How sizeof(array) works at runtime?

c sizeof

dark_shadow · Apr 9, 2012 · Viewed 10k times · Source

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

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

How sizeof(array) works at runtime?

Answer

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