How can I typedef a function pointer that takes a function of its own type as an argument?

Question

Example: A function that takes a function (that takes a function (that ...) and an int) and an int.

typedef void(*Func)(void (*)(void (*)(...), int), int);
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Answer 1

It's impossible to do directly. Your only options are to make the function pointer argument accept unspecified arguments, or to accept a pointer to a structure containing the function pointer, as Dave suggested.

// Define fnptr as a pointer to a function returning void, and which takes one
// argument of type 'pointer to a function returning void and taking
// an unspecified number of parameters of unspecified types'
typedef void (*fnptr)(void (*)());

Answer 2

You can wrap the function pointer in a struct:

struct fnptr_struct;
typedef void (*fnptr)(struct fnptr_struct *);
struct fnptr_struct {
  fnptr fp;
};

I'm not sure if this is an improvement on casting. I suspect that it's impossible without the struct because C requires types to be defined before they are used and there's no opaque syntax for typedef.