如何在另一个C ++名称空间的全局名称空间中定义好友? [英] How do I define friends in global namespace within another C++ namespace?

问题描述

我想在全局名称空间中定义一个二进制运算符.运营商 在另一个命名空间中定义的类上工作，并且运算符应获取 访问该类的私人成员.我的问题是我没有 在类定义中成为朋友时，知道如何确定该全局运算符的作用域.

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.

我尝试过类似的事情:

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;
    ...
}

编译器(g ++ 4.4)不知道如何处理.看来这行

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

friend A ::operator * ()

被评估为类似(伪代码)

is evaluated as something like (pseudo-code)

(A::operator)

代替

(A) (::operator)

如果在运算符的声明中省略::，则可以编译，但是运算符位于名称空间NAME中，而不位于全局名称空间中.

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.

在这种情况下如何限定全局名称空间?

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

推荐答案

首先，请注意您的运算符声明缺少A的命名空间限定:

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

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

然后决定性的诀窍是像您在伪代码"中所建议的那样，将括号添加到这样的朋友声明中

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);

要使其全部编译，则需要一些前向声明，以达到以下目的:

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;
}

但是，亚历山大是对的，您应该在与参数相同的名称空间中声明运算符.

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

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