在函数内部声明的相互C ++类

Question

How do I define classes inside of a function so that they "know" about each other? Here is a greatly dumbed down version of what I'm trying to understand. I'd like to do something like:

void foo () {

    struct A {
        static void bar () {
            B::hmm();
        }
    };

    struct B {
        static void hmm () {
            A::bar();
        }
    }
}

That doesn't work because A::bar() doesn't know anything about B yet. Outside of a function, I could defer the definition of A::bar() until after B is declared. Something like:

void foo () {

    struct A {
        static void bar ();
    };

    struct B {
        static void hmm () {
            A::bar();
        }
    }

    void A::bar () {
        B::hmm();
    }
}

But that doesn't seem to work inside of a function.

In lieu of some clever scoping declaration that makes this all work, I'd also welcome a definitive answer saying this is just not possible in C++ (reference please).

I know there are other ways to work around this (declaring the classes outside of foo, for instance), so I'm not looking for answers on how to do something similar...

Answer 1

You don't.

Member functions of local classes must be defined inside of the class definition.

Because the definitions of A::bar() and B::hmm() are mutually dependent and because member functions cannot be forward declared outside of a class definition, there is no way to order the definitions such that this can work.

(You said you aren't looking for workarounds, but I'll say anyway that, in my opinion, defining any remotely complex class, more or less multiple classes, inside a function definition, hinders code readability and is a bad idea.)