Question

void pass_arr(int arr[]);

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

    printf( "arr  = %p\n"
            "&arr = %p\n\n", arr, &arr);

    pass_arr(arr);
}

void pass_arr(int arr[])
{
    printf( "passed arr  = %p\n"
            "passed &arr = %p\n\n", arr, &arr);
}

---

Output:
arr = 0x28ccd0
&arr = 0x28ccd0

passed arr = 0x28ccd0
passed &arr = 0x28ccc0

---

Can someone explain why the value and adress of arr points to the same adress when evaluated in the block where arr was created, but when passed the value and adress point to two different adresses?

Answer 1

That’s because in the function arr is actually a pointer, not an array. Taking the address of a pointer does not yield the same address, the way it does for an array.