How can we use callbacks in ANSI-C?

Question

I have already tried it, see here:
https://code.sololearn.com/cg5LJ8X2P10Z/?ref=app

But this methode doesn't worked.
How I use callbacks in C?

Answer 1

You have to use function pointers instead of regular pointers.

A function pointer is defined as follows:
function_return_type (*variable_name) ( parameter_types, ... );

Might take a bit to get used to.
This would be the corrected program:

#include <stdio.h>

// function prototype
int sum( int, int );

int main() {
printSum(2,4, sum);
return 0;
}

int sum(int a, int b){
// No need for +=
return a+b;
}

void printSum(int a, int b, int (*doSum)(int, int)){
// doSum is the function pointer name
printf("%d",doSum(a,b));
}

Answer 2

@Hornetfly
The file is processed from top to bottom, assuming there is no prototype, when it reaches this line: printSum(2,4, sum);

C has to know what printSum is and what sum is.

Since the compiler hasn't come across sum yet it has no idea what that variable is. So with a prototype we tell C that sum is a function and that it exists somewhere.

But we don't have a prototype for printSum either and that still compiles? You may ask.

Well, C is a little weird.
When you call a function, C assumes the function already exists and has a int return type.
It's later the linkers job to find the correct function.

If you would output printSum's return type, you could, even though it returns void, C thinks it's an int.

So actually the correct program would also have a prototype for printSum ( which I forgot, sorry ) above main.
void printSum(int, int, int (*)(int,int));

C would give out a bunch of warnings without the prototypes.
But in C++ this wouldn't compile.

Answer 3

Dennis Thank you, that worked for me.
But why we need a function prototype?