Question

Possible Duplicate:
Is array name a pointer in C?

I was running the following code

#include <stdio.h>

int main()
{
  int a[4] = {1,2,3,4};
  int (*b)[4] = &a;
  int k = sizeof(a);
  printf("\n\n%d\n\n", k);
  printf("a = %u, b = %u, *b = %u, data = %d", a, b, *b, **b);
  return 0;
}

I got the following output

a = 3485401628, b = 3485401628, *b = 3485401628, data = 1

Here I am assigning the address of a to b, as the type of b is int**, but in the output I am getting that the address pointed by the a is the same as the address pointed by b.

It seems little confusing to me. What is the explanation?

Answer 1

The value of array a is the pointer to the first element of the array a.

The value of pointer &a (which is the same value and type as pointer b) is the pointer to the array a.

They have the same value (but different types) as they both start at the same memory address. There is no padding possible at the beginning of an array.

When evaluated a is of type int * and &a is of type int (*)[4].

Note that the correct way to print the address of a pointer is to use conversion specifier p. E.g.,

/* And p requires the argument to be a void *, so use a cast if
 * it is not the case
 */
printf("%p %p\n", (void *) a, (void *) &a);