首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >从Java读取Arduino太慢了

从Java读取Arduino太慢了
EN

Stack Overflow用户
提问于 2015-07-10 11:48:18
回答 1查看 490关注 0票数 1

我正在尝试编写一个简单的Java程序,在一些物理传感器上的用户的刺激下,显示输入到Arduino板的模拟输入。

如果我使用Arduino程序本身并查看串行监视器,我会看到基于用户输入的模拟读数立即更新。但是,当从我的Java程序读取输入时,在物理刺激和显示值之间有很大的延迟(5-6秒)。

上传到Arduino的代码非常简单,只需通过一堆模拟端口循环并打印数据:

代码语言:javascript
复制
void setup() {
  Serial.begin(9600);
  pinMode(30, OUTPUT);
}

void loop() 
  digitalWrite(30, HIGH);

  for (int i = 0; i < 10; i++) {
    int sensorValue = analogRead(i);
    Serial.print(String(i) + ":");
    Serial.println(sensorValue);
  }
}

我使用的Java代码几乎全部复制自here提供的示例,除了我更新了端口名称以适用于我的系统(MacOSX10.10.3)之外。

下面是Java代码:

代码语言:javascript
复制
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStream;
import gnu.io.CommPortIdentifier; 
import gnu.io.SerialPort;
import gnu.io.SerialPortEvent; 
import gnu.io.SerialPortEventListener; 
import java.util.Enumeration;


public class SerialTest implements SerialPortEventListener {

    SerialPort serialPort;

    /** The port we're normally going to use. */
    private static final String PORT_NAMES[] = { 
            "/dev/cu.usbmodem1411",
            "/dev/cu.usbmodem1451",
    };

    /**
    * A BufferedReader which will be fed by a InputStreamReader 
    * converting the bytes into characters 
    * making the displayed results codepage independent
    */
    private BufferedReader input;

    /** The output stream to the port */
    private OutputStream output;

    /** Milliseconds to block while waiting for port open */
    private static final int TIME_OUT = 2000;

    /** Default bits per second for COM port. */
    private static final int DATA_RATE = 9600;

    public void initialize() {
        CommPortIdentifier portId = null;
        Enumeration portEnum = CommPortIdentifier.getPortIdentifiers();

        //First, Find an instance of serial port as set in PORT_NAMES.
        while (portEnum.hasMoreElements()) {
            CommPortIdentifier currPortId = (CommPortIdentifier) portEnum.nextElement();
            for (String portName : PORT_NAMES) {
                if (currPortId.getName().equals(portName)) {
                    portId = currPortId;
                    break;
                }
            }
        }
        if (portId == null) {
            System.out.println("Could not find COM port.");
            return;
        }

        try {
            // open serial port, and use class name for the appName.
            serialPort = (SerialPort) portId.open(this.getClass().getName(),
                    TIME_OUT);

            // set port parameters
            serialPort.setSerialPortParams(DATA_RATE,
                    SerialPort.DATABITS_8,
                    SerialPort.STOPBITS_1,
                    SerialPort.PARITY_NONE);

            // open the streams
            input = new BufferedReader(new InputStreamReader(serialPort.getInputStream()));
            output = serialPort.getOutputStream();

            // add event listeners
            serialPort.addEventListener(this);
            serialPort.notifyOnDataAvailable(true);
        } catch (Exception e) {
            System.err.println(e.toString());
        }
    }

    /**
     * This should be called when you stop using the port.
     * This will prevent port locking on platforms like Linux.
     */
    public synchronized void close() {
        if (serialPort != null) {
            serialPort.removeEventListener();
            serialPort.close();
        }
    }

    /**
     * Handle an event on the serial port. Read the data and print it.
     */
    public synchronized void serialEvent(SerialPortEvent oEvent) {
        if (oEvent.getEventType() == SerialPortEvent.DATA_AVAILABLE) {
            try {
                String inputLine=input.readLine();
                System.out.println(inputLine);
            } catch (Exception e) {
                System.err.println(e.toString());
            }
        }
        // Ignore all the other eventTypes, but you should consider the other ones.
    }

    public static void main(String[] args) throws Exception {
        SerialTest main = new SerialTest();
        main.initialize();
        Thread t=new Thread() {
            public void run() {
                //the following line will keep this app alive for 1000 seconds,
                //waiting for events to occur and responding to them (printing incoming messages to console).
                try {Thread.sleep(1000000);} catch (InterruptedException ie) {}
            }
        };
        t.start();
        System.out.println("Started");
    }
}

该程序所做的全部工作就是打印出Arduino串行监视器的值,但是通过Java所花费的时间(5-6秒)要比Arduino程序本身所花费的时间长得多。

1)为什么会发生这种情况?

2)如何修复?

EN

回答 1

Stack Overflow用户

发布于 2015-07-18 02:15:05

对我有效的答案是在我的Arduino代码中添加一个delay方法。显然,Java已经被输入的速度压垮了。

当然,Java仍然可以以足够快的速度读取,这有任何实际的区别。

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/31332136

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档