Q: How can I declare a function that can return a pointer to a function of the same type? I'm building a state machine with one function for each state, each of which returns a pointer to the function for the next state. But I can't find a way to declare the functions--I seem to need a function returning a pointer to a function returning a pointer to a function returning a pointer to a function..., ad infinitum.
A: You can't quite do it directly. One way is to have the function return a generic function pointer (see question 4.13), with some judicious casts to adjust the types as the pointers are passed around:
typedef int (*funcptr)(); /* generic function pointer */
typedef funcptr (*ptrfuncptr)(); /* ptr to fcn returning g.f.p. */
funcptr start(), stop();
funcptr state1(), state2(), state3();
void statemachine()
{
ptrfuncptr state = start;
while(state != stop)
state = (ptrfuncptr)(*state)();
}
funcptr start()
{
return (funcptr)state1;
}
(The second ptrfuncptr typedef hides some particularly
dark
syntax;
without it,
the state variable
would have to be declared as funcptr (*state)()
and the call would contain a bewildering cast of the form
(funcptr (*)())(*state)().)
Another way (suggested by Paul Eggert, Eugene Ressler, Chris Volpe, and perhaps others) is to have each function return a structure containing only a pointer to a function returning that structure:
struct functhunk {
struct functhunk (*func)();
};
struct functhunk start(), stop();
struct functhunk state1(), state2(), state3();
void statemachine()
{
struct functhunk state = {start};
while(state.func != stop)
state = (*state.func)();
}
struct functhunk start()
{
struct functhunk ret;
ret.func = state1;
return ret;
}
(Note that these examples use
the older, explicit style
of calling via function pointers;
see question 4.12.
See also question 1.17.)
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