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B mn Cng ngh phn mm Khoa Cng ngh thng tin Trng i hc Khoa hc T nhin

1

ThS. ng Bnh Phng [email protected]

K THUT LP TRNH

D LIU KIU CON TR (C BN)

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Ni dung

D liu kiu con tr (c bn)

Khi nim v cch s dng 1

Cc cch truyn i s cho hm 2

Con tr v mng mt chiu 3

Con tr v cu trc 4

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3 3

Kin trc my tnh

B nh my tnh B nh RAM cha rt nhiu nh, mi nh

c kch thc 1 byte. RAM dng cha mt phn h iu hnh,

cc lnh chng trnh, cc d liu Mi nh c a ch duy nht v a ch ny

c nh s t 0 tr i. V d

RAM 512MB c nh a ch t 0 n 229 1 RAM 2GB c nh a ch t 0 n 231 1


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Quy trnh x l ca trnh bin dch Dnh ring mt vng nh vi a ch duy nht

lu bin . Lin kt a ch nh vi tn bin. Khi gi tn bin, n s truy xut t ng n

nh lin kt vi tn bin. V d: int a = 0x1234; // Gi s a ch 0x0B

Khai bo bin trong C


0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

a

34 12 00 00

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5 5

Khi nim con tr

Khi nim a ch ca bin l mt con s. Ta c th to bin khc lu a ch ca

bin ny Con tr.


0A

34

0B

12

0C

00

0D

00

0E 0F 10 11 12 13 14 15 16 17

a pa

0B 00 00 00

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6 6

Khai bo con tr

Khai bo Ging nh mi bin khc, bin con tr mun

s dng cng cn phi c khai bo

V d

ch1 v ch2 l bin con tr, tr ti vng nh kiu char (1 byte). p1 l bin con tr, tr ti vng nh kiu int (4

bytes) cn p2 l bin kiu int bnh thng. D liu kiu con tr (c bn)

*;

char *ch1, *ch2;

int *p1, p2;

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7 7

Khai bo con tr

S dng t kha typedef

V d

Lu khi khai bo kiu d liu mi Gim bi ri khi mi tip xc vi con tr. Nhng d nhm ln vi bin thng.


typedef *; ;

typedef int *pint;

int *p1;

pint p2, p3;

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8 8

Con tr NULL

Khi nim Con tr NULL l con tr khng tr v u c. Khc vi con tr cha c khi to.


NULL

int n;

int *p1 = &n;

int *p2; // unreferenced local variable

int *p3 = NULL;

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9 9

Khi to kiu con tr

Khi to Khi mi khai bo, bin con tr c t a

ch no (khng bit trc). cha gi tr khng xc nh tr n vng nh khng bit trc. t a ch ca bin vo con tr (ton t &)

V d


= &;

int a, b;

int *pa = &a, *pb;

pb = &b;

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S dng con tr

Truy xut n nh m con tr tr n Con tr cha mt s nguyn ch a ch. Vng nh m n tr n, s dng ton t *.

V d


int a = 5, *pa = &a; printf(%d\n, pa); // Gi tr bin pa printf(%d\n, *pa); // Gi tr vng nh pa tr n printf(%d\n, &pa); // a ch bin pa

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

a pa

0B 00 00 00 05 00 00 00

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Kch thc ca con tr

Kch thc ca con tr

Con tr ch lu a ch nn kch thc ca mi con tr l nh nhau: Mi trng MD-DOS (16 bit): 2 bytes (64KB) Mi trng Windows (32 bit): 4 bytes (4GB)


char *p1;

int *p2;

float *p3;

double *p4;

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Cc cch truyn i s

Truyn gi tr (tham tr)


#include

void hoanvi(int x, int y);

void main()

{

int a = 3; b = 6;

hoanvi(a, b);

printf(a = %d, b = %d, a, b); }

void hoanvi(int x, int y)

{

int t = x; x = y; y = t;

}

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int t = x; x = y; y = t;

Truyn gi tr (tham tr)


05 00 00 00 06 00 00 00

int a = 3

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

int b = 6

int x int y hoanvi

18 19 1A 1B 1C 1D 1E 1F 20 21 22 23 24 25

int x int y

05 00 00 00 06 00 00 00

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Cc cch truyn i s

Truyn a ch (con tr)


#include

void hoanvi(int *x, int *y);

void main()

{

int a = 3; b = 6;

hoanvi(&a, &b);

printf(a = %d, b = %d, a, b); }

void hoanvi(int *x, int *y)

{

int t = *x; *x = *y; *y = t;

}

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int t = *x; *x = *y; *y = t;

Truyn a ch (con tr)


int a = 3

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

int b = 6

int *x int *y hoanvi

18 19 1A 1B 1C 1D 1E 1F 20 21 22 23 24 25

int *x int *y

0B 00 00 00 0F 00 00 00

05 00 00 00 06 00 00 00

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Cc cch truyn i s

Truyn tham chiu (C++)


#include

void hoanvi(int &x, int &y);

void main()

{

int a = 3; b = 6;

hoanvi(a, b);

printf(a = %d, b = %d, a, b); }

void hoanvi(int &x, int &y)

{

int t = x; x = y; y = t;

}

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int t = x; x = y; y = t;

Truyn tham chiu (C++)


int a = 3

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

int b = 6

int &x int &y hoanvi

05 00 00 00 06 00 00 00

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Mt s lu

Mt s lu Con tr l khi nim quan trng v kh nht

trong C. Mc thnh tho C c nh gi qua mc s dng con tr. Nm r quy tc sau, v d int a, *pa = &a;

*pa v a u ch ni dung ca bin a. pa v &a u ch a ch ca bin a.

Khng nn s dng con tr khi cha c khi to. Kt qu s khng lng trc c.


int *pa; *pa = 1904; // !!!

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Con tr v mng mt chiu

Mng mt chiu

Tn mng array l mt hng con tr khng th thay i gi tr ca hng ny. Gi tr ca array l a ch phn t u tin

ca mng array == &array[0]


int array[3];

array

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

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Con tr n mng mt chiu



int array[3], *parray;

parray = array; // Cch 1

parray = &array[0]; // Cch 2

array

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

18 19 1A 1B 1C 1D 1E 1F

parray

0B 00 00 00

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Php cng (tng) + n + n * sizeof() C th s dng ton t gp += hoc ++

+2

Php ton s hc trn con tr


p = array

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

+1

int array[3];

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Php tr (gim) n n * sizeof() C th s dng ton t gp = hoc



p = &array[2]

1 2

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

int array[3];

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p2 = &array[2] p1 = array

Php ton tnh khong cch gia 2 con tr *p1, *p2; p1 p2 cho ta khong cch (theo s phn t)

gia hai con tr (cng kiu)



0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

int array[3];

p1 p2= (0B 13)/sizeof(int) = 2

p2 p1= (13 0B)/sizeof(int) = +2

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Cc php ton khc Php so snh: So snh a ch gia hai con

tr (th t nh) == != > >= <

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Truy xut n phn t th n ca mng int array[3], n = 2, *p = array; array[n] == p[n] == *(p + n)



p

0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17

int array[3];

+ 2 ) ( *

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V d nhp mng


void main()

{

int a[10], n = 10, *pa;

pa = a; // hoc pa = &a[0];

for (int i = 0; i

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V d xut mng


void main()

{

int a[10], n = 10, *pa;

pa = a; // hoc pa = &a[0]; for (int i = 0; i

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Truyn mng 1 chiu cho hm

Ch ! Mng mt chiu truyn cho hm l a ch ca

phn t u tin ch khng phi ton mng.


10 11 13 14 15 16 17 18 19 20 21 22 23 24

int array[3];

int a[3] int n xut

int a[] int *a

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i s mng truyn cho hm

V d


void xuat(int a[10], int n)

{

for (int i = 0; i

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Lu Khng thc hin cc php ton *, /, %. Tng/gim con tr n n v c ngha l

tng/gim gi tr ca n n*sizeof() (bytes) Khng th tng/gim bin mng (con tr

hng). Hy gn mt con tr n a ch u ca mng v tng/gim con tr . i s mng mt chiu truyn cho hm l a

ch phn t u tin ca mng.



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Con tr v cu trc

Truy xut bng 2 cch

V d


typedef struct

{

int tu, mau;

} PHANSO;

PHANSO ps1, *ps2 = &ps1; // ps2 l con tr

ps1.tu = 1; ps1.mau = 2;

ps2->tu = 1; ps2->mau = 2;

(*ps2).tu = 1; (*ps2).mau = 2;

-> (*).

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Bi tp

Bi 1: Cho on chng trnh sau: float pay;

float *ptr_pay;

pay=2313.54;

ptr_pay = &pay;

Hy cho bit gi tr ca: a. pay

b. *ptr_pay

c. *pay

d. &pay


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Bi tp

Bi 2: Tm li


#include

#include

void main()

{

int *x, y = 2;

*x = y;

*x += y++;

printf("%d %d", *x, y);

getch();

}

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Bi tp

Bi 3: Cho on chng trnh sau: int *pint;

float f;

char c;

double *pd;

Hy chn pht biu sai c php: a. f = *pint;

b. c = *pd;

c. *pint = *pd;

d. pd = f;


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Bi tp

Bi 4: Ton t no dng xc nh a ch ca mt bin?

Bi 5: Ton t no dng xc nh gi tr ca

bin do con tr tr n? Bi 6: Php ly gi tr gin tip l g?


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Bi tp

Bi 7: Cc phn t trong mng c sp xp trong b nh nh th no?

Bi 8: Cho mng mt chiu data. Trnh by 2 cch ly a ch phn t u tin ca mng ny.


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Bi tp

Bi 9: Trnh by 6 loi php ton c th thc hin trn con tr?

1. 2. 3. 4. 5. 6.


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Bi tp

Bi 10: Cho con tr p1 tr n phn t th 3 cn con tr p2 tr n phn t th 4 ca mng int th p2 p1 = ?

Bi 11: Ging nh cu trn nhng i vi mng float?


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Bi tp

Bi 12: Trnh by khai bo con tr pchar tr n kiu char.

Bi 13: Cho bin cost kiu int. Khai bo v khi

to con tr pcost tr n bin ny. Bi 14: Gn gi tr 100 cho bin cost s dng

hai cch trc tip v gin tip. Trc tip: Gin tip:


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Bi tp

Bi 15: In gi tr ca con tr v gi tr ca bin m n tr ti.

Bi 16: S dng con tr lm li cc bi tp v

mng mt chiu. Nhp/Xut mng Tm phn t tha yu cu Tnh tng/m cc phn t tha yu cu Sp xp tng/gim


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