Answer / susie

Answer :

1

Explanation:

Arrays cannot be passed to functions as arguments and only
the pointers can be passed. So the argument is equivalent to
int * array (this is one of the very few places where [] and
* usage are equivalent). The return statement becomes,
sizeof(int *)/ sizeof(int) that happens to be equal in this
case.

Answer / manoj kumar

Arrays cannot be passed to functions as arguments and only
the pointers can be passed.

main() { int i=-1; -i; printf("i = %d, -i = %d \n",i,-i); }

1 Answers  

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{ int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }

4 Answers  

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Is this code legal? int *ptr; ptr = (int *) 0x400;

1 Answers  

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main( ) { static int a[ ] = {0,1,2,3,4}; int *p[ ] = {a,a+1,a+2,a+3,a+4}; int **ptr = p; ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *++ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); ++*ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); }

2 Answers   Persistent,

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#include<stdio.h> main() { char s[]={'a','b','c','\n','c','\0'}; char *p,*str,*str1; p=&s[3]; str=p; str1=s; printf("%d",++*p + ++*str1-32); }

1 Answers  

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void main() { void *v; int integer=2; int *i=&integer; v=i; printf("%d",(int*)*v); }

1 Answers   Honeywell,

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PROG. TO PRODUCE 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

1 Answers  

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What is the output for the following program main() { int arr2D[3][3]; printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) ); }

1 Answers  

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main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }

1 Answers  

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main() { show(); } void show() { printf("I'm the greatest"); }

2 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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#define f(g,g2) g##g2 main() { int var12=100; printf("%d",f(var,12)); }

3 Answers  

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