Arduino 使用几分钟后停止工作

Posted 2023-04-15

【中文标题】Arduino 使用几分钟后停止工作

【英文标题】：Arduino stops working after being just minutes in use

【发布时间】：2020-02-27 22:53:30

【问题描述】：

当我出于某种原因在我的 arduino nano 上运行这段代码时，它似乎停止了一段时间。我试过启动它并等待 5 分钟，但这似乎还没有冻结它。

有谁知道 waitforkey() 是否可以在等待太久之后停止处理器运行？还是我可能有某种内存泄漏？

#include<Keypad.h>
const byte ROWS = 4;
const byte COLS = 3;
char keys[ROWS][COLS] = 
  '1', '2', '3',
  '4', '5', '6',
  '7', '8', '9',
  '0', 'E', 'C'
;
byte rowPins[ROWS] = 8, 7, 6, 9;
byte colPins[COLS] = 2, 3, 4;
char command[4];
char p0;
char p1;
char p2;
char p3;
int count = 0;
int relayPin = 10;
bool solved = false;

Keypad keypad = Keypad( makeKeymap(keys), rowPins, colPins, ROWS, COLS );

void setup() 
  keypad.setHoldTime(0);               // Default is 1000mS
  keypad.setDebounceTime(0);           // Default is 50mS
  pinMode(relayPin, OUTPUT);
  pinMode(A0, OUTPUT);
  pinMode(A1, OUTPUT);
  pinMode(A2, OUTPUT);
  pinMode(A3, OUTPUT);

void loop() 
  ledCompare(count);
  if (solved == true) 
    digitalWrite(relayPin, HIGH);
    digitalWrite(A0, HIGH);
    command[3] = 'E';//makes statement untrue
    char key = keypad.waitForKey();
    count = 0;
    solved = false;
    ledsOff();
  
  char key = keypad.waitForKey();
  if (key == 'C') 
    count = 0;
  
  else 
    command[count] = key;
    count++;
  
  if (count == 4) 
      if (command[0] == p0 && command[1] == p1 && command[2] == p2 && command[3] == p3) 
        solved = true;
      
      else if(command[0] == p0 && command[1] == p1 && command[2] == p2 && command[3] == p3)
        reprogram();
      
      else 
        ledsOff();
        digitalWrite(relayPin, LOW);
        solved = false;
        count = 0;
      
    


void ledsOn() 
  digitalWrite(A0, HIGH);//Meest rechter
  digitalWrite(A1, HIGH);
  digitalWrite(A2, HIGH);
  digitalWrite(A3, HIGH);


void ledsOff() 
  digitalWrite(A0, LOW);//Meest rechter
  digitalWrite(A1, LOW);
  digitalWrite(A2, LOW);
  digitalWrite(A3, LOW);


void ledCompare(int x)
  if (count >= 1) 
    digitalWrite(A3, HIGH);
  
  if (count >= 2) 
    digitalWrite(A2, HIGH);
  
  if (count >= 3) 
    digitalWrite(A1, HIGH);
  


void reprogram()
  ledsOff();
  char key = keypad.waitForKey();
  if(key != 'C')
    p0 = key;
    digitalWrite(A3, HIGH);
  
  else
    reprogram();
  
  key = keypad.waitForKey();
  if(key != 'C')
    p0 = key;
    digitalWrite(A2, HIGH);
  
  else
    reprogram();
  
  key = keypad.waitForKey();
  if(key != 'C')
    p0 = key;
    digitalWrite(A1, HIGH);
  
  else
    reprogram();
  
  key = keypad.waitForKey();
  if(key != 'C')
    p0 = key;
    digitalWrite(A0, HIGH);
    solved = true;
  
  else
    reprogram();
  

【问题讨论】：

幸运的是，reprogram() 永远不会被调用，无论未定义的 p1、p2、p3 具有哪些值。

【参考方案1】：

由于递归 reprogram() 调用，您很可能会耗尽内存。

我什至不明白为什么您首先需要递归调用。在内存如此有限的设备上，递归调用几乎总是一个坏主意。

另见：

https://arduino.stackexchange.com/questions/355/how-much-can-i-recurse-how-much-can-i-recurse-how-much-caqfsdrfw

【讨论】：

reprogram(); 永远不应该被调用，因为那个愚蠢的if (cond) else if (cond) /* impossible to reach */