Question on Function Pointers

Hello,

I was looking on using function pointers in my code and it would be great if you guys can point me in the right direction or give hints.

I basically want to know how to invoke different functions ( functions with different arg numbers and return type) using a single function pointer.

example

Code:

void func1();

int func2(int,float);

main()
{
    void (*funcptr) ();

   functptr = & func1
   funcptr();

   funcptr = & func2  // i know that i need a different funcptr here, but is there a way i can type cast or something?

  funcptr(2,2.5);

return;

}

Thanks!

Originally Posted by rebelstar

I basically want to know how to invoke different functions ( functions with different arg numbers and return type) using a single function pointer.

I do not think you can do that, at least not within the realm of well defined behaviour.

Thanks for the reply

Function Pointers Tutorial index. Take a look there, should have all the information you need and more.

i know that i need a different funcptr here, but is there a way i can type cast or something?

Sure, you can use a cast to convert one function type into your "generic" function pointer, and then cast it back to the original type and it's guaranteed to work. This gains you nothing, however, because you have to know the type of the function you're converting at compile-time, not run-time, so you may as well just create a proper type.

You can sort of achieve what you want with a lot of hacking:

Code:

struct function_pointer
{
  enum { VOID_VOID, VOID_INT, VOID_INTP } type;
  void (*fp)(void);
};
#define MAKE_FP(fp) ( (void(*)(void))(fp) )

void x(void);
void y(int);
void z(int *);

struct function_pointer fp;
fp.type = VOID_VOID; fp.fp = MAKE_FP(x);
fp.type = VOID_INT; fp.fp = MAKE_FP(y);
fp.type = VOID_INTP; fp.fp = MAKE_FP(z);

switch(fp.type)
{
  case VOID_VOID:
    fp.fp();
    break;
  case VOID_INT:
    ((void(*)(int))fp.fp)(10);
    break;
  case VOID_INTP:
    ((void(*)(int *))fp.fp)(NULL);
    break;
}

But all this is good at demonstrating, I think, is that C is not exactly Haskell when it comes to polymorphism.

Thanks for the hint cas and thanks Andrew.
This solution is pretty good,.