Question

我正在尝试打开和关闭3个LED。因此，基本上将1输入到串行监视器中，并且LED指示灯将亮起，我完成了这一部分，但是当我再次按下1时，LED指示灯将熄灭。第二部分是我遇到的问题，我需要创建某种切换。

这是我的代码

const int greenPin = 2;
const int yellowPin = 3;
const int redPin = 4;

void setup()

{
pinMode(greenPin, OUTPUT);
  pinMode(yellowPin, OUTPUT);
  pinMode(redPin, OUTPUT);


  Serial.begin(9600);

  while (!Serial);

  Serial.println("Input 1 to Turn LED on and 2 to off");

}

void loop() {

  if (Serial.available())

  {

    int state = Serial.parseInt();

    if (state == 1)

    {

      digitalWrite(greenPin, HIGH);
      digitalWrite(yellowPin, LOW);
       digitalWrite(redPin, LOW);
      Serial.println("Command completed LED turned ON");

    }

    if (state == 2)

    {
  digitalWrite(greenPin, LOW);
      digitalWrite(yellowPin, HIGH);
       digitalWrite(redPin, LOW);

      Serial.println("Command completed LED turned OFF");

    }

      if (state == 3)

    {
     digitalWrite(greenPin, LOW);
      digitalWrite(yellowPin, LOW);
       digitalWrite(redPin, HIGH);

      Serial.println("Command completed LED turned OFF");

    }




  }

}

Answer 1

digitalWrite(ledPin,  !digitalRead(ledPin));

那样，您总是会反转当前的引脚状态。

如何使用arduino串行监视器打开和关闭LED

1 个答案: