Unpacking Arrays Question

According to the Arrays section in Ultimate C++ Part 2: Intermediate course I am told I can’t assign one array to another because the array name is essentially a hexadecimal number (Pointer), which is the address of the array in memory. However, when I use the unpacking technique, I am able to use the array name to assign its elements to separate int variables. Why is that?

I believe the part you are referring to Mosh is telling you that assigning one array to another is not going to create a copy of that array. You will just have two variables that point to the same array.

int a[2] = {1, 2};
int b[2] = a;

If we do that, the two variables point to the same array. We can copy the array instead:

int b[2]; 
copy(begin(a), end(a), begin(b));

Now the two variables point to separate arrays which have the same elements.

In contrast, unpacking is just syntactic sugar to simplify a common operation. If the number of elements in an array is small and known you can use the unpacking technique to copy the elements of the array into individual variables. I think this actually uses an array copy operation under the surface, but either way the syntax gets you a copy of the value.

auto [x, y] = a;

Now x has the value 1, y has the value 2, and both are completely disconnected from a.

You can even use a special unpacking syntax to bind the variable as a pointer which is equivalent to the one that the array elements are pointing to (so updating the array value updates the same element the variable pointer is pointing to).

auto& [px, py] = a;

Now px is a pointer to a[0] and py is a pointer to a[1]. Changes to the array will update the values pointed to by the pointers:

a[0] = 3;
// *px == 3
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