C++ 函数模板偏特化? [英] C++ function template partial specialization?

问题描述

我知道下面的代码是一个类的部分特化:

I know that the below code is a partial specialization of a class:

template <typename T1, typename T2> 
class MyClass { 
  … 
}; 


// partial specialization: both template parameters have same type 
template <typename T> 
class MyClass<T,T> { 
  … 
}; 

我也知道 C++ 不允许函数模板部分特化(只允许完整).但是我的代码是否意味着我已经部分地为一个/相同类型的参数专门化了我的函数模板?因为它适用于 Microsoft Visual Studio 2010 Express！如果不是，那么您能否解释一下部分专业化的概念?

Also I know that C++ does not allow function template partial specialization (only full is allowed). But does my code mean that I have partially specialized my function template for one/same type arguments? Because it works for Microsoft Visual Studio 2010 Express! If no, then could you please explain the partial specialization concept?

#include <iostream>
using std::cin;
using std::cout;
using std::endl;

template <typename T1, typename T2> 
inline T1 max (T1 const& a, T2 const& b) 
{ 
    return a < b ? b : a; 
} 

template <typename T> 
inline T const& max (T const& a, T const& b)
{
    return 10;
}


int main ()
{
    cout << max(4,4.2) << endl;
    cout << max(5,5) << endl;
    int z;
    cin>>z;
}

推荐答案

按照标准，函数部分特化是不被允许的.在示例中，您实际上是重载 &不专门 max 函数.
如果允许，它的语法应该看起来有点像下面这样:

Function partial specialization is not yet allowed as per the standard. In the example, you are actually overloading & not specializing the max<T1,T2> function.
Its syntax should have looked somewhat like below, had it been allowed:

// Partial specialization is not allowed by the spec, though!
template <typename T> 
inline T const& max<T,T> (T const& a, T const& b)
{                  ^^^^^ <--- [supposed] specializing here
  return 10;
}

在函数模板的情况下，C++ 标准只允许完全特化， -- 不包括编译器扩展！

In the case of a function templates, only full specialization is allowed by the C++ standard, -- excluding the compiler extensions!

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