c++ - 回归虚无？

Question

我不明白为什么这段代码编译没有错误:

#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，还是编译器容忍的？

Answer 1

draft C++11 standard 看起来有效，如果我们看一下5.2.3部分显式类型转换(功能表示法)第2段说(强调我的 ):

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。

这在 constexpr 中没问题，即使第 7.1.5 段 3 说:

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

并包括此项目符号:

its return type shall be a literal type;

and void 不是 C++11 中的 literal 根据 3.9 段落 10，但是如果我们再看第6段，它给出了一个适合这种情况的异常(exception)，它说:

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.

作为凯西 noted在 C++14 draft standard void 是一个 literal，这是 3.9 部分 Types 段落 10 说:

A type is a literal type if it is:

包括:

— void; or