在类之外为模板类声明非模板朋友功能 [英] Declare non-template friend function for template class outside the class

问题描述

以下非模板代码效果很好:

struct A { };

struct B
{
    B() {}
    B(const A&) {}
    friend B operator+(const B&) { return B(); }    
};

B operator+(const B&);

int main()
{
    A a;
    B b;
    +b;
    +a;
}

但是，如果我在此代码中将类作为模板:

But if I make classes in this code templated:

template <class T>
struct A { };

template <class T>
struct B
{
    B() {}
    B(const A<T>&) {}
    friend B operator+(const B&) { return B(); }    
};

template <class T>
B<T> operator+(const B<T>&); // not really what I want 

int main()
{
    A<int> a;
    B<int> b;
    +b;
    +a;
}

某种出现问题:

错误:"operator +"(操作数类型为"A< int>")不匹配

error: no match for 'operator+' (operand type is 'A<int>')

是否可以为类外的模板类声明非模板朋友功能(就像我对上面的非模板类所做的那样)?

Is it possible to declare non-template friend function for template class outside the class (as I did for non-template classes above)?

我可以通过为A<T>参数添加模板运算符来解决此问题，在里面调用friend函数，但这并不有趣.

I can solve the problem by adding template operator for A<T> argument and call friend function inside, but it is not interesting.

UPD:

另一种解决方法(受 R Sahu的回答启发)是为类A添加friend声明:

Another workaround (inspired by R Sahu's answer) is add friend declaration for class A:

template <class T>
struct A { 
    friend B<T> operator+(const B<T>&);
};

但是此给出了警告，我不知道如何解决正确.

but this gives a warning, and I don't know how to fix it correctly.

推荐答案

是否可以在类之外为模板类声明非模板朋友功能(就像我对上面的非模板类所做的那样)?

Is it possible to declare non-template friend function for template class outside the class (as I did for non-template classes above)?

是的，有可能.但是，您可能还需要一个功能模板，并确保operator+<int>是A<int>的friend，operator+<double>是A<double>的friend，等等.

Yes, it is possible. However, what you probably need is a function template too, and make sure that operator+<int> is a friend of A<int>, operator+<double> is a friend of A<double>, etc.

请参见 https://stackoverflow.com/a/35854179/434551 了解如何完成此操作.

See https://stackoverflow.com/a/35854179/434551 to understand how that can be done.

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