数组的地址是否等于它在 C 中的第一个元素

Question

Assume this array:

int arr[5] = { 1, 2, 3, 4, 5 };

Here arr has two purpose - it is the name of the array and it acts as a pointer pointing towards the first element in the array, so arr is equal to &arr[0] by default.

My questions is, what about &arr ? is this also equal to the above?

Answer 1

An array is in fact an extent of memory. So the address of an extent is the address of the stored array in this extent and of its first element.

As a result an array designator used in expressions with rare exceptions is converted to pointer to its first element.

Using the example provided by you

int arr[5] = { 1, 2, 3, 4, 5 };

these expressions &arr and &arr[0] have the same value but are of different types.

The expression &arr has the type int( * )[5] while the expression &a[0] has the type int * .

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    int arr[5] = { 1, 2, 3, 4, 5 };

    printf( "&arr == &arr[0] is %s\n", 
            ( void * )&arr == ( void * )&arr[0] ? "true" : "false" );

    printf( "sizeof( *&arr )  = %zu\n", sizeof( *&arr ) );          
    printf( "sizeof( *&arr[0] )  = %zu\n", sizeof( *&arr[0] ) );            
    return 0;
}

Its output is

&arr == &arr[0] is true
sizeof( *&arr )  = 20
sizeof( *&arr[0] )  = 4