我使用Arduino Uno和ATMega328P来简单地控制几个LED,开关由用户控制。然而,在我的主循环中大约56次迭代(或大约16秒)后,我的程序重置。我怀疑它与看门狗计时器有关,但即使通过wtd_disable()禁用了它;在我的设置中,问题仍然存在。程序确实进入了一个循环,只有当用户拨动开关时才能退出。有什么建议吗?
//Don't worry, I have all necessary libraries and variables set up.
void setup()
{
Serial.begin(9600); // start serial for output
Serial.println(i2c_init());
wdt_disable();
//pinMode(22,INPUT_PULLUP);
//pinMode(23,INPUT_PULLUP);
pinMode(wakePin, INPUT);
pinMode(ACPin, INPUT);
pinMode(PowerPin, INPUT);
pinMode(PowerLED, OUTPUT);
pinMode(ACLED, OUTPUT);
pinMode(Battery1LED, OUTPUT);
pinMode(Battery2LED, OUTPUT);
pinMode(WifiLED, OUTPUT);
pinMode(TesterLED, OUTPUT);
pinMode(EnableLED, OUTPUT);
attachInterrupt(0, wakeUpNow, LOW);
}
void loop()
{
digitalWrite(PowerLED, LOW);
// digitalWrite(ACLED, LOW); Exclude AC Power LED
digitalWrite(Battery1LED, LOW);
digitalWrite(Battery2LED, LOW);
digitalWrite(WifiLED, LOW);
digitalWrite(TesterLED, LOW);
digitalWrite(EnableLED, LOW);
Enable = 0;
Serial.println("Reset Complete");
int ACPower = digitalRead(ACPin);
digitalWrite(ACLED, ACPower);
int v1 = fetchWord(deviceAddress1, VOLTAGE);
int v2 = fetchWord(deviceAddress2, VOLTAGE);
int BatteryVoltage = max(v1,v2);
Serial.print("Highest Battery Voltage: ");
Serial.println(BatteryVoltage);
delay(250);
if((BatteryVoltage >= 7000) | (ACPower == 1)){
int PowerOK = digitalRead(PowerPin);
if (PowerOK == 0){
loop();
}else {
bulk();
}
}else{
loop();
}
}
//This is the main part of my code that is constantly looped through,
//and after 16 seconds, the program resets, going back to loop()
void bulk()
{
Enable = 1;
digitalWrite(EnableLED, HIGH);
int ACPower = digitalRead(ACPin);
digitalWrite(ACLED, ACPower);
//int Battery1State = BatteryState(deviceAddress1);
int Battery1State = 2; // Simulating low battery
switch (Battery1State){
case 1:
digitalWrite(Battery1LED, HIGH);
break;
case 2: //I can't run parallel code to control the blinking LED,
//so I toggle the LED every pass through. Case 2 blinks slowly
if(i >= 4){
toggleLED(Battery1LED);
i = 0;
} else {
i++;
}
break;
case 3: //case 3 blinks quickly
toggleLED(Battery1LED);
break;
}
int Battery2State = 3; // simulating a very low battery
switch (Battery2State){
case 1:
digitalWrite(Battery2LED, HIGH);
break;
case 2:
if(j >= 4){
toggleLED(Battery2LED);
j = 0;
} else {
j++;
}
break;
case 3:
toggleLED(Battery2LED);
break;
}
buttonState = digitalRead(wakePin); //button is HIGH by default
if(buttonState == HIGH){
Serial.println(count);
if(count == 0){
starttime = millis();
} else if (count == 54){
endtime = millis();
runtime = endtime - starttime;
Serial.print("System Run Time: ");
Serial.println(runtime);
}
count++;
int PowerOK = digitalRead(PowerPin);
digitalWrite(PowerLED, PowerOK);
delay(250);
//Repeat this code if power switch is on, restart if power is turned off
if(PowerOK == 0){
loop();
}else {
bulk();
}
我怀疑——对于这个网站上发布的错误来说,这是多么合适——堆栈溢出。我还没有真正尝试理解您的整个代码,但据我所见,两个函数(loop和bulk)都在其末尾调用loop()或bulk()。从本质上讲,这些功能永远不会结束。
对于初学者,请尝试删除代码中对loop()的所有调用:
更改循环函数末尾的代码来自:
if((BatteryVoltage >= 7000) | (ACPower == 1)){
int PowerOK = digitalRead(PowerPin);
if (PowerOK == 0){
loop();
}else {
bulk();
}
}else{
loop();
}
至:
if((BatteryVoltage >= 7000) | (ACPower == 1)){
int PowerOK = digitalRead(PowerPin);
for ( ; PowerOK != 0 ; )
bulk();
}
并完全删除bulk()函数末尾的以下代码:
delay(250);
//Repeat this code if power switch is on, restart if power is turned off
if(PowerOK == 0){
loop();
}else {
bulk();
}
背景:
当你用C、C++和大多数其他语言调用一个函数时,返回地址(即被调用函数结束后代码中应该继续执行的位置)被放在内存中一个称为堆栈的特殊部分上。当被调用的函数返回时,返回地址将从堆栈中删除,一切正常。如果在第一个函数返回之前调用了另一个函数,则会向堆栈中添加一个新的返回地址。如果一个函数重复调用自己而不返回,那么最终整个堆栈空间(有限的资源)都会用完,就会发生不好的事情。这就是代码中发生的情况:函数loop()和bulk()永远不会返回,相反,它们会做自己的事情,无限地调用自己或另一个。
在arduino中,有一个隐式main()函数,大致如下:
void main(void)
{
// system initialisation code
...
...
...
// user code
setup() ;
for( ; ; )
loop() ;
}
也就是说,loop()是连续调用的。没有理由在它自己的一端再次调用它。
希望这能有所帮助。
没关系,我相信这是串行端口溢出。当我在代码末尾删除了一些我认为无关紧要的"Serial.println"时,问题自行解决了。
对于arduino,当启动setup()时会被调用一次,然后循环会被重复调用(每当它完成时)。。如前所述,循环不需要调用自己。。