Is there any difference between address of array and array itself?

Question

The following code snippet proofs that both are the same:

int a[4];
printf("a: %p\n&a: %p", a, &a);
"0x12345678"
"0x12345678"

But the compiler will warn in case of:

int a[4], *p;
p = &a;

assignment from incompatible pointer type

Which pointer type does &a have ?

@haccks sir, in this particular question, casting the arguments to `void *` is a required part. I would like to ask you for a reconsideration of marking this as dupe. Thank you. :-) — Sourav Ghosh, Jul 02 '15 at 11:11
@haccks Thanks for the edit, but that does not invalidate my earlier comment, either. :-) — Sourav Ghosh, Jul 02 '15 at 11:37

score 3 · Answer 1 · edited May 23 '17 at 11:58

3

To be correct, first, your print statement should look like

 printf("a: %p\n&a: %p", (void *)a, (void *)&a);

because %p expects a void * argument. Please, note, printf() being a variadic function, implicit conversion (cast) won't take place, so, the casting is required.

Now, that said, a being an array,

(but, both will return the same address, FWIW.) You can also see this answer.

OTOH, in your case, p is of type int *.

That's why your compiler warns you about the type mismatch in the assignment.

edited May 23 '17 at 11:58

Community

answered Jul 02 '15 at 11:08

Sourav Ghosh

@Quentin sadly, the dupe won't address the casting part required in `printf()` in this case. – Sourav Ghosh Jul 02 '15 at 11:10
1

`&a[0]` `a` and `&a` all have the same address but of different types – Gopi Jul 02 '15 at 11:10
@Gopi `a` is the same as `&a[0]` after decaying. – Quentin Jul 02 '15 at 11:11
`p=&a` fires a warning (but works). Why throws `p=a` not a warning since `int *` is not `int [4]`. Is `int (*)[4]` similar to a pointer to pointer type ? – mr.wolle Jul 02 '15 at 12:29
@mr.wolle I've updated my answer with a link, please check that. :-) – Sourav Ghosh Jul 02 '15 at 12:32

1 Answers1