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

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);

It explodes recursively where (...). Is there a fundamental reason this can't be done or is there another syntax? It seems to me it should be possible without a cast. I'm really trying to pass a dispatch-table but I could figure that out if I could just pass this one type.

From stackoverflow

• 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.

• 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 (*)());