Function pointer pointing to a function that takes a function pointer

Question

How do I declare a function pointer that points to a function taking the same function pointer as the argument?

I've tried the following without success:

typedef void (*fnptr)(void (*)());

void func(fnptr)
{
    /* ... */
}

void func2(fnptr)
{
    /* ... */
}

void main()
{
    fnptr fn = &func;
    func2(fn);
}

Is this possible?

Answer 1

I very much doubt it, but you can get the needed recursion by introducing a struct.

struct Rec;

typedef void (*RecFun)(const Rec&);

struct Rec {
    RecFun fun;
};

Example of use:

#include <iostream>

void nothing(const Rec& rec) {}
Rec recNothing = { nothing };

void f(const Rec& rec)
{
    std::cout << "f\n";
    rec.fun(recNothing);
}
Rec recF = { f };

void g(const Rec& rec)
{
    std::cout << "g\n";
    rec.fun(recNothing);
}
Rec recG = { g };

int main()
{
    recF.fun(recG);
}

Update: As per the suggestions of Chris, Vitus, and Johannes, here are some convenient implicit conversions (as in Herb Sutter's GotW #57):

struct Rec;

typedef void (*RecFun)(const Rec&);

struct Rec {
    RecFun fun;
    Rec(RecFun fun) : fun(fun) {}
    operator RecFun() const { return fun; }
};