Declaring a namespace as a friend of a class

c++ namespaces friend-function friend-class
kMaster picture kMaster · Dec 30, 2013 · Viewed 10.9k times · Source

I was wondering if there is a way such that we make all functions defined within a specific namespace friend with a class?

In particular, I have a class, for example:

class C {
    private:
        // ...
    public:
        // ...

        friend C* B::f1(C*);
        friend C* B::f2(C*);
        friend C* B::f3(C*);
        friend C* B::f4(C*);
        friend C* B::f5(C*);
};

and a namespace B as:

namespace B {
    C* f1(C* x);
    C* f2(C* x);
    C* f3(C* x);
    C* f4(C* x);
    C* f5(C* x);
};

Now, I would prefer to avoid writing 5 lines in the class definition to make all five functions of the namespace B friend with class C and just tell the compiler that all of the functions defined within the namespace B are friends with the class C (i.e. can access its private members).

A quick fix I guess is to change the namespace to a class and define the functions as its static members and then declare the class B as a friend of class C. However, out of curiosity I was wondering if such thing is possible with namespaces as well or not?

Thanks in advance.

Answer

Angew is no longer proud of SO picture Angew is no longer proud of SO · Dec 30, 2013

No, it's not possible befriend a namespace. If nothing else, it would constitute a "security breach," as namespaces can be extended anywhere. So anyone could add an arbitrary function to the namespace and access the class's non-public data.

The closest you can get is the solution you propose, making those functions static members of a class and befriending the class. But then again, why not make them static members of the original class (C in your code) in the first place?

As an aside, if I ran into a need for so many friend functions in my code, it would make think again about my design, hard; I'd take it as a sign I'm doing something wrong.

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