如何在C中使用typedef函数指针调用宏？

Question

Here, I have a typedef for a function pointer and then I also have a macro that points to the address 0x1234 that is of type "fptr". But I would like to know how can that macro be called with the arguments to call the resulting function? Thank you for your help.

 #include <stdio.h>
    
    typedef void (*fptr)(int, int);

   // 0x1234 will be the address where the reference to fptr will exist
   #define FPTR_MACRO ((fptr)0x1234) 
    
    void main(void){
       
      //How should I call the FPTR_MACRO with the arguments to call that function?
      //FPTR_MACRO(1,1);
    }

Answer 1

If you insist on using macro as is, it would be

(*FPTR_MACRO)(1, 1);

But it would be easier to use if rewritten as

#define FPTR_CALL(x, y) ((*(fptr)0x1234)(x, y))

And than used as

FPTR_CALL(1, 1);

Answer 2

Given #define FPTR_MACRO ((fptr)0x1234) , you can call the function with FPTR_MACRO(1, 1) .

However, this will only work if 0x1234 is a correct address for the function. If that is not actually the address of the function, but rather is “the address where the reference to fptr will exist”, meaning that 0x1234 is the address where a pointer to the function is, then the macro is incorrect. Instead, you need to get the pointer from the location, as with * (fptr **) 0x1234 , and then you can call the function using that pointer, as with (* (fptr **) 0x1234) (1, 1) .

Note that the expression used to call a function is a pointer, not a function. When we have f(x, y) , the operand f should actually be a pointer to a function, not a function, per C 2018 6.5.2.2 1. When we do write a function name there, such as pow(x, y) , the function is automatically converted to a pointer, as if we had written (&pow)(x, y) . This is what allows us to write function calls “naturally.” What this means is that, when you already have a pointer, such as FPTR_MACRO , you do not need anything special to call the function. You already have the required pointer, so you can call it with just FPTR_MACRO(1, 1) .