c - Arduino 不等待每个模式(模式)结束我怎样才能重新启动循环？

Question

我目前正在使用 Arduino Uno 进行交通灯项目。当我按下外部开关时，我应该在不同模式之间切换。有 4 种模式，模式 0:所有 LED 熄灭，模式 1:红色 LED 闪烁 1 秒亮 1 秒灭，模式 2:黄色 LED 闪烁 1 秒亮 1 秒灭，但是在模式 3 中:我有电位器相关的延迟。 (最多 10 秒)

我的问题是，如何在按下开关并更改模式后立即停止模式，而无需等待当前模式完成。

// Change the speed with delay using potentiometer,
#include <EEPROM.h>

short rPin = 4; // red led connected to D4
short yPin = 5; // yellow led connected to D5
short gPin = 6; // green led connected to D6
short switchPin = 3; // where the switch is connected

short potPin = 2; // input pin for the PM
short val = 0;    // variable to store the value coming from the PM

short timeOut = 300; // switch debounce noise handling,
volatile short last_change_time = 0;

static short dTime = 0; // Static Variable, that holds the last change, effects
                        // every function.
static short d = 0; // current d value.

static short mode = 0;

// EEPROM current address
int addr = 0;

void pressed() {
  int difference = millis()-last_change_time;
  if(difference > timeOut || last_change_time == 0)
  {
    if (mode >= 3){
    mode = 0;
      } 
    else {
      mode = mode + 1;
    }
  }
  else
  {
    //Serial.println("NO");
  }
  last_change_time = millis();
}

void setup()
{
  pinMode(rPin, OUTPUT);
  pinMode(yPin, OUTPUT);
  pinMode(gPin, OUTPUT);
  pinMode(switchPin, INPUT);
  Serial.begin(9600);
  attachInterrupt(1,pressed,FALLING); // digital pin 2 // pressed
}

void loop()
{
  if (mode == 3) mode3();  
  if (mode == 2) mode2();
  if (mode == 1) mode1();
  if (mode == 0) mode0();
  val = analogRead(potPin);
  if (val < 170) {
    d = 1;
  }
  else if ( (val >= 170) && (val < 340))
  {
    d = 2;
  }
  else if ((val >= 340)&&(val < 510))
  {
    d = 3;
  }
  else if ((val >= 510)&&(val < 680))
  {
    d = 4;
  }
  else if ((val >= 680) && (val < 850))
  {
    d = 5;
  }
  else if ((val >= 850) && (val < 1023))
  {
    d = 6;
  }
}


void serialEvent() {
  while (Serial.available()) {
  int value = Serial.parseInt();
  if (value == 1){
  Serial.print("d: ");
  Serial.print(d);
  Serial.print("  mode: ");
  Serial.print(mode);
  Serial.print("\n");
  }
}
}

void mode0() {
  while(mode == 0){
  digitalWrite(rPin, LOW);
  digitalWrite(yPin, LOW);
  digitalWrite(gPin, LOW);
  }
}

void mode1() {
  while(mode == 1){
  digitalWrite(yPin, LOW);
  digitalWrite(gPin, LOW);
  digitalWrite(rPin, HIGH);
  delay(1000);
  digitalWrite(rPin, LOW);
  delay(1000);
  }
}

void mode2() {
  while(mode == 2){
  digitalWrite(rPin, LOW);
  digitalWrite(gPin, LOW);
  digitalWrite(yPin, HIGH);
  delay(1000);
  digitalWrite(yPin, LOW);
  delay(1000);
  }
}


void mode3(){
  interrupts();
  while(mode == 3){
  digitalWrite(rPin, HIGH);
  delay(9*d*1000);
  digitalWrite(yPin, HIGH);
  delay(d*1000);
  digitalWrite(rPin, LOW);
  digitalWrite(yPin, LOW);
  digitalWrite(gPin, HIGH);
  delay(10*d*1000);
  digitalWrite(gPin, LOW);
  delay((d)*1000);
  digitalWrite(gPin, HIGH);
  delay(d*1000);
  digitalWrite(gPin, LOW);
  delay(d*1000);
  digitalWrite(yPin, HIGH);
  delay(d*1000);
  digitalWrite(yPin, LOW);
  }
}

Answer 1

想一想:一旦进入“模式”循环之一，您就再也不会检查开关(电位器)的状态。合理？ 有限状态机的建议很好，但为了保持简单，只需将检查开关和设置模式的代码拆分成它自己的函数，然后在每个模式循环中调用它。

void mode0() {
  while(mode == 0){
  digitalWrite(rPin, LOW);
  digitalWrite(yPin, LOW);
  digitalWrite(gPin, LOW);

  checkSwitch(); // this will potentially change the 'mode' so the while loop with exit if the mode has changed.
  }
}

---

更新:延迟 hack

不是很推荐，但我认为使用delay 方法并且不锁定所有交互的一种方法是将延迟降至最低并检查 for 循环中的输入 - 类似于:

void uglyDelayHack(int delayAmount, int lastMode){

 for (int i = 0; i < delayAmount; i++){

   delay(1); // 1 millisecond delay

   // call to a function that checks for input and changes mode 
   checkForInput(); 

   // get out of the fake delay if mode has changed
   // assumes "mode" is a global variable
   if (lastMode != mode) return;
 }
}