推导用户定义值模板参数（C ++ 2a，P0732R2） [英] Deducing a user-defined-value template argument (C++2a, P0732R2)

问题描述

我正在尝试推导用户定义的类的模板参数的值（ http：// wg21。 link / p0732r2 ），使用带有-std = c ++ 2a的GCC 9.1。

I am trying to get the value of a template parameter of a user-defined class deduced (http://wg21.link/p0732r2), using GCC 9.1 with -std=c++2a.

struct user_type {
   int a;
   constexpr user_type( int a ): a( a ){}
};

template< user_type u > struct value {};

template< user_type u > void f( value< u > arg ){}  

void g(){
  f( value< user_type( 0 ) >() ); // error here
}

编译器浏览器： https://godbolt.org/z/6v_p_R

我收到错误消息：

source>:8:30: note:   template argument deduction/substitution failed:
<source>:11:33: note:   couldn't deduce template parameter 'u'
   11 |    f( value< user_type( 0 ) >() );

我做错了吗？我原以为这样的值是可以扣除的。

Am I doing something wrong? I had expected such a value to be deductible.

按照Nikita的建议，我在用户类型中添加了==和！=运算符，但这没什么区别。

As suggested by Nikita I added == and != operators to user-type, but that made no difference.

struct user_type {
   int a;
   constexpr user_type( int a ): a( a ){}
   constexpr bool operator==( const user_type & arg ) const {
      return a == arg.a;
   }
   constexpr bool operator!=( const user_type & arg ) const {
      return a != arg.a;
   }
};

推荐答案

这应该格式错误：

struct user_type {
   int a;
   constexpr user_type( int a ): a( a ){}
};

template< user_type u > struct value {};

要成为模板非类型参数，您需要满足 [temp.param] / 4 ：

In order to be a template non-type parameter, you need to satisfy [temp.param]/4:

非类型模板参数应具有以下（可选的cv限定）类型之一：

A non-type template-parameter shall have one of the following (optionally cv-qualified) types:

• 具有强等式结构的文字类型（[class.compare.default]），


• [...]

需要强大的结构平等性的地方，来自 [class.compare.default] / 3 ：

Where strong structural equality requires, from [class.compare.default]/3:

类型C具有很强的结构相等性如果给定const C类型的glvalue x，则：

A type C has strong structural equality if, given a glvalue x of type const C, either:

• C是非类类型，并且[...]


• C是类类型，其中在C的定义中默认定义了==运算符，x == x很好-在上下文转换为bool时形成，所有C的基类子对象和非静态数据成员都具有很强的结构相等性，并且C没有可变或不稳定的子对象。

• C is a non-class type and [...], or
• C is a class type with an == operator defined as defaulted in the definition of C, x == x is well-formed when contextually converted to bool, all of C's base class subobjects and non-static data members have strong structural equality, and C has no mutable or volatile subobjects.

关键是我们需要默认的 == 类型...而我们没有一个，因此我们的类型没有强大的结构相等性，因此不能用作模板非类型参数。

The key is that we need a defaulted == in the type... and we don't have one, so our type doesn't have strong structural equality, so it cannot be used as a template non-type parameter.

但是，gcc 尚未允许您声明这样的运算符，因此您无法解决问题。

However, gcc doesn't let you declare such an operator yet, so you can't fix the problem.

这只是新功能的不完全实现。

This is just an incomplete implementation of a new feature.

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