通过Perl和Java套接字的Byte数组中的值错误

Question

Perl
x=1
y=222

java
x=257
y=222

I understand that I can only put an integer between 0 and 256 in a byte. How to send an integer higher than 256 in a pack(C*) or byte[][] ?

$data = $n->read($data2, 6);
@arr =  unpack("C*", $data2);

Sometimes when I send a value from Perl to Java, I catch a negative value in Java side, the issue is that I want to keep byte array only.

This is the java code from MousePressed on swing (I want to send to the server the current click)

public void mousePressed(MouseEvent e) {
        Point p = e.getPoint();
        byte[] buff = new byte[]{02,00,(byte)p.x,(byte)p.y,00,00};
                //write buff on my socket

Thanks

Answer 1

Java bytes are signed and they will hold their sign when you try to convert back to an integer. Therefore, if you want to extract an integer from a length-4 byte array, you need to do something like

int num = 0;
for(int i=0;i<4;i++){
    num <<= 8;
    num |= byteArray[i] & 255;
}
return num;

If you leave out the "& 255", you probably won't get the number you were expecting

Answer 2

You can send 32-bit ints in the following manner.

DataOutputStream dos = ...
dos.writeByte(2);
dos.writeByte(0);
dos.writeInt(p.x);
dos.writeInt(p.y);
dos.writeByte(0);
dos.writeByte(0);
dos.flush(); // assuming you use buffering.