问题描述
我正在尝试用 C++ 为我的 ESP32 构建一个转速计。当我在条件之外取消注释 Serial.printf("outside rev: %d \n",rev);
时,它可以工作,但是当我对其进行注释时,我得到的值比它们应有的数量级大(没有 700 转,有 7 转)。我最好的猜测是打印语句减慢了 loop()
的速度,刚好足以让 incrementRevolutions()
在下一个循环之前将全局变量 passedMagnet
从真切换为假。这是有道理的,因为更新 passedMagnet 的延迟将允许 newRevCount++;
被多次触发。但鉴于竞争条件的时间敏感性,这显然是我无法通过打印语句或逐步调试进行调试的。
bool passedMagnet = true;
int incrementRevolutions(int runningRevCount,bool passingMagnet)
{
// Serial.printf("passedMagnet: %d,passingMagnet %d,runningRevCount: %d \n",passedMagnet,passingMagnet,runningRevCount);
int newRevCount = runningRevCount;
if (passedMagnet && passingMagnet)
{ //Started a new pass of the magnet
passedMagnet = false;
newRevCount++;
}
else if (!passedMagnet && !passingMagnet)
{ //The new pass of the magnet is complete
passedMagnet = true;
}
return newRevCount;
}
unsigned long elapsedtime = 0;
unsigned long intervalTime = 0;
int rev = 0;
void loop()
{
intervalTime = millis() - elapsedtime;
rev = incrementRevolutions(rev,digitalRead(digitalPin));
// Serial.printf("outside rev: %d \n",rev);
if (intervalTime > 1000)
{
Serial.printf("rev: %d \n",rev);
rev = 0;
elapsedtime = millis();
}
}
这是已知的 Arduino 或 C++ 编程问题吗?我该怎么做才能修复它?
解决方法
我认为应该归咎于测试。我不得不重新命名和移动一些东西以可视化逻辑,对此感到抱歉。
bool magStateOld = false; // initialize to digitalRead(digitalPin) in setup()
int incrementRevolutions(int runningRevCount,bool magState)
{
int newRevCount = runningRevCount;
// detect positive edge.
if (magState && !magStateOld) // <- was eq. to if (magState && magStateOld)
// the large counts came from here.
{
newRevCount++;
}
magStateOld = magState; // record last state unconditionally
return newRevCount;
}
你也可以把它写成...
int incrementRevolutions(int n,bool magState)
{
n += (magState && !magStateOld);
magStateOld = magState;
return n;
}
但最经济(也是最快)的方式是:
bool magStateOld;
inline bool positiveEdge(bool state,bool& oldState)
{
bool result = (state && !oldState);
oldState = state;
return result;
}
void setup()
{
// ...
magStateOld = digitalRead(digitalPin);
}
void loop()
{
// ...
rev += (int)positiveEdge(digitalRead(digitalPin),magStateOld);
// ...
}
它是可重用的,并且节省了堆栈空间和不必要的分配。
如果您无法从传感器获得清晰的转换(正边缘和负边缘上的噪声,您需要使用计时器对信号进行一些去抖动。
示例:
constexpr byte debounce_delay = 50; // ms,you may want to play with
// this value,smaller is better.
// but must be high enough to
// avoid issues on expected
// RPM range.
// 50 ms is on the high side.
byte debounce_timestamp; // byte is large enough for delays
// up to 255ms.
// ...
void loop()
{
// ...
byte now = (byte)millis();
if (now - debounce_timestamp >= debounce_delay)
{
debounce_timestamp = now;
rev += (int)positiveEdge(digitalRead(digitalPin),magStateOld);
}
// ...
}