BufferedOutputStream没有将所有内容写入文件

Question

我试图创建一个TFTP服务器，但是当它接收到文件时，似乎并不是所有文件都保存到了服务器上（有些字节丢失了）。 文件创建良好并且写入了大部分数据，但是由于文件不完整，因此被分类为损坏且无法打开。 有人知道如何解决此问题吗？

主班

            WRQ WRQ = new WRQ();
            ACK ACK = new ACK();
            DatagramPacket outPacket;
            BufferedOutputStream bufferedOutput = new BufferedOutputStream(new FileOutputStream(filename));
            byte[] bytes;
            byte[] fileOut;
            outPacket = WRQ.firstPacket(packet);
            socket.send(outPacket);

            socket.receive(packet);

            while (packet.getLength() == 516){

            bytes = WRQ.doWRQ(packet);
            bufferedOutput.write(bytes);

            outPacket = ACK.doACK(packet);
            socket.send(outPacket);

            socket.receive(packet); 

            }

            bytes = WRQ.doWRQ(packet);
            bufferedOutput.write(bytes);

            bufferedOutput.close();

            outPacket = ACK.doACK(packet);
            socket.send(outPacket);

WRQ类

public class WRQ {

public DatagramPacket firstPacket(DatagramPacket packet) throws IOException{

    ACK ACK = new ACK();
    DatagramPacket ACKpacket = ACK.doACK(packet);

    //takes ACK packet and sets block # as 0 to signal that this is the first packet in a WRQ
    byte[] ACKcontents = new byte[3];
    ACKcontents = ACKpacket.getData();
    ACKcontents[2] = 0;
    ACKcontents[3] = 0;
    ACKpacket.setData(ACKcontents);

    return ACKpacket;

}

public byte[] doWRQ(DatagramPacket packet){

    int length = packet.getLength();
    byte[] packetData = packet.getData();
    byte[] data = new byte[length - 4];
    data = Arrays.copyOfRange(packetData, 4, length - 4);

    return data;

}

}

Answer 1

这段代码对我来说非常可疑：

byte[] packetData = packet.getData();
byte[] data = new byte[length - 4];
data = Arrays.copyOfRange(packetData, 4, length - 4);

您的输出数组（数据）的长度为length - 4 ，但您仅将length - 8复制length - 8个字节。 如果packetData中要忽略的字节是前4个字节，则应为

data = Arrays.copyOfRange(packetData, 4, length);

因为最后一个参数不是长度，而是to索引（不包括）。 有关详细信息，请参见javadoc 。