Difference between array name and pointer to the first element in c

Question

As I have learnt,array name acts like a pointer to first element.But:

int c[]={0,1,2};
printf("%d \t %d",c,&c[0]); //Different values,Why?

Also then why *c=0?

*c is the same as c[0] and you have set the first element of c to be 0 so that explains that part. — Seth Carnegie
– Seth Carnegie, Commented May 5, 2012 at 19:15
You should print pointers with %p, not %d. Sometimes it doesn't matter much, on some platforms it does a lot. — ugoren
– ugoren, Commented May 5, 2012 at 19:16
This is undefined behaviour. &c[0] is an int*, but %d expects an int. You should cast the pointer to void* and use %p, or (assuming your platform supports this) cast to uintptr_t and use one of the macros from <inttypes.h>. — Kerrek SB
– Kerrek SB, Commented May 5, 2012 at 19:17

Fred Foo · Accepted Answer · 2012-05-05 19:26:19Z

11

Just a guess: you're on a platform with 64-bit pointers and 32-bit int. Your code passes two pointer values to printf, which then interprets these as int values; that might print the two halves of a 64-bit pointer as two separate integers.

You should print pointers with %p, not %d, after casting them to void*.

edited May 5, 2012 at 19:26

answered May 5, 2012 at 19:17

Fred Foo

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