Passing a function pointer with it's own parameters in C

Question

#include <stdio.h>

void runner(void (* function)(int in)){
  (*function)(in); 
}

void print(int in){
 printf("%d\n", in);
}

int main(){
 // 
 return 0;
}

Above code is not compiling and shows this error: 'in' undeclared.

Do I have to pass the parameters of a function pointer separately like this?

void runner(void (* function)(int), int in){
  (*function)(in); 
}

i.e form a syntax point of view, there is no way to pass the function pointer and its argument all at once?

Answer 1

Parameters are not passed. It is arguments that are passed.

So this function declaration

void runner(void (* function)(int in)){
  (*function)(in); 
}

has only one parameter: a pointer to a function, But if you want to call the pointed function that expects an argument then you need to supply an argument.

In this declaration of a function pointer

void (* function)(int in)

the function parameter in has the function prototype scope..

You may declare the function parameter without its identifier like

void (* function)(int)

So you have to declare the function with two parameters like

void runner(void (* function)(int), int in ){
  function(in); 
}

Pay attention to that to dereference the pointer to function is redundant.

All these calls as for example

( *function )( int );

or

( *****function )( in );

are equivalent to

function( in );