Question

所以，我在这里有我的程序，我正在尝试制作一个程序，我们手动将导线拉出并放回去，每次将其拉出时，它是1次迭代，然后再放回1次迭代中。只要将其连接到端口，外部LED就会变暗；将其拔出时，外部LED就会变亮。我应该能够执行10次迭代，并在到达那里后停止。

问题是，当我包含while（true）时；语句中，我的外部LED不起作用，但是如果没有该语句，该程序将按照我想要的方式运行，除了它在10次迭代后不会停止工作之外，对此不胜感激！

#include<EEPROM.h>
const int LED = 12;
const int SWITCH = 4;

void setup() {
  // put your setup code here, to run once:
  Serial.begin(115200);

  pinMode(LED, OUTPUT);           //LED is always outputting information
  pinMode(LED_BUILTIN, OUTPUT);   //Built in LED is always outputting information
  pinMode(SWITCH, INPUT_PULLUP);  //Switch inputs value when in/out of ground
}

void loop() {
  // put your main code here, to run repeatedly:
  int addr = 0;     //Declaring variables
  int count = 0;
  int seconds;

  if (digitalRead(SWITCH) == LOW) {             //If wire starts in ground, record values
    Serial.println("----Recording----");

    while (count <= 10) {                       //While count value is less than or equal to 10
      if (digitalRead(SWITCH) == LOW) {         //When wire is connected to 4
        count = count + 1;                      //Add one to count in each iteration
        digitalWrite(LED, LOW);                 //LED light is off in this position
        delay(50);                              //Checks switch state every 0.05 seconds
      }
      else if (digitalRead(SWITCH) == HIGH) {   //When wire isnt connected to 4
        count = count + 1;                      //Add one to count in each iteration
        digitalWrite(LED, HIGH);                //LED light is on in this position
        delay(50);                              //Checks switch state every 0.05 seconds
      }
          while (true);
    }
  }
}

Answer 1

为什么不起作用？让我们稍微重写一下代码：

    while (count <= 10) {
      count = count + 1;
      digitalWrite(LED, digitalRead(SWITCH));
      delay(50);

      while (true); // aaand kill it
    }

如果第一个数字读取返回HIGH，然后第二个数字返回LOW（它不会等待50ms停止），则我忽略了竞争条件

我希望像指望针脚变化这样的东西：

    while (count <= 10) {
      byte = digitalRead(SWITCH);
      if (byte != digitalRead(LED))
      { 
        count = count + 1;
        digitalWrite(LED, byte);
      }
      delay(50);
    }
    while (true); // aaand kill it

如何使程序在迭代次数后停止

1 个答案: