[edit] Dubious Statement

From the page:

> This struct has the same aggregate type semantics as a C-style array. The size and efficiency of array<T,N> for some number of elements is equivalent to size and efficiency of the corresponding C-style array T[N].

The standard does not guarantee that std::array<T, N> is the same size as T[N]. It doesn't guarantee layout compatibility between std::array and C-style arrays. Check section 9.2; layout compatibility only exists between structs/classes, unions, and enumerators. It doesn't exist between structs and arrays.

Therefore, the standard does not guarantee that sizeof(std::array<T, N>) == sizeof(T[N]). Korval 03:54, 17 May 2013 (PDT)

I agree, the only thing the Standard guarantees is that the pointer returned by data() and the reference returned by operator[](size_t) (as well as at() and *begin()) refer to something that is "compatible" to a C-style raw array (i.e. contiguous storage, like vector).-- 05:17, 17 May 2013 (PDT)

The standard requires std::array<T,N> to be an aggregate with N elements, and that these N elements occupy contiguous storage. That's a very strong requirement, layout compatibility with T[N] is one of its consequences when T is a standard-layout type. Let's let a few more comments to be posted to that SO discussion and edit accordingly. --Cubbi 06:24, 17 May 2013 (PDT)
Edited the page, thank you for bringing this up. --Cubbi 10:12, 17 May 2013 (PDT)