仅对某些类型使用enable_if专门化功能的最佳方法 [英] Best way to specialize function using enable_if for only some types

问题描述

我有这段代码专门用于两种类型的打印功能，并且对于其他任何类型都恢复为基本版本.

I have this code that specializes the print function for two types, and reverts back to the base version for any other types.

我的问题是，有没有办法在基本打印功能的enable_if中为所有专用版本键入否定大小写的方法?

My question is, is there a way to write this without having to type out the negative case for all the specialized versions in the enable_if of the base print function?

即有没有办法删除所有！std :: is_same 并且仍然具有明确的打印功能?

i.e. is there a way to remove all the !std::is_same and still have an unambiguous print function?

欢迎使用任何版本的C ++，但可以在c ++ 14中使用的版本会有所帮助.

Any versions of C++ welcome, but one that works in c++14 would be helpful.

#include <iostream>

template<typename T, std::enable_if_t<!std::is_same<T, int>::value && !std::is_same<T, double>::value, int> = 42>
void print(T data)
{
    std::cout << "base:" << data << std::endl;
}

template<typename T, std::enable_if_t<std::is_same<T, double>::value, int> = 42>
void print(T data)
{
    std::cout << "double:" << data << std::endl;
}

template<typename T, std::enable_if_t<std::is_same<T, int>::value, int> = 42>
void print(T data)
{
    std::cout << "int:" << data << std::endl;
}

int main()
{
    std::string foo("foo");
    double bar = 1.2;
    int baz = 5;
    print(foo);
    print(bar);
    print(baz);
}

推荐答案

对于您的用例，您可以根据需要简单地为 print 函数提供重载

For your use case, you can simply provide overloads for the print function as needed

#include <iostream>

template<typename T> 
void print(T data)
{
    std::cout << "base:" << data << std::endl;
}

void print(double data)
{
    std::cout << "double:" << data << std::endl;
}

void print(int data)
{
    std::cout << "int:" << data << std::endl;
}

但是，如果对 T 的约束比较复杂，那么在没有显式提供默认"情况下约束的否定的情况下，就无法专门化 print .

However, if you have more complicated constraints on T, then you can't specialize print without explicitly providing the negation of the constraints in the "default" case.

如果可以访问c ++ 17，则可以在单个函数中编写此代码.通常的if-else逻辑意味着仅在没有专业化"的情况下才触发基本情况.这样避免了必须指定否定项.

If you have access to c++17, you can write this in a single function. The usual if-else logic means that the base case is only triggered if the "specializations" are not. This avoids having to specify the negations.

template<typename T> 
void print(T data)
{
 if constexpr(std::is_same<T, double>{})
   std::cout << "double:" << data << std::endl;
 else if constexpr(std::is_same<T, int>{})
   std::cout << "int:" << data << std::endl;
 else // if not same as int or double
   std::cout << "base:" << data << std::endl;
}

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