Error void value not ignored as it ought to be

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При компиляции возникает ошибка «void value not ignored as it ought to be». Помогите, ибо я в программировании стартер. Писал программу не я, так что за неё не ручаюсь. Библиотеку RF24 скачал, дело не в ней. Подскажите что куда написать, чтобы всё заработало. Буду ооооочень благодарен =)

Программа следующая :

#include <SPI.h>
#include <nRF24L01.h> 
#include <RF24.h> //Скачиваем и устанавливаем библиотеку RF24 для радио модуля!!!!!!!!!
#include <Servo.h>
Servo myservo1;
Servo myservo2;
Servo myservo3;
#define CE_PIN   2
#define CSN_PIN 9
int pwm = 0;
int pwm2 = 0;
const uint64_t pipe = 0xE8E8F0F0E1LL;
RF24 radio(CE_PIN, CSN_PIN); 
int joystick[9];
//unsigned long time;
void setup()   
{
  delay(50);
  radio.begin();
  radio.setChannel(9);
  radio.setDataRate(RF24_250KBPS);        // Установка минимальной скорости;
  radio.setPALevel(RF24_PA_HIGH);         // Установка максимальной мощности;
  radio.openReadingPipe(1,pipe);
  radio.startListening();
  myservo1.attach(4);
  myservo2.attach(8); 
  myservo3.attach(3);
  pinMode(7, OUTPUT);
  pinMode(10, OUTPUT);
  digitalWrite(10,LOW);
  pinMode(14, OUTPUT);
  pinMode(15, OUTPUT);
  pinMode(16, OUTPUT);
  pinMode(17, OUTPUT);
  pinMode(18, OUTPUT);
  pinMode(19, OUTPUT);
}


void loop()  
{
  if ( radio.available() )
  {
    bool done = false;
    while (!done)
    {
      done = radio.read( joystick, sizeof(joystick) );
      myservo1.write(joystick[0]);
      myservo2.write(joystick[2]);
      myservo3.write(joystick[4]);
      digitalWrite(14, !joystick[5]);
      digitalWrite(15, !joystick[6]);
      digitalWrite(16, !joystick[7]);
      digitalWrite(7, !joystick[8]);  
      
     if(joystick[1]>500 && joystick[1]<524)
{    digitalWrite(17,LOW);
     digitalWrite(10,LOW);
}
             if(joystick[1]>524)
             {   pwm = map(joystick[1], 524,1024,0,255);
                 analogWrite(5,pwm);
                 digitalWrite(17,LOW);
                 digitalWrite(10,HIGH);
             }
                         if(joystick[1]<500)
                         {   pwm = map(joystick[1],500,0,0,255);
                             analogWrite(5,pwm);
                             digitalWrite(17,HIGH);
                             digitalWrite(10,LOW);
                         }
                         
if(joystick[3]>500 && joystick[3]<524)
{    digitalWrite(18,LOW);
     digitalWrite(19,LOW);
}
             if(joystick[3]>524)
             {   pwm2 = map(joystick[3], 524,1024,0,255);
                 analogWrite(6,pwm2);
                 digitalWrite(18,LOW);
                 digitalWrite(19,HIGH);
             }
                           if(joystick[3]<500)
                           {   pwm2 = map(joystick[3],500,0,0,255);
                               analogWrite(6,pwm2);
                               digitalWrite(18,HIGH);
                               digitalWrite(19,LOW);
                           }          
  }
  }
  else
  {       
  }
  }

1. 02-20-2018

   
   

   #1

   

   
   Registered User
   

   ---

   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 ;
      }

   ---

2. 02-20-2018

   
   

   #2

   

   
   and the hat of int overfl
   

   

   ---

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

   ---

3. 02-20-2018

   
   

   #3

   

   
   Registered User
   

   ---

   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 ;
          
          
      }

   ---

4. 02-20-2018

   
   

   #4

   

   
   and the hat of int overfl
   

   

   ---

   1. Your result should be int.
   2. You need a return statement.

   ---