使用SFINAE检测constexpr [英] Detecting constexpr with SFINAE

问题描述

我正在努力升级一些C ++代码以利用C ++ 11中的新功能。我有一个trait类与几个函数返回基本类型，这在大多数时候，但不总是，返回一个常量表达式。我想根据函数是否 constexpr 做不同的事情。我想出了以下方法：

  template< typename Trait> 
 struct test 
 {
 template< int Value = Trait :: f（）> 
 static std :: true_type do_call（int）{return std :: true_type（）; } 
 
 static std :: false_type do_call（...）{return std :: false_type（）; } 
 
 static bool call（）{return do_call（0）; } 
}; 
 
 struct trait 
 {
 static int f（）{return 15; } 
}; 
 
 struct ctrait 
 {
 static constexpr int f（）{return 20; } 
}; 
 
 int main（）
 {
 std :: cout< regular：<< test< trait> :: call（）<< std :: endl; 
 std :: cout<< constexpr：< test< ctrait> :: call（）<< std :: endl; 
} 
  

额外的 int / ... 参数是这样，如果两个函数在 SFINAE 后可用，第一个通过重载分辨率来选择。

使用 Clang 3.2 编译和运行此操作显示：

  regular：0 
 constexpr：1 
  

所以这看起来工作，但我想知道代码是否合法C ++ 11。

解决方案

p> 注意：我在此处打开了一个问题，说明OPs代码是否实际有效。以下重写的示例将适用于任何情况。

---

但我想知道代码是否合法C ++ 11

这是，虽然可以考虑默认的模板参数有点不寻常。我个人喜欢下面的风格更好，这是类似于你（读：我）写一个trait到检查函数的存在< a>，只需使用非类型模板参数并省略 decltype ：

  #include< type_traits> 
 
命名空间详细信息{
 template< int> struct sfinae_true：std :: true_type {}; 
 template< class T> 
 sfinae_true<（T :: f（），0）> check（int）; 
 template< class> 
 std :: false_type check（...）; 
} // detail :: 
 
 template< class T> 
 struct has_constexpr_f：decltype（detail :: check< T>（0））{}; 
  

Live example。 a>

---

说明时间〜

^†，因为默认模板参数的实例化点是其函数模板的实例化点，在你的情况下，在 main

§14.6.4.1[temp.point] p2

如果一个函数模板[...]被调用，使用该函数模板的默认参数定义[...] ，默认参数的实例化点是函数模板[...]的实例化点。

它只是通常的SFINAE规则。

---

†至少我认为，这不是完全标准。

I'm working on upgrading some C++ code to take advantage of the new functionality in C++11. I have a trait class with a few functions returning fundamental types which would most of the time, but not always, return a constant expression. I would like to do different things based on whether the function is constexpr or not. I came up with the following approach:

template<typename Trait>
struct test
{
    template<int Value = Trait::f()>
    static std::true_type do_call(int){ return std::true_type(); }

    static std::false_type do_call(...){ return std::false_type(); }

    static bool call(){ return do_call(0); }
};

struct trait
{
    static int f(){ return 15; }
};

struct ctrait
{
    static constexpr int f(){ return 20; }
};

int main()
{
   std::cout << "regular: " << test<trait>::call() << std::endl;
   std::cout << "constexpr: " << test<ctrait>::call() << std::endl;
}

The extra int/... parameter is there so that if both functions are available after SFINAE, the first one gets chosen by overloading resolution.

Compiling and running this with Clang 3.2 shows:

regular: 0
constexpr: 1

So this appears to work, but I would like to know if the code is legal C++11. Specially since it's my understanding that the rules for SFINAE have changed.

解决方案

NOTE: I opened a question here about whether OPs code is actually valid. My rewritten example below will work in any case.

---

but I would like to know if the code is legal C++11

It is, although the default template argument may be considered a bit unusual. I personally like the following style better, which is similar to how you (read: I) write a trait to check for a function's existence, just using a non-type template parameter and leaving out the decltype:

#include <type_traits>

namespace detail{
template<int> struct sfinae_true : std::true_type{};
template<class T>
sfinae_true<(T::f(), 0)> check(int);
template<class>
std::false_type check(...);
} // detail::

template<class T>
struct has_constexpr_f : decltype(detail::check<T>(0)){};

Live example.

---

Explanation time~

Your original code works^† because a default template argument's point of instantiation is the point of instantiation of its function template, meaning, in your case, in main, so it can't be substituted earlier than that.

§14.6.4.1 [temp.point] p2

If a function template [...] is called in a way which uses the definition of a default argument of that function template [...], the point of instantiation of the default argument is the point of instantiation of the function template [...].

After that, it's just usual SFINAE rules.

---

† Atleast I think so, it's not entirely clear in the standard.

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