为什么函数模板不能被部分专门化？ [英] Why function template cannot be partially specialized?

问题描述

我知道语言规范禁止功能模板的部分专业化。

I know the language specification forbids partial specialization of function template.

我想知道为什么它禁止的理由？

I would like to know the rationale why it forbids it? Are they not useful?

template<typename T, typename U> void f() {}   //allowed!
template<> void f<int, char>()            {}   //allowed!
template<typename T> void f<char, T>()    {}   //not allowed!
template<typename T> void f<T, int>()     {}   //not allowed!

推荐答案

AFAIK在C ++ 0x 。

我想这只是一个疏忽（考虑到你可以总是获得部分专业化效果与更详细的代码，通过将函数作为 static 一个类的成员）。

I guess it was just an oversight (considering that you can always get the partial specialization effect with more verbose code, by placing the function as a static member of a class).

您可以查找相关的DR（缺陷报告） 。

You might look up the relevant DR (Defect Report), if there is one.

EDIT ：检查这一点，我发现其他人也相信，但没有人能够在标准草案。 此SO线程似乎表明不支持功能模板的部分专门化在C ++ 0x 中。

EDIT: checking this, I find that others have also believed that, but no-one is able to find any such support in the draft standard. This SO thread seems to indicate that partial specialization of function templates is not supported in C++0x.

编辑2 ：只是一个例子， code> static 类的成员：

EDIT 2: just an example of what I meant by "placing the function as a static member of a class":

#include <iostream>
using namespace std;

// template<typename T, typename U> void f() {}   //allowed!
// template<> void f<int, char>()            {}   //allowed!
// template<typename T> void f<char, T>()    {}   //not allowed!
// template<typename T> void f<T, int>()     {}   //not allowed!

void say( char const s[] ) { std::cout << s << std::endl; }

namespace detail {
    template< class T, class U >
    struct F {
        static void impl() { say( "1. primary template" ); }
    };

    template<>
    struct F<int, char> {
        static void impl() { say( "2. <int, char> explicit specialization" ); }
    };

    template< class T >
    struct F< char, T > {
        static void impl() { say( "3. <char, T> partial specialization" ); }
    };

    template< class T >
    struct F< T, int > {
        static void impl() { say( "4. <T, int> partial specialization" ); }
    };
}  // namespace detail

template< class T, class U >
void f() { detail::F<T, U>::impl(); }    

int main() {
    f<char const*, double>();       // 1
    f<int, char>();                 // 2
    f<char, double>();              // 3
    f<double, int>();               // 4
}

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