DataOutputSteam的writeInt为什么要这么实现呢?

Question

看thingking in java的IO部分时提到了DataOutputSteam, 于是查看了一下源码, 发现DataOutputStream在实现writeLong, writeInt ,writeChar和writeShort都采用了这种方法实现:

public final void writeChar(int v) throws IOException {

    out.write((v >>> 8) & 0xFF);

    out.write((v >>> 0) & 0xFF);

    incCount(2);

}

public final void writeInt(int v) throws IOException {

    out.write((v >>> 24) & 0xFF);

    out.write((v >>> 16) & 0xFF);

    out.write((v >>>  8) & 0xFF);

    out.write((v >>>  0) & 0xFF);

    incCount(4);

}

public final void writeLong(long v) throws IOException {

    writeBuffer[0] = (byte)(v >>> 56);

    writeBuffer[1] = (byte)(v >>> 48);

    writeBuffer[2] = (byte)(v >>> 40);

    writeBuffer[3] = (byte)(v >>> 32);

    writeBuffer[4] = (byte)(v >>> 24);

    writeBuffer[5] = (byte)(v >>> 16);

    writeBuffer[6] = (byte)(v >>>  8);

    writeBuffer[7] = (byte)(v >>>  0);

    out.write(writeBuffer, 0, 8);

    incCount(8);

}

我想知道的是为什么要先无符号右移，然后和00ff进行与运算呢？