用于显式特化模板类中的模板函数的 C++ 语法?

2021-12-13 00:00:00 gcc templates c++

我的代码在 VC9(Microsoft Visual C++ 2008 SP1)中有效，但在 GCC 4.2(Mac 上)中无效:

I have code which works in VC9 (Microsoft Visual C++ 2008 SP1) but not in GCC 4.2 (on Mac):

struct tag {};

template< typename T >
struct C
{   
    template< typename Tag >
    void f( T );                 // declaration only

    template<>
    inline void f< tag >( T ) {} // ERROR: explicit specialization in
};                               // non-namespace scope 'structC<T>'

我知道 GCC 希望我将我的显式专业化转移到课程之外，但我无法弄清楚语法.有什么想法吗?

I understand that GCC would like me to move my explicit specialization outside the class but I can't figure out the syntax. Any ideas?

// the following is not correct syntax, what is?
template< typename T >
template<>
inline void C< T >::f< tag >( T ) {}

推荐答案

如果不显式特化包含类，就不能特化成员函数.
但是，您可以做的是将调用转发到部分专用类型的成员函数:

You can't specialize a member function without explicitly specializing the containing class.
What you can do however is forward calls to a member function of a partially specialized type:

template<class T, class Tag>
struct helper {
    static void f(T);   
};

template<class T>
struct helper<T, tag1> {
    static void f(T) {}
};

template<class T>
struct C {
    // ...
    template<class Tag>
    void foo(T t) {
        helper<T, Tag>::f(t);
    }
};

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