Thread: Function pointer issue

Function pointer is use to call function by it's address Why program doesn't print statement ?

Code:

#include <stdio.h>

void colour ()
{
    printf("Green colours");
}


int main ()
{
    void (*fun_ptr)(void) = &colour; 
    return 0;
}

Originally Posted by Salem

It no more calls the function, than saying

You need to call your function via it's pointer if you want something to happen.

oh Thank you What's wrong with code

Code:

 #include <stdio.h> 
void addition (int a, int b)
{
    int c = a + b; 
    
    printf ("c = %d /n", c);
}
 
 
int main ()
{
    addition (2, 5);
    
    void (*fun_ptr)(int) = &addition; 
    fun_ptr(); 
    return 0;
}

main.c:24:5: error: too few arguments to function ‘fun_ptr’
fun_ptr();
^~~~~~~

Originally Posted by laserlight

The error message is telling you exactly what's wrong: too few arguments to function ‘fun_ptr’

But I am passing two actual argument in main function
gone through following links
Too few arguments to function (C language Error)

Originally Posted by Salem

> But I am passing two actual argument in main function
Yeah, but only when you're calling addition directly.

> void (*fun_ptr)(int) = &addition;
This is just a bunch of lies - addition takes two int parameters.

> fun_ptr();
More lies - you don't even supply one parameter.

Code:

#include <stdio.h> void addition (int a, int b)
{
    int c = a + b; 
     
    printf ("c = %d \n", c);
}
  
  
int main ()
{
    addition (2, 5);
     
    void (*fun_ptr)(int, int) = &addition; 
//    fun_ptr(2, 5); 
    return 0;
}

c = 7

There is no need to use this fun_ptr(2, 5);

Code:

void (*fun_ptr)(void) = &colour;

You can leave off the & when assigning the colour function to the pointer. In C, a function name is a pointer to the function's code.