为什么不能定义一元运算符，然后在MSVC的模板类中声明具有相同名称的友元二进制运算符？

Question

Consider such code:

template<typename S>
class C;

template<typename S>
C<S> operator-(C<S> lhs, C<S> rhs);

template<typename S>
class C
{
public:
    C operator-() { return *this; }
    friend C operator-<S>(C lhs, C rhs);//error on this line
};

template<typename S>
C<S> operator-(C<S> lhs, C<S> rhs) { return C<S>(); }

int main(int argc, char** argv)
{
    C<int> a,b;
    a-b;
    return 0;
}

This gives me 6 errors in MSVC. But if I move the definition of C operator-() after the friend declaration, it compiles. If I change the class to an untemplated class, it compiles. And this seems to compiles in g++ too. (I have no g++ installed, based on https://wandbox.org/ )

So what's wrong here?

Answer 1

I have found some useful information data here . The problem is that the member operator unary minus hides the global binary subtract. So I have to declare the friend with

friend C<S> (::operator-<S>)(C<S> lhs, C<S> rhs);

to explicitly ask for operator- in the global scope: in c++ name lookup happens before type check.

Besides, as the friend is not part of the class (I suppose that's the reason), it is a must to write template argument after the return type and parameter type. Without them (ie friend C (::operator-<S>)(C lhs, C rhs); ) it still works in MSVC, but not with g++.

Answer 2

This line is messing your code:

C operator-() { return *this; }

Remove it and your code compiles (well, change a+b to ab ).

It tries to overload an arithmetic operator with no operands . Arithmetic operators have one operand if declared inside a class, or two operands if declared outside.

From operator overloading doc :

It is not possible to change the precedence, grouping, or number of operands of operators.

The friend declaration, despite of being declared inside the class is not a member function of the class.

For much more info about operator oveloading see this question