What's the difference between a and a + 0 as C pointers?

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Look at the C code.

#include <stdio.h>
int main(void)
    int v[3] = {1, 2, 3};
    printf("%d\n", v);
    printf("%d\n", v + 0);
    printf("%zu\n", sizeof(v));
    printf("%zu\n", sizeof(v + 0));
    printf("%zu\n", sizeof(v + 1));
    printf("%zu\n", sizeof(v + 2));
    return 0;

Here's one of the outputs:


I think v is the same as v + 0.

Both are pointers pointing to the first element in array v[3].

Therefore, v and v+0 have the same value.

But why don't they hold different bytes? (sizeof(v) and sizeof(v + 0) are different )

answered question

1 Answer


An array identifier decays to a pointer to the first element of the array, in most cases. But, when an array is the operand of the sizeof operator, this conversion does not take place, and the operator yields the size of the array in bytes.

So, sizeof(v) yields the size of the array v[] in bytes, which is 12 bytes. But with sizeof (v + 0) the type of the expression v + 0 determines the result yielded by the sizeof operator. In the expression v + 0, the array v[] decays to a pointer to the first element of v[], and then 0 is added according to the rules of pointer arithmetic. The result is a pointer to int (since &v[0] is itself a pointer to int), so the type of the expression v + 0 is int *. The sizeof operator thus yields the size of a pointer to int in this case. The same holds for the similar expressions sizeof (v + 1), etc.

As an aside, note that you must print addresses with the %p conversion specifier after first casting to void * to avoid undefined behavior:

printf("%p\n", (void *)v);
printf("%p\n", (void *)(v + 0));

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