Arduino通过串行OutofBoundsException处理Long数据类型

时间:2017-08-10 13:23:18

标签: arduino serial-port processing indexoutofboundsexception long-integer

我正在尝试通过串行从Arduino传输长值到处理,并具有以下代码。

Arduino的:

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

void loop() {
  long int randomno = random(0, 1520);
  unsigned char buf[sizeof(long int)];
  memcpy(buf,&randomno,sizeof(long int));
  Serial.write(buf,sizeof(buf));
  delay(50);
}

处理:

import processing.serial.*;
Serial myPort;

void setup() {
  size(1920, 1080);
  myPort = new Serial(this, "COM3", 9600);
}

void draw(){
  background(0,0,0);
   long value;
   byte[] inBuffer = new byte[4];

  if (myPort.available() > 0) {
    try{
    inBuffer = myPort.readBytes();
    if (inBuffer != null) {
         value = byteAsULong(inBuffer[0]) << 0 | 
         (byteAsULong(inBuffer[1])) << 8 | 
         (byteAsULong(inBuffer[2])) << 16| 
         (byteAsULong(inBuffer[3])) << 24;
      println(value);
    }
    } catch(RuntimeException e) {
      e.printStackTrace();
    }
  }
}

public static long byteAsULong(byte b) {
    return ((long)b) & 0x00000000000000FFL; 
}

当我运行它时,我得到一些值,但随后不断向我抛出ArrayIndexOutofBounds Exceptions。现在我已经通过使用printstacktrace catch来克服它,但我想知道问题是什么。

1 个答案:

答案 0 :(得分:0)

Arduino中的long是32位无符号或有符号。值0x00000000000000FF是64位数,在Arduino体系结构中无法理解。当你可以简单地执行::

时,该功能似乎没有用处
value = inBuffer[0] | 
        inBuffer[1] << 8 | 
        inBuffer[2] << 16| 
        inBuffer[3] << 24;