If LED is already on Serial Monitor must display "LED is already on"

Projects Discussion and Showcase Home Automation and Networked Objects
#1

I want to turn LED ON/OFF through serial monitor by sending 1,0 Respectively. If LED is already OFF and i send "0" Serial Monitor must reply "LED is already OFF" same in the case of LED ON. here's my Code.

int led = 13;
int x;
int y;
int s;


void setup() {

  Serial.begin(9600);
  pinMode(led, OUTPUT);
  digitalWrite(led, LOW);
  Serial.println("READY");


}
void loop() {


  if (Serial.available() > 0) {
    x = Serial.read();
    y = digitalRead(led);




    if (x == '1') {
      digitalWrite(led, HIGH);
      Serial.println("LED is ON");
      delay(10);
      
    }


    if (x == '0') {
      digitalWrite(led, LOW);
      Serial.println("LED is OFF");
      delay(10);
    }

    if (x == 's') {
      Serial.println(y);
    }



  }
}
#2

Do you think "&&" might be part of your answer?
Or maybe just a simple "else"?

#3

why can't you turn off the led in void setup:

int pin = 13;
int pin_state = 0;

void setup(){

Serial.begin(9600);

digitalWrite(pin, pin_state);

Serial.println("--Ready--");

}

void loop(){

if(Serial.available() > 0){

if(Serial.read() == '1'){

pin_state = 1;

}else{

pin_state = 0;

}

digitalWrite(pin, pin_state);

delay(10);

}

}

#4