Why can I use an address of address without any errors? [duplicate]

Question

I found codes that uses address operator to address. See below example codes.

int arr[4] = { -1, -1, -1, -1 };
int arr2[4] = { 0, 0, 0, 0 };

memcpy(arr, &arr2, sizeof(arr2)); // & operator to arr2

When I check values of (arr, &arr) with printf, both addresses are equal.

And I have no idea why this codes are valid. (I'm using C++ 14)

&arr2 is the address of the array as a whole. arr2 by itself decays to the address of the array's first element. It so happens that the array and its first element are located at the same address. — Igor Tandetnik
– Igor Tandetnik, Commented Feb 23, 2023 at 0:54
Arrays naturally decay to pointers to their first element. So e.g. arr is the same as &arr[0], and has the type int *. Same when using plain arr2. But &arr2 is a pointer to the array itself, and will have the type int (*)[4], which really is incompatible with the type int *. With that said, both &arr[0] and &arr "luckily" happens to be the same location. — Some programmer dude
– Some programmer dude, Commented Feb 23, 2023 at 0:59
Also note that because of the array to pointer decay, it's very hard to pass arrays to functions. As an argument the declaration int arr[] (or even using a size like int arr[4]) is treated as int *arr. If you then use the pointer-to operator on the argument variable, you will get a pointer to that variable (which have the type int **), not to the original array. — Some programmer dude
– Some programmer dude, Commented Feb 23, 2023 at 1:02
And in C++, please don't use plain arrays if you can avoid it. If the size is fixed and known at compile-time (and not too big) then use std::array (as in std::array<int, 4>). Otherwise use std::vector. This will solve a lot of problems you might encounter when using plain C-style arrays. — Some programmer dude
– Some programmer dude, Commented Feb 23, 2023 at 1:04

eerorika · Accepted Answer · 2023-02-23 00:58:42Z

3

Why can I use an address of address without any errors?

Because std::memcpy accepts arguments of type void* (const qualifier aside), and all object pointers are implicitly convertible to void*.

Although arr itself isn't a pointer, it is an array, and arrays implicitly convert to pointer to first element, which implicitly converts to void*.

&arr2 is a pointer to an array which implicitly converts to void*.

When I check values of (arr, &arr) with printf, both addresses are equal.

The memory location of the first element of an array is the same as the memory location of the array.

answered Feb 23, 2023 at 0:58

eerorika

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