Is `std::array` default constructible where `T` is not default constructible?

Question

Consider the code below:

#include 

struct T
{
    T() = delete;
};

int main()
{
    std::array a;
    a.size();
}

<

Answer 1

Since there's no elements, no constructor of T should be called.
Does non-default-constructible T prevent std::array from being default-constructible?

The standard doesn't specify what layout std::array should have for us to answer that. The zero sized array specialization is only said to behave as follows:

[array.zero]

1 array shall provide support for the special case N == 0.
2 In the case that N == 0, begin() == end() == unique value. The return value of data() is unspecified.
3 The effect of calling front() or back() for a zero-sized array is undefined.
4 Member function swap() shall have a non-throwing exception specification.

The behavior you note is most probably due to differences in implementation alone.