Answer / susie

Answer :

i = 1 j = 1 k = 1

Explanation:

Since static variables are initialized to zero by default.

Answer / mittu

0 ,0 ,0

It gives 0 0 0 for all three.
Cause at compile time compiler assigns memory to the static variable in HEAP. So these are automatically initialized to 0.

So First Allocation and then Initialization takes place.

At compile time it allocates memory to i,j,k.
And then initialization phase first is assigns 0 to 'i','j' and 'k' and
then it uses this ++ for increment the value at "Increment Buffer".
So i,j,k are initialized by 0 and value of i , j,k is incremented by 1 in "Increment Buffer".

THIS IS DONE AT COMPILE TIME.

Now at run time it refers the value of i,j and k from HEAP.
And at run time it skips "static statements".

So in HEAP value of i , j, and k is 0(zero).

Answer / ankur

illegal initialization

main() { int x=5; clrscr(); for(;x==0;x--) { printf("x=%d\n”", x--); } } a. 4, 3, 2, 1, 0 b. 1, 2, 3, 4, 5 c. 0, 1, 2, 3, 4 d. none of the above

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#include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d..%d",*p,*q); }

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main() { register int a=2; printf("Address of a = %d",&a); printf("Value of a = %d",a); }

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void main() { int const * p=5; printf("%d",++(*p)); }

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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); }

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