Passing pointer to function as parameter

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?

Do you want mm() to return the function pointer, or want to define mm as the type HandlerFunc? — Alvin Wong
– Alvin Wong, Commented Jun 20, 2012 at 10:42

Alok Save · Accepted Answer · 2012-06-20 10:50:59Z

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.

zmbq · Accepted Answer · 2012-06-20 10:40:12Z

No, no, you've confused the types. mm is a function that returns a function pointer of the appropriate type. However, mm itself is not of the appropriate type - it doesn't accept any parameters.

You should pass mr in main, or pass mm() (that is - call mm and pass the return value)

Scroog1 · Accepted Answer · 2012-06-20 10:40:03Z

The code in main should be:

sample(1,2,mm());

as mm returns a HandlerFunc, it isn't one itself.

rashok · Accepted Answer · 2012-06-21 11:29:38Z

HandlerFunc mm() { return mr; }

This means nm is function which will not receive any agruments(void), and it will reutrn a pointer to function of type void (*)(int, int)

so that only you are getting that error.

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