回归虚无? [英] Returning a void?

问题描述

我不明白为什么这段代码编译没有错误:

I do not understand why this code compiles without error:

#include <iostream>

template <class T>
struct Test
{
    static constexpr T f() {return T();} 
};

int main()
{
    Test<void> test;
    test.f(); // Why not an error?
    return 0;
}

按照标准是可以的，还是编译器的容忍度?

Is it ok according to the standard, or is it a compiler tolerance?

推荐答案

这看起来有效 草案 C++11 标准，如果我们查看 5.2.3 部分 显式类型转换(功能符号)em> 段落 2 说(强调我的):

This looks valid by the draft C++11 standard, if we look at section 5.2.3 Explicit type conversion (functional notation) paragraph 2 says (emphasis mine):

表达式 T()，其中 T 是简单类型说明符或非数组完整对象类型的类型名称说明符 或(可能是 cv 限定的)void 类型，创建指定类型，其值是由值初始化产生的(8.5) 类型 T 的对象；void() 没有初始化案例.[...]

The expression T(), where T is a simple-type-specifier or typename-specifier for a non-array complete object type or the (possibly cv-qualified) void type, creates a prvalue of the specified type, whose value is that produced by value-initializing (8.5) an object of type T; no initialization is done for the void() case.[...]

措辞非常相似 pre C++11 也是如此.

the wording is pretty similar pre C++11 as well.

这在 constexpr 中没问题，尽管 7.1.5 段 3 说:

This okay in a constexpr even though section 7.1.5 paragraph 3 says:

constexpr 函数的定义应满足以下条件约束:

The definition of a constexpr function shall satisfy the following constraints:

并包括此项目符号:

它的返回类型应该是一个文字类型；

its return type shall be a literal type;

和 void 不是 C++11 中的 文字，如 3.9 部分 10，但是如果我们再看6段，它给出了一个适合这种情况的例外，它说:

and void is not a literal in C++11 as per section 3.9 paragraph 10, but if we then look at paragraph 6 it gives an exception that fits this case, it says:

如果一个 constexpr 函数的实例化模板特化类模板的模板或成员函数将无法满足constexpr 函数或 constexpr 构造函数的要求，该特化不是 constexpr 函数或 constexpr构造函数.[ 注意:如果函数是成员函数，它将仍然是常量，如下所述.—end note ] 如果没有专业化模板将产生一个 constexpr 函数或 constexpr构造函数，程序格式错误；无需诊断.

If the instantiated template specialization of a constexpr function template or member function of a class template would fail to satisfy the requirements for a constexpr function or constexpr constructor, that specialization is not a constexpr function or constexpr constructor. [ Note: If the function is a member function it will still be const as described below. —end note ] If no specialization of the template would yield a constexpr function or constexpr constructor, the program is ill-formed; no diagnostic required.

正如凯西在C++14 草案标准 void 是一个文字，这是 3.9 Types 段 10 说:

As Casey noted in the C++14 draft standard void is a literal, this is section 3.9 Types paragraph 10 says:

一个类型是文字类型，如果它是:

A type is a literal type if it is:

并包括:

——无效；或

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