Simplest way to get memory size of std::array's underlying array?

Question

Is this the simplest/shortest way to get size in memory of the content of what std::array::data() returns?

arr.size() * sizeof(arr.value_type)

Edit: My question wasn't precise. By "size in memory" I mean size of all elements (themselves) contained in the array so if e.g. they are pointers pointing to structures, I want the size of the pointers alone, not the structures pointed to. I also don't want to include the size of any possible overhead of the std::arr implementation. Just the array elements.

Some people suggested sizeof(arr). This: What is the sizeof std::array<char, N>? begs to disagree. And while it seems to work on my machine I want to know what the standard guarantees.

Answer 1

You can use the sizeof operator directly on your std::array instance:

sizeof(arr)

Example:

struct foo
{
    int a;
    char b;
};

int main()
{
    std::array<foo, 10> a;
    static_assert(sizeof(foo) == 8);
    static_assert(sizeof(a) == 80);
}

live example on wandbox

---

From cppreference:

std::array is a container that encapsulates fixed size arrays.

This container is an aggregate type with the same semantics as a struct holding a C-style array T[N] as its only non-static data member.

Answer 2

There's no guarantee that sizeof(std::array<T,N>) == N*sizeof(T), but it is guaranteed that sizeof(std::array<T,N>) >= N*sizeof(T). The extra size might be named (but unspecified) members and/or unnamed padding.

The guarantee follows from the fact that the wrapped T[N] array must be the first member of std::array<T,N>, but other members aren't specified.

Answer 3

Since no one has posted anything better than my first guess in question and sizeof(arr) is most likely NOT guaranteed to not include the size of any possible additional std::array's fields I'm choosing this as the accepted answer.

arr.size() * sizeof(arr.value_type)

Should anyone come up with anything better I'd be happy to accept their answer instead.