Why is sizeof(a) zero? where a is int a[0]?

[kakultech]

$ cat VLA.c

#include<stdio.h>
void main()
{
        int a[0];
        int *b;

        /*As we say a[0] is equivalent to *(a+0) */
        printf("\n a = %d b = %d",sizeof(a),sizeof(b));
}

$gcc VLA.c
$ ./a.out

 a = 0 b = 4

Why sizeof(a) is zero ?

[Gobi Sakthivel]

u have declared int a[0], which means declaring an array which contains no elements, tats y it is returning as 0.Also u cant do anything with it, it is of no use, to declare an array with 0 elements.

[kakultech]

Thanks for reply.
How should we differentiate int a[0] and int *b ? As we say a[0] is equivalent to *(a+0), which is nothing but *a.

[Gobi Sakthivel]

int *b, is a pointer to an integer and int a[n], is an array of n integers. Dont get confused by the two.


if we declare int *b; 4 bytes of memory will be allocated
for it (considering sizeof int as 4).

consider b points to memory location 1000 , so b[0]==*(b+0) points to 1000,
b[1]==*(b+1) points to 1004
and so on.

if we declare int a[0];
intially no memory gets allocated for it.
but when do some operations like a[0]=2; then 'a' will points to some memory location say '2000'
no a[1] will points 2004 and goes on.

[donbock]

By the way, line 2 should be

int main(void)

As a consequence of making this change, you will need to return a value from main.

[swapnali143]

for any pointer size is 4 Bytes because.. Pointer is used to store memory address and every memory address in Computer are integer valuse so sizeof((int*a)) is always 4 bytes...