By running this program on my computer, I'm getting same addresses. I'm for case of array and &array[0] I understand that name of array points to the address of first item in the array. And both of them are same.

But I'm unable to understand why name of array and &array points to the same address. What comes in my mind about this is that it will print the address of that pionter in which address of first item in array is stored.

Code

#include <stdio.h>

int main()
{
    char arr[3];

    printf("array = %p\n", arr);
    printf("&array[0] = %p\n", &arr[0]);
    printf("&array = %p\n", &arr);

    return 0;
}

Output

array = 0061FF1D
&array[0] = 0061FF1D
&array = 0061FF1D

CodePudding user response：

It's important to note that while there are some similarities between arrays and pointers, they are not the same.

The name array does not point to the address of the first item. It is in fact the entire array. The confusion comes from the fact that in most contexts, the name of an array decays to a pointer to its first member.

Two notable places where array-to-pointer decay does not happen are when the array is the operand of either the sizeof operator or the address-of operator &. Regarding the latter, an array is a contiguous set of objects, and (like any object) it has an address. The address of an array is the same as the address of its first member, so array (when used in an expression, or equivalently &array[0]) and &array will have the same value but different types.

In your particular case, &array[0] has type char * while &array has type char (*)[3] (i.e. a pointer to an array of char of size 3).

CodePudding user response：

1 2 and 1.0 2.0 both have the same value sum, yet of different types. printf("%p\n", arr); printf("%p\n", &arr); printf("%p\n", &arr[0]); printing the same does not mean arr and &arr are the same thing - anymore than 3 and 3.0 differ in some attributes.