Answer / susie

Answer :

0

Explanation:

After ‘ptr’ reaches the end of the string the value pointed
by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of
‘least’. So the value of ‘least’ finally is 0.

main( ) { void *vp; char ch = ‘g’, *cp = “goofy”; int j = 20; vp = &ch; printf(“%c”, *(char *)vp); vp = &j; printf(“%d”,*(int *)vp); vp = cp; printf(“%s”,(char *)vp + 3); }

1 Answers  

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struct Foo { char *pName; }; main() { struct Foo *obj = malloc(sizeof(struct Foo)); clrscr(); strcpy(obj->pName,"Your Name"); printf("%s", obj->pName); } a. Your Name b. compile error c. Name d. Runtime error

3 Answers   HCL,

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#define SQR(x) x * x main() { printf("%d", 225/SQR(15)); } a. 1 b. 225 c. 15 d. none of the above

3 Answers   HCL,

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Finding a number multiplication of 8 with out using arithmetic operator

8 Answers   Eyeball, NetApp,

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main() { char *a = "Hello "; char *b = "World"; clrscr(); printf("%s", strcat(a,b)); } a. Hello b. Hello World c. HelloWorld d. None of the above

3 Answers   HCL,

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main() { char *p; p="%d\n"; p++; p++; printf(p-2,300); }

1 Answers  

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#include<stdio.h> main() { int i=1,j=2; switch(i) { case 1: printf("GOOD"); break; case j: printf("BAD"); break; } }

1 Answers  

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struct aaa{ struct aaa *prev; int i; struct aaa *next; }; main() { struct aaa abc,def,ghi,jkl; int x=100; abc.i=0;abc.prev=&jkl; abc.next=&def; def.i=1;def.prev=&abc;def.next=&ghi; ghi.i=2;ghi.prev=&def; ghi.next=&jkl; jkl.i=3;jkl.prev=&ghi;jkl.next=&abc; x=abc.next->next->prev->next->i; printf("%d",x); }

1 Answers  

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#include<stdio.h> int main() { int a=3,post,pre; post= a++ * a++ * a++; a=3; pre= ++a * ++a * ++a; printf("post=%d pre=%d",post,pre); return 0; }

3 Answers  

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main() { int x=5; clrscr(); for(;x<= 0;x--) { printf("x=%d ", x--); } } a. 5, 3, 1 b. 5, 2, 1, c. 5, 3, 1, -1, 3 d. –3, -1, 1, 3, 5

2 Answers   HCL,

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Write a C program to print ‘Campus Force training’ without using even a single semicolon in the program.

3 Answers   Wipro,

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writte a c-programm to display smill paces

2 Answers  

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