是`std :: array <T, 0> `默认可构造的，而`T`不是默认可构造的？

Question

Consider the code below:

#include <array>

struct T
{
    T() = delete;
};

int main()
{
    std::array<T, 0> a;
    a.size();
}

We default initialize a 0-sized array. Since there's no elements, no constructor of T should be called.

However, Clang still requires T to be default constructible, while GCC accepts the code above.

Note that if we change the array initialization to:

std::array<T, 0> a{};

Clang accepts it this time.

Does non-default-constructible T prevent std::array<T, 0> from being default-constructible?

Answer 1

Since there's no elements, no constructor of T should be called.
Does non-default-constructible T prevent std::array<T, 0> from being default-constructible?

The standard doesn't specify what layout std::array<T, 0> 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.

Answer 2

Thanks to @TC, as pointed out in his comment , it's addressed in LWG 2157 , which is still an open issue as of this writing.

The proposed resolution adds this bullet point (emphasis mine):

The unspecified internal structure of array for this case shall allow initializations like:

 array<T, 0> a = { }; 

and said initializations must be valid even when T is not default-constructible .

So it's clear that the intended behavior is to have std::array<T, 0> default constructible even when T is not.

Answer 3

This question explains what happens with clang and std::array Deleted default constructor. Objects can still be created... sometimes

But with gcc the difference comes from the library code. There is indeed a specific implementation detail in the gcc codebase that is relevant to this question as @StoryTeller mentioned

gcc has a special case for std::array with a size of 0, see the following code from their <array> header (from gcc 5.4.0 )

template<typename _Tp, std::size_t _Nm>
struct __array_traits
{
  typedef _Tp _Type[_Nm];

  static constexpr _Tp&
  _S_ref(const _Type& __t, std::size_t __n) noexcept
  { return const_cast<_Tp&>(__t[__n]); }

  static constexpr _Tp*
  _S_ptr(const _Type& __t) noexcept
  { return const_cast<_Tp*>(__t); }
};

template<typename _Tp>
struct __array_traits<_Tp, 0>
{
 struct _Type { };

 static constexpr _Tp&
 _S_ref(const _Type&, std::size_t) noexcept
 { return *static_cast<_Tp*>(nullptr); }

 static constexpr _Tp*
 _S_ptr(const _Type&) noexcept
 { return nullptr; }
};

as you can see, there is a specialization of __array_traits (which is used in std::array for the underlying array) when the array size is 0, that doesn't even have an array of the type it's templated on. The type _Type is not an array, but an empty struct!

That is why there are no constructors invoked.