How sizeof(array) works at runtime?

Question

I have read that sizeof operator in C is interpreted at compile time and since at compile time compiler knows the array size and its type,sizeof is abled to compute the number of bytes occupied by array.But how is sizeof working for the following code :

 #include<stdio.h>
 #include<string.h>
 int main()
 {
    int n;
    scanf("%d",&n);
    int a[n];
    int s=sizeof(a);
    printf("%d",s);
    return 0;
 }

Here array size is not known at compile time,then how is it working properly ?

Answer 1

sizeof is always computed at compile time in C89. Since C99 and variable length arrays, it is computed at run time when a variable length array is part of the expression in the sizeof operand.

Same for the evaluation of the sizeof operand: it is not evaluated in C89 but in C99 if the operand is of variable length array type it is evaluated. For example:

int n = 5; sizeof (int [n++]);   // n is now 6 

Answer 2

Since you are applying sizeof to a variable-length array, whose size is not fully known at compile time, the compiler must generate code to do part of it at runtime.

gcc 4.6.3's high level optimizers convert the code you showed to

scanf ("%d", &n);
t12 = (long unsigned int) n;
t13 = t12 * 4;
__builtin_alloca (t13);
t16 = (unsigned int) t12;
t17 = t16 * 4;
s18 = (int) t17;
printf ("%d", s18);

Does that explain what is going on under the hood? (Don't be put off by the silly-looking number of temporary variables -- that's an artifact of the program being in static single assignment form at the point where I asked for an intermediate-code dump.)