std :: array< T，0>是否默认可构造，而T不是默认可构造? [英] Is `std::array<T, 0>` default constructible where `T` is not default constructible?

问题描述

考虑以下代码:

#include <array>

struct T
{
    T() = delete;
};

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

我们默认初始化一个0大小的数组.由于没有元素，因此不应调用T的构造函数.

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

但是， C语仍然要求T是默认可构造的，而

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

请注意，如果我们将数组初始化更改为:

Note that if we change the array initialization to:

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

Clang 这次接受了.

非默认构造的T是否阻止std::array<T, 0>被默认构造?

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

推荐答案

感谢@TC，正如他的

Thanks to @T.C., as pointed out in his comment, it's addressed in LWG 2157, which is still an open issue as of this writing.

建议的决议添加了这一要点(强调我的意思):

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

这种情况下未指定的数组内部结构应允许类似以下的初始化:

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

array<T, 0> a = { };

，并且所说的初始化必须有效，即使T不是默认可构造的.

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

因此，很明显，预期的行为是即使不是T，也要具有std::array<T, 0>默认构造值.

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

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