Question

Consider the following program.

#include <stdio.h>

int main()
{
  int a[10]={0};

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

  return 0;
}

a and &a are same. But *a and *(&a) are not. I am out of answers. Please help.

Answer 1

There is no “array base pointer”, i.e. there no pointer variable that points to the array. The name of the array refers to the array itself. Therefore you can not take an address of a, instead, &a is handled as a special case.

When you use the name of an array in an expression, it decays into a pointer that points to the first element of the array. However, & and sizeof operators are exceptions.

ANSI C specifies that &a means the address of the array itself, and its type is “pointer to array”, not “pointer to array element”.

In pre-ANSI compilers, &a would cause a warning.