Question

以下代码使Arduino和Processing通过＆＃39; A＆＃39;像烫手山芋一样来回：

处理事件＆＃39; A＆＃39;到ARDUINO

import processing.serial.*; //import the Serial library
Serial mySerial;  //the Serial port object

void setup() 
{
  mySerial = new Serial(this, Serial.list()[7], 9600); //Pick whichever works
  mySerial.bufferUntil('\n');
}

void serialEvent( Serial mySerial) 
{
  println("writing A...");
  delay(1000);                   //How I defined this is not relevant. Trust me.
  mySerial.write('A');        //Send 'A' to Arduino
}

ARDUINO SENDS＆＃39; A＆＃39;返回加工

void setup()
{
  Serial.begin(9600);
}

void loop()
{
  delay(1000);
  Serial.println('A');      //Send 'A' back to Processing
}

会发生什么？ 好吧......游戏适用于大约十次交流（我怎么知道？好吧，我得到了＃34;写了A ...＆＃34;消息五次，我使用bufferUntil（）告诉我每个这些消息是通过从Arduino接收Serial.println（＆＃39; A＆＃39;） BUT 然后处理中断而出现以下错误：

Error, disabling serialEvent() for /dev/tty.usbmodem1421
null
java.lang.RuntimeException: Error reading from serial port /dev/tty.usbmodem1421: Port not opened
    at processing.serial.Serial.serialEvent(Unknown Source)
    at jssc.SerialPort$LinuxEventThread.run(SerialPort.java:1299)
Could not run the sketch (Target VM failed to initialize).
For more information, read revisions.txt and Help → Troubleshooting.

发生了什么？

Answer 1

Port not opened

是错误消息中的关键字：串行端口随时随机关闭。解决办法是不断强制它在平局中打开：

Serial serial;
boolean serialInited;


void draw () {
    if (serialInited) {
        // serial is up and running
        try {
            byte b = serial.read();
            // fun with serial here...
        } catch (RuntimeException e) {
            // serial port closed :(
            serialInited = false;
        }
    } else {
        // serial port is not available. bang on it until it is.
        initSerial();
    }
}

void initSerial () {
    try {
        serial = new Serial(this, Serial.list()[0], BAUD_RATE);
        serialInited = true;
    } catch (RuntimeException e) {
        if (e.getMessage().contains("<init>")) {
            System.out.println("port in use, trying again later...");
            serialInited = false;
        }
    }
}

Arduino和Processing之间的一场比赛

1 个答案: