BufferedOutputStream沒有將所有內容寫入文件

Question

我試圖創建一個TFTP服務器，但是當它接收到文件時，似乎並不是所有文件都保存到了服務器上（有些字節丟失了）。 文件創建良好並且寫入了大部分數據，但是由於文件不完整，因此被分類為損壞且無法打開。 有人知道如何解決此問題嗎？

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類

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

這段代碼對我來說非常可疑：

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

您的輸出數組（數據）的長度為length - 4 ，但您僅將length - 8復制length - 8個字節。 如果packetData中要忽略的字節是前4個字節，則應為

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

因為最后一個參數不是長度，而是to索引（不包括）。 有關詳細信息，請參見javadoc 。