CODEWITHSUNDEEP
c++pointers
user1468913
user1468913

Reputation: 23

Passing pointer to function as parameter

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?

Upvotes: 2

Views: 411

Answers (4)

rashok
rashok

Reputation: 13414

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.

Upvotes: 0

Alok Save
Alok Save

Reputation: 206508

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.

Upvotes: 5

zmbq
zmbq

Reputation: 39013

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)

Upvotes: 4

Scroog1
Scroog1

Reputation: 3579

The code in main should be:

sample(1,2,mm());

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

Upvotes: 0

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