Question

I have the following code:

#include<iostream>
using namespace std;

typedef void (*HandlerFunc)(int, int);
HandlerFunc  mr;
HandlerFunc mm()
{
    return mr;
}

void sample(int a, int b, HandlerFunc func)
{
}

void main()
{
     sample(1, 2, mm);
}

Here I’m trying to pass a function of type HandlerFunc to another function, but I am getting an error:

Error :*: cannot convert parameter 3 from 'void (__cdecl *(void))(int,int)' to 'void (__cdecl *)(int,int)'

If I type cast as sample(1, 2, (HandlerFunc)mm); everything works fine.
Can anyone tell what is the way to solve the error issue?

Answer 1

HandlerFunc mm()
{...}

should be:

void mm(int, int)
{...}

Your function sample() takes a pointer to function(you typedefd as HandlerFunc) as last argument.
The address of the function you pass as this argument must match the type of the function to which it is a pointer.