返回空格？ [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：

  #include< iostream> 
 
 template< class T> 
 struct Test 
 {
 static constexpr T f（）{return T（）;} 
}; 
 
 int main（）
 {
 Test< void>测试; 
 test.f（）; //为什么不报错？ 
 return 0; 
} 
  

根据标准是否正确， / p>

解决方案

草案C ++ 11标准，如果我们看一下 5.2.3 ： >

表达式T（），其中T是非数组完整对象类型的简单类型说明符或
typename-specifier <
（可能是cv限定的）void类型创建
指定类型的prvalue，其值是通过值初始化
（8.5）生成的类型对象T;没有为void（）
case。[...]进行初始化。

href =http://www.open-std.org/Jtc1/sc22/wg21/docs/papers/2005/n1804.pdf =nofollow> pre C ++ 11 以及。 / p>

即使 7.1.5 段落<$ c $ c> 3 说：

constexpr函数的定义应满足以下
约束：

并包含此项目：

其返回类型应为文字类型;

和 void 不是< ， 我们然后看看段落 6 它给出了适合这种情况的例外，它说：

constexpr函数的模板专门化
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
专业化不是constexpr函数或constexpr
构造函数。 [注意：如果函数是一个成员函数，它将
仍然是const，如下所述。 -end note] 如果没有特殊化的
模板将产生一个constexpr函数或constexpr
构造函数，程序是不成形的;无需诊断。

As Casey 在 C中注明 ++ 14草案标准 void 是一个字符，这是 3.9 / em>段落 10 说：

一个类型是文字类型，

并包括：

<或

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;
}

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

解决方案

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):

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.[...]

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

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

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

and includes this bullet:

its return type shall be a literal type;

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:

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.

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

A type is a literal type if it is:

and includes:

— void; or

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