Thread: error: void value not ignored as it ought to be

I am getting this error: void value not ignored as it ought to be in my program. What is this error and how to remove it?

Code:

#include <stdio.h>

void addition(int x, int y);


int main(void)
{
  	 int a = 10; int b = 20;
	 
	 int c;
	 
	 c = addition(10, 20);
	 
	 printf("addition = %d \n",c);
	 
    return 0;
}


void addition(int x, int y)
   {
	   unsigned result;
	   result = x +y ;
   }

Originally Posted by Salem

By making your function return an int, and not void.

I understood where I was confuse
addition with no return type but passing value

Code:

#include <stdio.h> 
void addition(int x, int y);
 
 
int main(void)
{
     int a = 10; int b = 20;
      
     addition(10, 20);
      
    return 0;
}
 
void addition(int x, int y)
   {
       unsigned result;
       result = x +y ;
       
       printf("addition = %d",result);
   }

addition with return type and passing value

Code:

#include <stdio.h> 
int addition(int x, int y);
 
 
int main(void)
{
   
     int c;
      
     c = addition(10, 30);
     
     printf("addition = %d",c);     
    
     return 0;
}
 
int addition(int x, int y)
   {
       unsigned result;
       result = x +y ;
       
       
   }