Thread: sizeof

The argument to sizeof() must be a known type at compile time......It should not be executable statement....and it take one argument only...that is why it is called compile time unary operator

Well since C99 this is not correct anymore. Some (propably most) sizeof-statements are evaluated during compile time. Others are not, since C99 introduced VLAs (variabel length arrays) so the sizeof operator had to be modified to support VLAs.

Originally Posted by TactX

Well since C99 this is not correct anymore. Some (propably most) sizeof-statements are evaluated during compile time. Others are not, since C99 introduced VLAs (variabel length arrays) so the sizeof operator had to be modified to support VLAs.

This is a new concept for me...plz explain it with some example

Code:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    unsigned int x;
    printf("Enter array size: ");
    fflush(stdout);
    scanf("%u", &x); /* check for valid input in real code */
    int k[x]; /* C99 variable array, C99 also allows declarations anywhere */
    printf("sizeof(k) computed at runtime: %zd\n", sizeof k);
    return 0;
}

My output:

Code:

Enter array size: 100
sizeof(k) computed at runtime: 400

Edit: change array from char to int, to illustrate better.

Amazing.....it worked on dev C++..but didn't worked on VC6.0(i guess its old)......

Originally Posted by TactX

Well since C99 this is not correct anymore. Some (propably most) sizeof-statements are evaluated during compile time. Others are not, since C99 introduced VLAs (variabel length arrays) so the sizeof operator had to be modified to support VLAs.

i searched a bit more on this.....VLA's are pointers in C99.
That is, sizeof(VLA)==sizeof(Type *)==4 for a 32 bits compiler!....so my statement was perfectly right