是否可以调用注入的朋友模板函数? [英] Is it possible to invoke an injected friend template function?

问题描述

为了与类中的其他(非模板)函数保持一致，我想定义并调用一个朋友模板函数.

To be consistent with other (non-template) functions in a class I wanted to define and invoke a friend template function.

我可以毫无问题地定义它(请参见下面的功能t).

I can define it with no problem (see function t below).

namespace ns{
struct S{
    void m() const{}
    friend void f(S const&){}
    template<class T>
    friend void t(S const&){}
};
template<class T>
void t2(S const& s){}
}

但是以后我无法以任何方式调用此t函数吗?

However later I am not able to invoke this t function in any way?

int main(){
    ns::S s;
    s.m();
    f(s);
//  t<int>(s); // error: ‘t’ was not declared in this scope (I was expecting this to work)
//  ns::t<int>(s); // error: ‘t’ is not a member of ‘ns’
//  ns::S::t<int>(s); // error: ‘t’ is not a member of ‘ns::S’
}

即使根本不可能，我也很惊讶能够定义它.

Even if it is not possible at all, I am surprised that I am allowed to define it.

我用gcc 8和clang 7进行了测试.

I tested this with gcc 8 and clang 7.

推荐答案

要实现此目的，需要几个前向声明.

What you need for this to work are a couple of forward declarations.

以下两行代码应位于名称空间ns之前.

The below two lines of code should come before the namespace ns.

struct S; //Needed because S is used as a parameter in the function template
template<class T> void t(S const&);

然后这种形式的通话将在main内部起作用.

And then this form of call will work inside main.

t<int>(s);

请参见演示此处.

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