如何在另一个C ++命名空间中的全局命名空间中定义朋友？

Question

I'd like to define a binary operator on in the global namespace. The operator works on a class that is defined in another namespace and the operator should get access to the private members of that class. The problem I have is that I don't know how to scope that global operator when making it a friend in the class definition.

I tried something like:

namespace NAME
{
    class A {
        public:
            friend A ::operator * (double lhs, const A& rhs);
        private:
            int private_var;
    };
}

A operator * (double lhs, const A& rhs)
{
    double x = rhs.private_var;
    ...
}

The compiler (g++ 4.4) didn't know what to do with it. It seems that the line

friend A ::operator * ()

is evaluated as something like (pseudo-code)

(A::operator)

instead of

(A) (::operator)

If I leave out the :: in the declaration of the operator the compiling works but the operator is then in namespace NAME and not in the global namespace.

How can I qualify the global namespace in such a situation?

Answer 1

First, note that your operator declaration was lacking a namespace qualification for A:

NAME::A operator * (double lhs, const NAME::A& rhs)

and then the decisive trick is to add parentheses to the friend declaration like this, just as you proposed in your "pseudo-code"

friend A (::operator *) (double lhs, const A& rhs);

To make it all compile, you then need some forward declarations, arriving at this:

namespace NAME
{
    class A;
}

NAME::A operator * (double lhs, const NAME::A& rhs);

namespace NAME
{
    class A {
        public:
            friend A (::operator *) (double lhs, const A& rhs);
        private:
            int private_var;
    };
}

NAME::A operator * (double lhs, const NAME::A& rhs)
{
    double x = rhs.private_var;
}

Alexander is right, though -- you should probably declare the operator in the same namespace as its parameters.

Answer 2

This compiles, I assume without testing that it also works. Note the use of parentheses:

namespace NAME {class A; }
NAME::A operator * (double lhs, const NAME::A& rhs);

namespace NAME
{
    class A {
        public:
            friend A (::operator *) (double lhs, const A& rhs);
        private:
            int private_var;
    };
}

NAME::A operator * (double lhs, const NAME::A& rhs)
{
    double x = rhs.private_var;
    return rhs;
}

int main() {}

As Alexander mentions, though, your question doesn't explain why the operator isn't in namespace NAME. Either way it can be called as 1.0 * some_A_instance . So you may be creating needless trouble for yourself.

Answer 3

I don't know exact answer to your question.

But it's terrible bad idea to define operator outside namespace of its parameters (now you cut argument dependent lookup which is very usable for operators).

Answer 4

It is possible - you can enclose the declarator in parentheses: friend A (::operator * (double lhs, const A& rhs));

You also need to forward-declare both the class and the function.

namespace NAME {class A;}
NAME::A operator *(double lhs, const NAME::A& rhs);

// ...

namespace NAME
{
    class A {
        public:
            friend A (::operator * (double lhs, const A& rhs));
        private:
            int private_var;
    };
}

NAME::A operator *(double lhs, const NAME::A& rhs) {
  //...
}

But I agree with Andreas that it would be better to define both in the same namespace if possible.