数组和函数指针

Question

How do pass an array of pointers to functions as a argument to a function? If the types of the functions are void and so is the pointer, how do you access the functions?

int main() {

  void(*fun_ptr[])(void) = {fun1, fun2, fun3};
  pickFun(fun_ptr);
  return 0;
}

All 3 functions only print a message and are declared as void

void pickFun(void *function_ptr[])
{
   int pick = 0;
   while(pick != 0)
   {
     scanf("%i", &pick);
    *(function_ptr + (pick - 1));
   }
 }

I can never get the function's printf statements to appear when a function is chosen. The program just loops and never prints the message contained in each function. Any suggestions or clues?

Answer 1

Function parameter void *function_ptr[] is of type pointer to pointer to void`. You need to change it to

void pickFun(void (*function_ptr[])(void));  

Also note that, as others pointed out in comments, program never enters the loop body. Better to use do while loop here

void pickFun(void *function_ptr[])
{
   do
   {
        scanf("%i", &pick);
        //*(function_ptr + (pick - 1));
        function_ptr[pick-1]();     //Function call. pick should not be greater than 3
   } while(pick != 0)
 }

Answer 2

Also, you can call your functions with:

(*function_ptr[pick-1])();

or (yuck):

(*(function_ptr + (pick-1)))();

or:

function_ptr[pick-1]();

or:

(function_ptr + (pick-1))();