Arduino阅读大量序列数据而没有损失

Question

在Arduino系列中，我需要阅读这类行＆lt; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16＆gt; 尽可能地不阻止。 目标是在不延迟的情况下设置数组补偿的值[]。 在此示例中损失为96％，然后是95％……因此，我需要我的代码每次显示1000（100％）。

这就是我已经拥有的：

static int compensastion[64];

int passage=1;
const byte numChars = 1000;
char receivedChars[numChars];
boolean newData = false;

int score = 0;
int mal_score = 0;

void setup() {
  Serial.begin(115200);
  Serial.println();
}

void loop() {
  read_inputs();
  // get the percentage of success for this simple message from 1 to 16
  Serial.print(double(double(score) / double(mal_score+score))*10000);
  Serial.println();
}

void read_inputs(){
  recvWithStartEndMarkers();
  check();
}

// Get the full line    <1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16>
void recvWithStartEndMarkers() {
  static boolean recvInProgress = false;
  static byte ndx = 0;
  char startMarker = '<';
  char endMarker = '>';
  char rc;

  if (Serial.available() > 0 && newData == false) {
    rc = Serial.read();

    if (recvInProgress == true) {
      if (rc != endMarker) {
        receivedChars[ndx] = rc;
        ndx++;
        if (ndx >= numChars) {
          ndx = numChars - 1;
        }
      }
      else {
        receivedChars[ndx] = '\0';
        recvInProgress = false;
        ndx = 0;
        newData = true;
      }
    }

    else if (rc == startMarker) {
      recvInProgress = true;
    }
  }
}

// Parse the line in digits 
void check(){
 if (newData == true) {
    newData = false;
    
    int marge = 0;
    passage=1;

    for(int j = 0; j < numChars; j=j+1){
      int my_int = -1;
      char buf[4];
      if(isDigit(receivedChars[j]) &&marge == 0 ){
        if(!isDigit(receivedChars[j+1])){
          my_int = receivedChars[j] - '0';
          marge=marge+1;
        }

        if(isDigit(receivedChars[j+1]) && !isDigit(receivedChars[j+2]) ){
          my_int = receivedChars[j] - '0';
          my_int = my_int*10 + (receivedChars[j+1] - '0');
          marge=marge+2;
        }

        if(isDigit(receivedChars[j+1]) && isDigit(receivedChars[j+2]) && !isDigit(receivedChars[j+3]) ){
          my_int = receivedChars[j] - '0';
          my_int = my_int*10 + (receivedChars[j+1] - '0');
          my_int = my_int*10 + (receivedChars[j+2] - '0');
          marge=marge+3;
        }
      }
      if (isDigit(receivedChars[j]) && my_int != -1){
          if(passage == my_int){
            score=score+1;
          }else{
            mal_score= mal_score+1;
          }

          compensastion[(passage-1)*2] = my_int;
          passage=passage+1;


          if(passage+1 >= 16+2){
            passage=1;
          }
      }
      if(marge > 0){
        marge=marge-1;
      }
    }
  }
}

Answer 1

根据我的经验，处理大量串行数据的最有效方法之一是使用中断和有限状态机器（FSM）。

请仔细阅读这是惊人的尼克·加蒙（Nick Gammon）的指南。在处理密集处理任务时，FSM是最通用的方法。 （不要与 rtos ）

关于串行通信，我将通过serialevent（）沿着沿着以下示例的行：

char serialString[] = "    "; // Empty serial string variable
bool stringFinished = false; // Flag to indicate reception of a string after terminator is reached

void setup(){
    previousEncoderTime = 0;

}

void loop(){

    unsigned long now = millis();

   
    if (stringFinished){ // When the serial Port has received a command

        stringFinished = false;
       // Implement your logic here

    }

}

void serialEvent()
{
    int idx = 0;

    while (Serial.available())
    {
        char inChar = (char)Serial.read();

        if (inChar == '\n')    // The reading event stops at a new line character
        {
            serialTail = true;
            serialString[idx] = inChar;
        }

        if (!serialTail)
        {
            serialString[idx] = inChar;
            idx++;
        }

        if (serialTail)
        {
            stringFinished = true;
            Serial.flush();
            serialTail = false;
        }
    }
}

最后，尝试将单用函数中的所有必要数据处理划分。这样，您的代码将具有适当的流量，您将降低用繁重的任务阻止系统的风险。

编辑：我忘了提到serial.available（）被视为一种高度不可靠且耗时的轮询方法。坚持中断，您不会遇到巨大的问题