使用不同的 enable_if 条件选择成员函数

问题描述

我正在尝试根据类模板参数确定调用哪个版本的成员函数.我试过这个:

I am trying to determine which version of a member function gets called based on the class template parameter. I have tried this:

#include <iostream>
#include <type_traits>

template<typename T>
struct Point
{
  void MyFunction(typename std::enable_if<std::is_same<T, int>::value, T >::type* = 0)
  {
    std::cout << "T is int." << std::endl;
  }

  void MyFunction(typename std::enable_if<!std::is_same<T, int>::value, float >::type* = 0)
  {
    std::cout << "T is not int." << std::endl;
  }
};

int main()
{
  Point<int> intPoint;
  intPoint.MyFunction();

  Point<float> floatPoint;
  floatPoint.MyFunction();
}

我认为这是说如果 T 是 int，则使用第一个 MyFunction，如果 T 不是 int，则使用第二个 MyFunction，但是我收到编译器错误，说错误:'struct std:: 中没有名为'type'的类型:enable_if'".谁能指出我在这里做错了什么?

which I thought is saying "use the first MyFunction if T is int, and use the second MyFunction if T is not int, but I get compiler errors saying "error: no type named ‘type’ in ‘struct std::enable_if’". Can anyone point out what I am doing wrong here?

推荐答案

enable_if 有效，因为 模板参数的替换导致错误，因此替换从重载决议集中被删除，编译器只考虑其他可行的重载.

enable_if works because the substitution of a template argument resulted in an error, and so that substitution is dropped from the overload resolution set and only other viable overloads are considered by the compiler.

在您的示例中，实例化成员函数时没有发生替换，因为模板参数 T 那时已经知道.实现您正在尝试的最简单方法是创建一个默认为 T 的虚拟模板参数，并使用它来执行 SFINAE.

In your example, there is no substitution occurring when instantiating the member functions because the template argument T is already known at that time. The simplest way to achieve what you're attempting is to create a dummy template argument that is defaulted to T and use that to perform SFINAE.

template<typename T>
struct Point
{
  template<typename U = T>
  typename std::enable_if<std::is_same<U, int>::value>::type
    MyFunction()
  {
    std::cout << "T is int." << std::endl;
  }

  template<typename U = T>
  typename std::enable_if<std::is_same<U, float>::value>::type
    MyFunction()
  {
    std::cout << "T is not int." << std::endl;
  }
};

<小时>

正如 HostileFork 在评论中提到的，原始示例留下了用户为成员函数明确指定模板参数并得到错误结果的可能性.以下应防止编译成员函数的显式特化.

As HostileFork mentions in the comments, the original example leaves the possibility of the user explicitly specifying template arguments for the member functions and getting an incorrect result. The following should prevent explicit specializations of the member functions from compiling.

template<typename T>
struct Point
{
  template<typename... Dummy, typename U = T>
  typename std::enable_if<std::is_same<U, int>::value>::type
    MyFunction()
  {
    static_assert(sizeof...(Dummy)==0, "Do not specify template arguments!");
    std::cout << "T is int." << std::endl;
  }

  template<typename... Dummy, typename U = T>
  typename std::enable_if<std::is_same<U, float>::value>::type
    MyFunction()
  {
    static_assert(sizeof...(Dummy)==0, "Do not specify template arguments!");
    std::cout << "T is not int." << std::endl;
  }
};

这篇关于使用不同的 enable_if 条件选择成员函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！