon' tde l e t ei t , i twi l lr e ma i na f t e rt hepr ogr a me xi t s f r o mt hef unc t i on:
7. 15 Thi spoi nt er Thi spoi nt e ri sas pe c i a lt ypeo fpoi nt e ri nC++whi c hc ont a i nt hea ddr e s sofac l a s s obj e c tort hec l a s si ns t a nc e, t ha tc a na c c e s st heme mbe rf unc t i onso ft hec l a s s .Not e t hef ol l owi ngpoi nt sf ort hi spoi nt e r
t hi spoi nt e rs t or et headdr e s soft hec l a s si ns t a nc e, t oe na bl epoi nt e ra c c e s soft he me mb e r st ot heme mb e rf unc t i o nso ft hec l a s s. t hi spoi nt e ri snotc ount e df orc al c ul at i ngt hes i zeoft heobj e c t .
t hi spoi nt e r sa r enota c c e s s i bl ef ors t a t i cme mbe rf unc t i ons .
t hi spoi nt e r sa r enotmodi fia bl e .
As
weknow t hatC++kee psonl yonec opyofeac hmemberf unct i onandt hedat a me mbe r sa r ea l l oc a t e dme mor yf ora l loft he i ri ns t a nc e s . ,e,-er %n&tions are 9o,,on or all o-;e&ts etdat a()
-;e&t 1 o-1
setdat a()
-;e&t 2 o-2
-;e&t 3 o-3
Cons i derac l as sABCwhi c hi sdefinedbel ow c l as sABC { i ntda t a ; publ i c : i ntge t da t a ( ) / /f unc t i onnotus i ngt hi spoi nt e r { r e t ur nda t a ; } voi ds e t da t a ( i ntne wva l ) //f unc t i onus i ng t hi spoi nt e r { t hi s >da t a=ne wv al ; } } ; voi dmai n( ) { ABC ob1 , ob2; ob1 . ge t da t a ( ) ; } Thi sc l a s swhi c hha st wome mbe rf unc t i onsg e t da t a ( )a nds e t da t a ( ). Thef unc t i onc a l l ob1 . ge t da t a ( )wi l ls e tt hepoi nt e rt hi st ot headdr e s sofobj e c tob1. Thi suni quepoi nt e ri s a ut oma t i c a l l ypa s s e dt oame mbe rf unc t i onwhe ni ti sc a l l e d. Thec a l lob2 . ge t da t a ( )wi l l s e tt heva l ueof‘ t hi s ’ poi nt e rt ot headdr e s sofob2obj e c t .
ob1 . ge t da t a ( ) : ,e,-er %n&tions are 9o,,on or all o-;e&ts etdat a() -;e&t 1 o-1 1010
101 0
-;e&t 2 o-2 1020
setdat a() -;e& t3 o-3 1030
t hi spoi nt e r
ob2 . ge t da t a ( ) : ,e,-er %n&tions are 9o,,on or all o-;e&ts etdat a() -;e&t 1 o-1 1010
102
-;e&t 2 o-2 1020
setdat a() -;e& t3 o-3 1030
t hi spoi nt e r
t hepr i v at eva r i a bl eoft he‘ da t a ’ofc l a s sABCc a nbea c c e s s e di ni t sme mbe rf unc t i on s e t dat a( ): t hi s >da t a =ne wv al;
t hes a mes t a t e me ntc a nal s ober e f e r r e da sda t a =ne wv al ; Weha venotbe e nus i ng‘ t hi s ’ poi nt e rf orr e f e r r i ngt opr i va t eda t aoft hec l a s si nt he i r me mbe rf unc t i ons. Onei mpor t a nta ppl i c a t i onoft het hi spoi nt e ri st or e t ur nt he a ddr e s soft hec ur r e ntobj e c ti tpoi nt st o. r e t r un* t hi s
wi l lr e t ur nt heobj e c tt ha tha sc a l l e dt ha tpa r t i c ul a rme mbe rf unc t i on. Thi spoi nt eri sus ef ulwhenwewantt ocompar et woormor eobj ec t si ns i deamember f unc t i ona ndr e t ur nt hei nv oki ngobj e c ta sar e s ul t . c o ns i de ra no t he re x a mpl e
Ex a mp l e: 7 . 2 6 i n c e n t i v e s .
T ofin do utt h ed e t a i l so ft hes a l e s p e r s o nwh oi sh av i n gma x i mum
# i nc l ude # i nc l ude c l a s ss a l e s pe r s on { c h a rna me [ 2 0 ] ; floa ti nc e nt i ve ; publ i c: s a l e s pe r s on(c ha r* s, floa ta ) { s t r c py( na me , s ) ; i nc e nt i v e=a ; }
(ote that how the aress of the current object is returne here usin' this pointer .
s a l e s pe r s o n&s a l e s pe r s o n: : ma x i mum( s a l e s pe r s o n&s ) { i f( s . i nc e nt i ve >=i nc e nt i ve ) r e t ur ns ; e l s e r e t ur n* t hi s ;
%his function call return the object s1 ! if the incenti&e of the salesperson S1 is 'reater than that of s# !other wise it will return the object s#usin'
t
s.
}
v oi ddi s pl a y( ) { c out <<" \n\nName: "<
7 . 1 5Se l f r e f e r e n t i a l s t r u c t u r e s I ti se x ac t l ywha ti ts o undsl i ke :As t r uc t ur ewhi c hc o nt a i nsar e f e r e nc et oi t s e l f .Whe na me mbe ro fas t r uc t ur ei sde c l a r e da sapoi nt e rt ot hes t r uc t ur ei t s e l ft he nt hes t r uc t ur ei s c a l l e da ss el fr e f e r e nt i a ls t r uc t ur e . .Ge ne r a l l ynode sofl i nke dl i s t sa r ee xa mpl e sofs e l f r e f e r e nt i a ls t r uc t ur e s-whi c hr e f e rt ome mor yl oc a t i onofs a met ypebys t or i ngas e l f t y pepo i nt e ra same mb e rEa c hno dene e dsar e f e r e nc et ot hene x tno dei nt hec h a i n.
s t r uc tnode { i ntdat a; node* ne xt ; } ;
/ /<-s e l fr e f e r e nc e
Theda t ame mbe r‘ ne x t‘oft hes t r uc t ur ec a npoi ntt os t r uc t ur eva r i a bl eoft ypenode c o ns i de ra no t he re x a mpl ei nwhi c hs t r uc t ur epo i ntc o nt a i name mbe rv ar i a bl e‘ ne x t ’ whi c hi spoi nt e rt ot hes t r uc t ur epoi nti t s e l f . s t r uc tP oi nt { flo a tx ; flo a ty ; Poi nt *ne xt ; } ; The s el i nke ds t r uc t ur eha sgr e a tpo t e nt i a la ndus e di ni mpl e me nt i ngda t as t r uc t ur e sl i ke l i nke dl i s t, s t ac ks, que ue s,t r e s sdynami c al l y . .
7 . 1 6Re f e r e nc eva r i a bl e s Ar e f e r e nc ei sno t hi ngmo r et ha na na l i a s . Ana l i a sme a nsa no t he rna mef o rt hes a me v a r i a bl e. An yt hi ngy oudowi t har e f e r e nc ey oua r ea c t ua l l ydoi ngt ot heob j e c ty oua r e r e f e r e nc i ng.Re f e r e nc e sa r ede c l a r e db yus i nga na mpe r s a nd( &)b e t we e nt her e f e r e nc e ore x a mpl e:a na l i a syf ort heva r i a bl exc a nbec r e a t e da s t y pea n dt hev a r i a b l en a me .f i ntx; i nt&y=x; x=15
Thea mpe r s a ndi nt hi sc o nt e x tdo e sno tme a n“ a ddr e s so f ” , i tme a ns“ r e f e r e nc et o ” . Now t her ear et wonamesdefinedf ort hes amememor yl oc at i on100 3ass howni nt hedi agr am.
c o ns i de rt hef ol l o wi ngpr o gr a mc o de: c o ut<<” a ddr e s so fxi s:“< <&x ; cout <<”addr es sofyi s: ”<<&y i twi l ldi s pl a y a ddr e s sofxi s1 0 0 3 a ddr e s sofyi s1 0 0 3 t hi ss ho wsb ot ht hea l i a sv a r i a bl e sha st hes a mea ddr e s s .
x=1 5 ;
/ /xi sno w1 5
y=7 ;
/ /xi snow 7
c out <
/ / di s pl a ys7
x++;
/ /xi si nc r e me nt e dt o8
c out<
/ / di s pl a ys8
Thi ss ho wsa n yc h ang ema dei nt hexi sa l s or e fl e c t e db ac ki nt heya l s oa ndvi c ev e r s a . yi sj us t t hedi ffe r e ntna met ot hev ar i a bl ex .
Re me mbe r
Re f e r e nc e sa l wa ys“ poi nt ”t ov al i dobj e c t s , a ndc a nne v e rbepoi nt e dt ode a ll oc a t e d me mor y , r e f e r e nc e sa r es a f e rt ous et ha npoi nt e r s .
Commone r r or s i ntVa l ue=5 ; Re f e r e nc e smus tbeg i v e nv a l ue i nt&Re f va l ue=Va l ue ;/ /v al i dr e f e r e nc e upondecl a r a t i o n i nt& va l
/ / a not he rr e f e r e nc eva r i a bl ewhi c hi snoti ni t i a l i z ed
Commone r r or s i ntv al ue 1=5 ; i ntv al ue 2 =6 i nt&r e f v al ue=va l ue 1; / /va l i dr e f e r e nc e r e f v al ue =va l ue 2;
The reference can not be redirected/ to another variable
/ / as s i g nv al ue6t or e f v al ueb utd oe sn otc ha ng et her e f e r e n c e
$emember
Gecause references always HpointI to valid ob
Ques t i ons Mul t i pl ec hoi c eque s t i ons 1 . As s umet hev a r i a b l ede c l a r a t i o ns : i ntF=0; i nt* pt r=&F; Whi c hoft hef ol l o wi ngs t a t e me nt swi l lc h a ng et heva l ueofFt o1 ? a. pt r ++; b. F++; c . ( * F) ++; d. ( * pt r ) ++; F orq ue s t i o ns2t hr o ug h3 , a s s umet ha tPa ndQa r epo i nt e r so ft hes a met y pe , a nd t ha te a c hha sb e e na s s i gne dav al ue .
2.Whatc ompar i s onwoul ddet er mi newhet herPandQ havet ar get swi t ht hes ame v a l u e ? a. &P==&Q b. * P==* Q c . P==Q d. Al loft he m 3. Whatc ompar i s onwoul dde t e r mi newhet herPandQ havet hes amev al ue? a)&P==&Q b)* P==* Q c )P==Q d)Al loft he m 4 .Co ns i de ri mpl e me nt i ngaf unc t i o nt odyna mi c a l l ya l l o c a t ea na r r a yo fi nt e g e r s a nds e ta l li t se l e me nt st oz er o : v o i dZe r oI t (_ _ __ _ __ _ _A,c ons ti ntS i z e ) { A=ne wi nt [ S i z e ] ; f or( i ntI dx=0 ;I dx i ntma i n( ) { char *p=NULL; p=ne wc har ; * p=' x' ; r e t ur n0; } Wha ti st hev a l ueo f( * p) ++; a. x b. y
c . d.
x+1; p+1
6 .Whi c hc o def r a gme nt ( s )c o ul dbei ns e r t e di nt hebl a nki nor de rt os a f e l yi ni t i a l i z e e a c he l e me n to fAt oz e r o ? *p=&A[ 0] ; f or( i ntI dx=0 ;I dx
pt r+=3; r e t ur n( pt r ) ; } i ntma i n( ) { c ha r* x ,* y ; x=" HELLO" ; y=myFunc( x) ; c out<
b. c . d. e .
Ther ewi l lbeamemor yl eak. The r ewi l lbeas e gme nt a t i onf a ul t . No te no ughs p ac ei sa l l o c a t e db yt hene w. I twi l lnotc ompi l e .
12. i nty[ 4] ={ 6 ,7, 8, 9} ; i nt* pt r=y+2 ; c out <
/* pt r +1==pt r [ 1] * /
Whati spr i nt edwhent hes ampl ec odeabovei se xec ut ed? a . 6 b. 7 c . 3 d. 9 e . Thec o dewi l lno tc o mpi l e . 1 3 . c ha r*dwa r v e s[ ]= { " S l e e p y " , " Do pe y "" Do c " , " Happy" , " Gr umpy"" Snee zy" , " Bas hf ul " , } ; How manyel ement sdoest hear r aydwar ves( dec l ar edabove)c ont ai n?As s ume t heCc o mpi l e re mpl o ye ds t r i c t l yc o mpl i e swi t ht her e q ui r e me nt so fS t a nda r dC. 4 a . b. 5 c . 6 d. 7 e . 8 14. c h ar* buffe r=" 0 1 2 3 45 6 7 8 9" ; c ha r* pt r=buffe r ; pt r+=5 ; c o ut<
Whatwi l lbepr i nt edwhent hes ampl ecodeabovei se xec ut ed? a .0 1 2 3 4 5 6 7 8 9 5 6 7 8 9 1 2 3 4 5 6 7 8 9 b.5 5 1 2 3 4 5 6 7 8 9 c .5 6 7 8 9 5 6 7 8 9 1 2 3 4 5 6 7 8 9 d.0 0 1 2 3 4 5 6 7 8 9 6 7 8 9 e .5 0 1 2 3 4 5 6 7 8 9
15.@onsider the following statements : -include iostream.h/ &oid main0 1 { int a4,(ptr4
56G , u
0b1
567 , t
0c1
56G , &
0d1
56H, u
16.Ehich of the following expression would correctly add ;%st 4 to a pointer DpB pointing to an element of type D struct abc D F 0a1
p?
0b1
0struct abc ( 100char (1p ? 1
0c1
0struct abc ( 100&oid (1p ? 1
0d1
3ore than one of the abo&e
17.>redict the output of the following code : - include iostream.h/ &oid func 0int (ptr1 { (ptr 4 G2 $ int main0 1 { int x 4 5" func 0
G2 5 he program generates a compilation error he >rogram generates a run time error
18.9tudy the following code :
@ns
main0 1 { char (p 4 8I;;" strcpy 0 p , D B1" coutp" $ Ehat will be displayed if the abo&e program runs : 0a1 8I;; 0b1 8othing will be displayed 0c1 @ompiler cannot recogniJe DB in the strcpy function 0d1 Kun ime error as DpB has not been allocated +J 4J ++J +4J
b Cd bCc a a
1J 7J +1J +7J
b c d a
2J J +2J +J
c a b b
3J 6J +3J
d d e
0'$T 2E0T4'50 1 . WHATWI I LBEOUTPUTOFFOLLOWI NG PROGRAM? # i nc l ude #i nc l ude voi dmai n( ) { c l r s c r ( ) ; i nts um( i nt ( * ) ( i nt ) , i nt ) ; i nts quar e ( i nt ) ; i ntc ube ( i nt ) ; cout <
5J +>J +5J
b d a
cout <
30 10 0 2 .“ P oi nt e r sa l wa y sc o nt a i ni nt e g e r s“Co mme n t .
Ans Poi nt e rv ar i a bl ea l wa yss t or ea ddr e s sofav a r i a bl ewhi c hi sa l wa ysa ni nt e ge ror he xa de c i ma li nt e ge r s, Sopoi nt e r sa l wa yss t or ei nt e ge r s . 3 .F i ndt heo ut puto ft hef o l l o wi ngpr o gr a m( 3 )
#i nc l ude # i nc l ude c l as ss t at e { c ha r*s t a t e _ na me ; i nts i ze ; publ i c :
s t at e () {s i ze =0 ; s t a t e _ n a me =ne wc h a r [ s i z e +1 ] ; } s t at e ( c har* s ) {s i ze=s t r l e n( s ); s t a t e _ na me=ne wc h a r [ s i z e +1 ] ; s t r c py( s t a t e _na me , s ) ; } voi ddi s pl a y( ) { cout < voi dmai n( )
{ i ntx [ 5 ] , * y ,z[ 5] f or( i=0;i<5;i++) { x[ i ]=i ; z [ i ]=i+3 ; y=z; x=y; } } Ans # i nc l ude voi dmai n( ) { i ntx [ 5 ] , * y , z [ 5 ]; f or( i=0;i<5;i++) { x[ i ]=i ; z [ i ]=i+3 ; y=z; x=y;
/ / //
Se mi c ol onmi s s i ng; var i abl e‘ i ' i snotde fine d
//val ueofxc annotbec hange d,a sxi saar r a yname ,whi c h // i sa poi nt e rc o ns t a nt
} }
5 . Re wr i t et hef ol l o wi ngc ode sa f t e rr e mo vi nger r or s , i fa ny ,i nt hef ol l o wi ngs ni ppe t . Ex pl a i n e a c h e r r o r . voi dmai n( ) { c ons ti nti=2 0; c o ns ti nt*c ons tpt r=&i ; ( * p t r ) ++; i ntj =1 5; pt r=&j ; } Ans voi dmai n( ) { c ons ti nti=2 0; c o ns ti nt*c o ns tpt r=&i ; ( * pt r ) ++; i ntj =1 5; pt r=&j ;
/ /Va l ueof* pt rc a nnotbemodi fie da si ti sac ons t a nt / / Va l ueo fpt rc a nno tbemo di fie da si ti sa l s oac o ns t a nt
}
6 . Gi vet heout putoft hef ol l owi ngpr ogr am: voi dmai n( ) { char* p=“Di fficul t ”; c h arc ; c=++* p++; cout <
v et heout puto ft hef o l l o wi ngpr o gr a m(As s umea l lne c e s s a r yhe a de rfil e sa r e 8 . Gi i nc l ude d): voi dmai n() { char*x=“t eAmI ndi a”; c ha rc ; c=++* x++; cout<
c=(* ( x +1 ))++; cout< voi dpr i nt( c ha r*p) { p=" pa s s " ; c o ut <<" v a l uei s" <
Out put va l uei spa s s ne wv al uei sBe s tofl uc k 1 1 .Gi v et heo ut puto ft hef o l l o wi ngpr o gr a m: voi dmai n( ) { i ntar r [ ]={ 2 ,3, 4,5} ; i nt* pt r=ar r ; i ntva l=* pt r ; cout<
Ans o u t p u t 2 2
3 4 v et heo ut puto ft hef o l l o wi ngpr o gr a m. 1 2 .Gi # i nc l ude # i nc l ude i nta=1 0 ; i ntma i n() { v oi dde mo(i nt&, i nt,i nt* ) ; c l r s c r () ; i nta=2 0 ; i ntb=1 0 ; de mo(: : a , a , &b) ; c o ut<<: : a<<" \t "<
20 20
400 20
42 0 42 0
HI GH ORDERTHI NKI NG QUESTI ONS 1 3 .Ans wert hef ol l owi ngg nt hede c l a r a t i onflo hepoi nt e rv a r i a bl efi spoi nt i ngt oava r i a bl eoft ype_ __ __ __ 1 .I at* ft I st hef ol l o wi ngde c l a r a t i onva l i d *pt 2 . r . a t et r ueorf a l s e:I nt hede c l a r at i oni ni nt e ge ri sbe i ngde c l a r e d. 3 . St nt*pt r 1a t a t et r ueorf a l s e:I sapoi nt e ri sno ti ni t i a l i z ed,t he r ei sahi ghc ha nc eoft hepr o gr a mc r a s hi ng. 4 .S st hede c l a r a t i on: sv al i d? 5 .I i ntx,*pt ri _ __ _ _ __ _ __ _ _ __ _ __ope r a t o ri su s e dt oge tt hea ddr e s soft hev ar i a bl e . 6 . The_ Ans 1. float 2 . No ,Whe nt hepo i nt e rv a r i a b l ei sde c l a r e dThet y pet owhi c hi ti spo i nt i ngmus tb ede c a l r e d 3 .F a l s e , Apo i nt e rt oi nt e g e ri sde c l a r e dno ti nt e g e r . 4. Tr ue 5 . Ye s , i ti sc o r r e c t.De c l a r a c t i onofapoi nt e rwi t ht hes a met ypeofva r i a bl e si sv al i d. 6. &
oi dmai n( ) 1 4 .v
{ c ons ti nti =2 0; c ons ti nt*c ons tpt r =&i ; ( * pt r ) ++; i ntj =20; pt r =&j ; } ( a)20 ( b )Er r o r Ca nno tMo di f yaCo ns tOb j e c t ( c )ga r b ageva l ue ( d)No neo ft he s e Ans ( b) i nc l ude 1 5 .# # i nc l ude voi dmai n( ) {i nti ; c h a r* p=” S t ude n t ” ; c ha rc ; f or( i =1; i <6; i ++) {i f( i %2 ==0 ) ++p; } c =* p; c o ut <
voi dmai n( ) { c ha r* s t r [ ] ={ “ Book” , ” mbd” , ” c omput e r ” } ; c out <
i ndt heo ut put: 1 7 .F # i nc l ude voi df unc ( i nt* * p) { c o ut <<* * p; } voi dmai n( ) { i nta [ 3 ] [ 4] ={1 , 2, 3, 4, 4 , 3 , 2 , 1 , 5 , 6 , 7 , 8 , } ; i nt* pt r ; pt r =&a [ 0 ] [ 0 ] ; f unc ( &pt r ) ; } ( a) 1 ( b)4 ( c )5 ( d)Noneoft hes e Ans( a) i ndt heo ut put: 1 8 .F voi dmai n( ) { c ha rf ar* S 1, * S 2; c o ut <
Thes i z e of ( a )=6asi tc o nt a i nsa r r a yac ont a i ns5e l e me nt spl uso nef ort henul lc ha r a c t e r s i z e of ( b)=2onl yheme mor ys pa c ef ort hepoi nt e ri sr e s e r v e d s i z eof ( * a )=1s i z eof ( * b) =1a st he ybot ha r er e f e r r i ngt oc ha r a c t e r s . 2 0 .. F i ndt heout putoft hef ol l owi ngpr ogr am :
# i nc l ude voi dmai n( ) { i ntNumbe r s [ ]={ 1 2, 4 , 1 0, 8 } ; i nt* pt r=Number s ; f or( i ntC=0 ;C<3 ;C++) { c out <<* pt r<<“@”; p t r + + ; } c out <
2 1 .Fi ndt heout putoft hef ol l owi ngpr ogram code:
( i )
c har* s1,* s 2; s1=“ BUTTER” ; s t r c py( s 1,s t r nc py(s 1,s 2,3) ) ;
s t r ns et (s 1,77,3) ; cout <
i nta[ ]={ 3,4,5,6} ; i nt* p,* * q,* s,* t,* * s s; p=a; s=p+1; q=&s; t=(* q+1) ; ss=&t ; cout<<* p<<“\t“<<* * q<<“ \t ”<
Ans 3
4
0x123456( He xadec i malv al ue)
i t et heout putoft hef ol l woi ngpr ogr am s egment 2 2 .Wr
# i nc l ude # i nc l ude # i nc l ude # i nc l ude voi dmai n( ) { c har* NAME=" ComPUt e R" ; c l r s c r ( ) ; f or ( i ntx =0 ; x
put s ( NAME) ; } Ans
c OMMuTEE
2 3 .Whatwi l lbet heout putoft hef ol l owi ngpr ogr am : # i nc l ude # i nc l ude # i nc l ude # i nc l ude v oi dc ha ng es t r i ng( c ha rt e xt [ ] ,i nt&c ount e r ) { c ha r* p t r=t e xt ; i ntl e ngt h=s t r l e n( t e xt ) ; f or ( ; c o unt e r
MouSeFun@8
ndt heo ut puto ft hef o l l o wi ngpr o gr a m: 2 4 .Fi # i nc l ude # i nc l ude c l a s ss t ude nt {char* name; i ntI; publ i c : s t ude nt (){ I=0 ;na me =ne wc h ar[I+1 ] ; } s t ude nt( c ha r* s ) {I=s t r l e n( s ) ;na me =ne wc ha r [ I +1 ] ;
s t r c p y( na me , s ) ; } v oi ddi s p l a y (){ c o ut < # i nc l ude voi dmai n() {i nt* P oi nt e r Ar r a y[ 1 0] ; i ntmar ks[ ]={ 75,6 8, 90,3 4, 0,10, 9 0,65 } ; c l r s c r ( ) ; f or( i ntI=0 ;ma r ks[ I ] ! =0 ;I ++) {Poi nt e r Ar r a y[ I ] =&ma r ks [ I ] ; *( Poi nt e r Ar r a y[ I ] )+=5 ; } i nti nde x=0 ; whi l e ( i nde x =6 0 ) c o ut <
Ans 8 0, 7 3, 9 5,
v et heo ut puto ft hef o l l o wi ngpr o gr a m: 2 6 .Gi # i nc l ude # i nc l ude mai n() { i nta =0 ; c l r s c r () ; c har* name; name=“I nt er netBr ows i ng”; f or ( a =0 ; a <=8 ; a ++) c out <
Out put nt e r ne tB B 65 4 0Gi v et heo ut puto ft hef o l l o wi ngpr o gr a m: # i nc l ude # i nc l ude # i nc l ude # i nc l ude mai n() { c l r s cr () ; c har* name; i ntl e n=0 ; name=" Obj ec tOr i ent edPr ogr ammi ng" ; l e n=s t r l e n( na me ) ;
l e n=l e n1 ; f or ( i nti =l e n; i >=0; i =i 2) { c o ut <
4 5Wha ti st heout puto ft hef o l l o wi ngpr o gr a m(As s umea l lne c e s s a r yhe a de rfil e sa r e i nc l ude d): # i nc l ude # i nc l ude # i nc l ude voi dmai n() { c har*s t r=" TEACHER" ; f or(i nti=s t r l e n( s t r )-1; i>=0; i) { f or(i ntx=0;x<=i;x++) c out <
4 6Wha ti st heout puto ft hef o l l o wi ngpr o gr a m(As s umea l lne c e s s a r yhe a de rfil e sa r e i nc l ude d): voi dmai n() { i nt*p[ 10] ; i nts c or e [ ]={ 80, 30, 74, 50, 34,67, 0, 98} ; f or(i nti=0; s c or e [ i ]! =0; i++) { p[ i ]=&s c or e [ i ] ; * ( p[ i ] )+=1 0; } i ntc o unt=0 ; whi l e (c o unt5 0) c out < # i nc l ude # i nc l ude # i nc l ude v oi dNe wT e xt ( c ha rs t r [] , i nt& pos ) { c ha r*p=s t r ; i ntl e ngt h=s t r l e n( p) ; f or (; pos
} 4 8Wha ti st heout puto ft hef o l l o wi ngpr o gr a m(As s umea l lne c e s s a r yhe a de rfil e sa r e i nc l ude d): #i nc l ude #i nc l ude # i nc l ude # i nc l ude c l a s sc ount r y { c har*c _name; i ntl e ngt h; publ i c : c o unt r y() { l e ng t h=0 ; c _name=newc har [ l engt h+1] ; } c o unt r y( c ha r*s ) { l e ngt h=s t r l e n( s ) ; c _name=newc har [ l engt h+1] ; s t r c py( c _ na me ,s ) ; } v oi ddi s pl a y( ) { c out <
c 3. di s pl a y() ; ge t c h() ; }
e1 Ehat will be the output of the following program: [*! -includeiostream.h/ -includectype.h/ -includeconio.h/ -includestring.h/ &oid >ointersLun0char ext[!, int ointersLun0Mata, position1"
coutMata DNB position" cout.write0Mata ? *, 1" $ 12Consi de rt hef o l l o wi ngs t a t e me nt s Cha rc h , c h 1=’ A’ ; Char* p, * p1=&ch1; *p1=ch1+1; *p=ch; S uppo s ee a c hc h ar a c t e ro c c u pi e s2b yt e so fme mo r y . I ft heva l uea s s i gne dt oc hi ss t o r e di n a ddr e s s0 x 22a ndt hev al uea s s i gne dt oc h 1i ss t o r e di na ddr e s s0 x 10 5t he n ( i )wha tv al uei sa s s i gne dt op1 ; ( i i )wha ti st heva l ueof* p1 ; ( i i i )wha tv a l uei sa s s i gne dt oc h? ( i v)Wha tv al uei sa s s i gne dt o* p?
e )F i ndt hec o r r e c tpos s i bl eo ut puto ft hef o l l o wi ngpr o gr a m (Cons i de ra l lt hehe a de r fil e sa r epr e s e nt ). s t r uc tc ar d{ c ha rs ui t [ 1 0] ; i ntdi gi t ; } ; car d*ca r ds=newcar d[ 52] ;//Al l ocat eMemory v oi dc r e a t e de c k( ) {
chart emp[ ] [ 10]={ " Cl ubs" , " Spades " , " Di amonds " , " Hear t s " } ; i nti , m=0 , c nt =1 ; f or ( i =1 ; i <=5 2; i ++) {
s t r c py( c ar ds [ i ] . s ui t , t emp[ m] ) ; c ar ds [ i ] . di gi t =c nt ; cnt++; i f ( i% 1 3==0 ) {m++;c nt =1; } }}
car ddr awcar d( i ntnum) {
i ntr ndnum; r a ndo mi z e ( ) ; r ndnum =r a ndom( num) +1; r e t ur n( c a r ds [ r ndnum] ) ;
} voi dmai n( ) {
c r e at e de c k( ) ; c a r dc ; c=dr a wc a r d( 3 9 ) ; i f ( c . di gi t>1 0||c . di gi t==1 ) {
s wi t c h( c . di gi t ) {
c as e11 :
c out <<" J ac kof" ; br e a k;
c as e12 :
c out <<" Que e nof" ;br e a k;
c as e13 :
c out <<" Ki ngof" ; br e ak;
c as e1 :
c out <<" Ac eof" ;
}} e l s e c out <
i )
Ki ndofSpade s
i i i ) Ac eofDi amond
i i )
Ac eofCl ubs
i v)
Que e nofHe a r t s
4 1 .Gi v et heo ut putoft hef o l l o wi ngpr o gr a m(As s umea l lne c e s s a r yhe a de rfil e sa r e i nc l ude d) : voi dmai n() { c har*x=“t eAmI ndi a”; c ha rc ; c=++(* ( x +1 )) ; cout<
17Whatwi l lbet heout putoft hef ol l owi wi ngpr ogr am: m: # i nc l ude # i nc l ude # i nc l ude # i nc l ude v oi dc ha ng es t r i ng( c ha rt e xt [ ] ,i nt&c ount e r ) { c ha r* p t r=t e xt ; i ntl e ngt h=s t r l e n( t e xt ) ; f or (; c o unt e r #include #include #include void main() { clrscr(); char * NAM ! "a ro$i%e"; &or (int ' ! ;' #include #include #include void main() { clrscr(); char *s ! "3445%678"; &or (int ' ! strlen(s)01 ; '>!; '00) { &or ( int y!; y
cout< #include #include #include void main() { int a ! 9: * ! a; char ch ! =: eco ! ch; eco ! a; * ! ch; cout << a << ?:? << ch << endl; 2 #include #include #include #include void Nete't(char @tring+,: int osition) { char *ointer ! @tring; int %ength ! strlen(@tring); &or ( ; osition<%ength 0 ; osition!: ointer) { *(ointerosition) ! toupper(*(ointer osition)); 2 2 void main() { clrscr(); int %ocation ! ; char Message+, ! "5ynamic Act"; Act"; Nete't(Message: Nete't(Message: %ocation); cout << Message << " # " << %ocation; 2 #include void main() int '+, ! {1: : 9: :2; int *p: **B: *t; p ! '; t ! ' 1; B ! t; cout<<*p<<":"<<**B<<":"<<*t; 2 #include #include struct 3ame { char Mag+,; int @core; 2; void main() { clrscr(); 3ame M!{"Ciger": 2; char *7hoice; 7hoice ! M.Mag; 7hoice+, ! ??; 7hoice+, ! ?%?; M.@core!; cout << M.Mag << M.@core << endl; 3ame N ! M;
{
N.Mag+, ! ?A?; N.Mag+9, ! ?D?; N.@core 0! 1; cout << N.Mag << N.@core << endl; 2
Q1( i ) Fi ndt hemi mi st ake,i fanyi nt hef ol l owi ngdefini t i on andma maket he necessary corr ect i ons: c har* pt r,s [ 10] ; pt r=s [ 6] ; ( i i )
Di s t i ngui shbet we en: i nt*pt r=ne wi nt ( 5) ; i nt* p t r=ne wi nt [ 5] ;
Q2
Gi v ent hef ol l owi ngdec l ar at i ons: i nt i v al =2048; i nt * i pt r ; doubl e* dpt r ; Whi c ho ft hef ol l o wi wi ngas s i g nme nt s, i fany,ar ei l l e g al? Expl ai nwhy? ( a)i val=* i pt r ;
Q3
( b) * i pt r=&i val
( c)* i pt r=i val ( d)dpt r=i pt r
I fARR i sanarr ayofi nt eg er s,whyi st heexpr es si on ARR++i s no tl e gal?
Q4
Di ffer ent i at ebe t wee nSt at i candDynami mi cal l oc at i onofmemo mor y .
Q5
Whatdoyou under s t and byme memor yl eaks?W Wh hatar epos si bl er eas onsf ori t? How canme memo mor yl eaksbeav oi ded?
Q6
Whati st hi spoi nt er?Wh Whati si tsi gni ficance?
3 8 .Wr i t eaf unc t i onwhi c ht a ke sp oi nt e r st ot wos t r i ngsa spa r a m me e t e ra ndc o mp mpa r et he s e s t r i ngs . Thef unc t i ons h oul dr e t ur n1i ft wos t r i ngsa r es a me meot he r wi s er e t ur n0 .
Q7
How doe st hef unct i oni ngofaf unct i ondi fferwhen ( i ) ( i i ) ( i i i )
anobj ecti spas sedbyval ue anobj ec ti spassedbyre f er ence an obj ecti spassedbypoi nt er
Q8
Expl ai nse l fr ef er ent i als t r uct ur eswi t hexampl e.
Q9
Whatdoest hi spoi nt erdo?Whatar et heappl i cat i onoft hi spoi nt er?
Q10 Definet hes t r i ngcl asshavi ngf ol l owi ngmemberf unct i ons: ( a) ( b) ( c) ( d) ( e)
upcas e() l owcase() l engt h() scopy() sconc at ()
: : : :
Thatconv er t st hes t r i ngi nt oupper cas e : Thatconve r tt hest r i ngt ol ower case Thatcal cul at est hel engt hofs t r i ng Tocopyt hes t r i ng Toconc at enat et wos t r i ngs
DEC (D6@C EN CN MEN6C@) 1.
Wr i t east at ementt hatdi spl ayst headdr essoft hev ari abl et es t var . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
2.
Thecont ent soft wopoi nt er st hatpoi ntt oadj acentvar i abl eoft ypefloat di fferby____ ____ __
3.
Apoi nt eri s
4.
1.an addr essoft hevar i abl e 2.an i ndi cat i onoft hevari abl et obeaccessednext 3.avari abl eofst ori ngaddr ess 4.Thedat at ypeofanaddr essvar i abl e. I nasel fr ef er ent i alst r uct ur e
5.
1.eachl i nkcont ai nsapoi nt ert ot henextl i nk 2.anarr ayofpoi nt er spoi ntt ot hel i nks 3.eachl i nkcont ai nsdat aorpoi nt ert odat a 4.t hel i nksarest or edi nan arr ay Thene w ope r at or
6.
1.r et ur nsapoi nt ert oavari abl e 2.cr eat esavari abl ecal l ednew 3.obt ai nsmemor yf oranew v ar i abl e 4.t el l show muchmemor yi sav ai l abl e Thefir s tel ementi nas t r i ngi s 1.Thenameo ft hes t r i ng 2.Thefir stcharact eri nt hest r i ng 3.Thel engt hoft hest r i ng
7.
4.t henameoft hearr ayhol di ngt hest r i ng Thee xpr ess i on* t es tcanbes ai dt o
8.
1.beapoi nt ert onext 2.r e f e rt ot hec o nt e nto ft e s t 3.der ef er encet est 4.r e f e rt ot hev al ueoft hev ar i abl epo i nt e dt obyt e s t Usi ngnew mayr es ul ti nl es s__ __ __ __ __ __ _ __ __ __memor yt hanusi ngan ar r ay .
9
Thedel et eoper at orr et ur ns__ __ __ __ __ __ __ __ __ __ _t ot heoper at i ngsy st em.
10.
Gi ve napoi nt erpt hatpoi nt st oanobj ectoft ypeupper cl ass,wr i t ean expr essi on t hatexecut est heexcl u()memberf unct i on i nt hi sobj ect . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
11.
Wr i t eadefini t i onf oranarr aynumpt r sofpoi nt er st ot hes t r i ngs“ One”, “ t wo”
and “Thr ee” . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
12.
Usi ngpoi nt ernot at i on wri t et hepr ot ot ypef oraf unct i on cal l edr evst r ()t hat r et ur nsast r i ngval ueandt akesoneargumentt hatr epr esent sast r i ng. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
13.
Usi ngpoi nt ernot at i onwr i t ecodet hatwi l lt r ansf er80charact er sf r om t he s t r i ngs1t ot hes t r i ngs 2. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
14.
Usi ngpoi nt ernot at i onwri t eapr ot ot ypedecl arat i onf oraf unct i oncal l ed f unc ()t hatr et urnst ypev oi d andt akesasi ngl ear gumentt hati san arr ayoft ype char. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
15. Oft het hr eewayst opassar gument st of unct i ons,onl ypassi ngby ____ ____ ____ _____ andpassi ng
by____ _____ ____ ____ _al l ow thef unct i on t o
modi f yt hear gumenti nt hecal l i ngpr ogr am.
