Java:BufferedReader 读取的内容超过一行?

发布于 2024-08-13 09:06:28 字数 4251 浏览 15 评论 0原文

我正在用 Java 编写一个带有套接字的程序。我可以向客户端发送命令,也可以从客户端向服务器发送命令。为了读取命令,我使用了 BufferedReader。要编写它们,需要一个PrintWriter,但现在我想通过那个套接字传输文件(不仅仅是创建第二个连接)
首先,我向输出流写入文件包含的字节数。例如 40000 字节。因此,我通过套接字写入数字 40000,但连接的另一端读取 78

所以我在想:BufferedReader 读取的不仅仅是行(通过调用readLine(),这样我就丢失了一些字节文件数据。因为它们位于 BufferedReader 的缓冲区中。
所以数字78是我要传输的文件的一个字节。

这种思维方式对还是不对。如果是这样,如何解决这个问题。
我希望我已经解释得很好了。


这是我的代码,但我的默认语言是荷兰语。所以有些变量名听起来很奇怪。

public void flushStreamToStream(InputStream is, OutputStream os, boolean closeIn, boolean closeOut) throws IOException {
    byte[] buffer = new byte[BUFFERSIZE];
    int bytesRead;
    if ((!closeOut) && closeIn) { // To Socket from File
        action = "Upload";
        os.write(is.available()); // Here I write 400000
        max = is.available();
        System.out.println("Bytes to send: " + max);
        while ((bytesRead = is.read(buffer)) != -1) {
            startTiming(); // Two lines to compute the speed
            os.write(buffer, 0, bytesRead);
            stopTiming(); // Speed compution
            process += bytesRead;
        }
        os.flush();
        is.close();
        return;
    }
    if ((!closeIn) && closeOut) { // To File from Socket
        action = "Download";
        int bytesToRead = -1;
        bytesToRead = is.read(); // Here he reads 78.
        System.out.println("Bytes to read: " + bytesToRead);
        max = bytesToRead;
        int nextBufferSize;
        while ((nextBufferSize = Math.min(BUFFERSIZE, bytesToRead)) > 0) {
            startTiming();
            bytesRead = is.read(buffer, 0, nextBufferSize);
            bytesToRead -= bytesRead;
            process += nextBufferSize;
            os.write(buffer, 0, bytesRead);
            stopTiming();
        }
        os.flush();
        os.close();
        return;
    }
    throw new IllegalArgumentException("The only two boolean combinations are: closeOut == false && closeIn == true AND closeOut == true && closeIn == false");
}

解决方案如下:
感谢詹姆斯的建议
我认为 laginimaineb anwser 是解决方案的一部分。

阅读命令。

DataInputStream in = new DataInputStream(is); // Originally a BufferedReader
// Read the request line
String str;
while ((str = in.readLine()) != null) {
    if (str.trim().equals("")) {
       continue;
    }
    handleSocketInput(str);
}

现在是flushStreamToStream:

public void flushStreamToStream(InputStream is, OutputStream os, boolean closeIn, boolean closeOut) throws IOException {
    byte[] buffer = new byte[BUFFERSIZE];
    int bytesRead;
    if ((!closeOut) && closeIn) { // To Socket from File
        action = "Upload";
        DataOutputStream dos = new DataOutputStream(os);
        dos.writeInt(is.available());

        max = is.available();
        System.out.println("Bytes to send: " + max);
        while ((bytesRead = is.read(buffer)) != -1) {
            startTiming();
            dos.write(buffer, 0, bytesRead);
            stopTiming();
            process += bytesRead;
        }
        os.flush();
        is.close();
        return;
    }
    if ((!closeIn) && closeOut) { // To File from Socket
        action = "Download";
        DataInputStream dis = new DataInputStream(is);
        int bytesToRead = dis.readInt();
        System.out.println("Bytes to read: " + bytesToRead);
        max = bytesToRead;
        int nextBufferSize;
        while ((nextBufferSize = Math.min(BUFFERSIZE, bytesToRead)) > 0) {
            startTiming();
            bytesRead = is.read(buffer, 0, nextBufferSize);
            bytesToRead -= bytesRead;
            process += nextBufferSize;
            os.write(buffer, 0, bytesRead);
            stopTiming();
        }
        os.flush();
        os.close();
        return;
    }
    throw new IllegalArgumentException("The only two boolean combinations are: closeOut == false && closeIn == true AND closeOut == true && closeIn == false");
}

Martijn。

I'm making a program in Java with Sockets. I can send commands to the client and from the client to the server. To read the commands I use a BufferedReader. To write them, a PrintWriter But now I want to transfer a file through that socket (Not simply create a second connection).
First I write to the outputstream how many bytes the file contains. For example 40000 bytes. So I write the number 40000 through the socket, but the other side of the connection reads 78.

So I was thinking: The BufferedReader reads more than just the line (by calling readLine()) and on that way I lose some bytes from the file-data. Because they are in the buffer from the BufferedReader.
So the number 78 is a byte of the file I want to transmit.

Is this way of thinking right, or not. If so, how to sovle this problem.
I hope I've explained well.


Here is my code, but my default language is Dutch. So some variable-name can sound stange.

public void flushStreamToStream(InputStream is, OutputStream os, boolean closeIn, boolean closeOut) throws IOException {
    byte[] buffer = new byte[BUFFERSIZE];
    int bytesRead;
    if ((!closeOut) && closeIn) { // To Socket from File
        action = "Upload";
        os.write(is.available()); // Here I write 400000
        max = is.available();
        System.out.println("Bytes to send: " + max);
        while ((bytesRead = is.read(buffer)) != -1) {
            startTiming(); // Two lines to compute the speed
            os.write(buffer, 0, bytesRead);
            stopTiming(); // Speed compution
            process += bytesRead;
        }
        os.flush();
        is.close();
        return;
    }
    if ((!closeIn) && closeOut) { // To File from Socket
        action = "Download";
        int bytesToRead = -1;
        bytesToRead = is.read(); // Here he reads 78.
        System.out.println("Bytes to read: " + bytesToRead);
        max = bytesToRead;
        int nextBufferSize;
        while ((nextBufferSize = Math.min(BUFFERSIZE, bytesToRead)) > 0) {
            startTiming();
            bytesRead = is.read(buffer, 0, nextBufferSize);
            bytesToRead -= bytesRead;
            process += nextBufferSize;
            os.write(buffer, 0, bytesRead);
            stopTiming();
        }
        os.flush();
        os.close();
        return;
    }
    throw new IllegalArgumentException("The only two boolean combinations are: closeOut == false && closeIn == true AND closeOut == true && closeIn == false");
}

Here is the solution:
Thanks to James suggestion
I think laginimaineb anwser was a piece of the solution.

Read the commands.

DataInputStream in = new DataInputStream(is); // Originally a BufferedReader
// Read the request line
String str;
while ((str = in.readLine()) != null) {
    if (str.trim().equals("")) {
       continue;
    }
    handleSocketInput(str);
}

Now the flushStreamToStream:

public void flushStreamToStream(InputStream is, OutputStream os, boolean closeIn, boolean closeOut) throws IOException {
    byte[] buffer = new byte[BUFFERSIZE];
    int bytesRead;
    if ((!closeOut) && closeIn) { // To Socket from File
        action = "Upload";
        DataOutputStream dos = new DataOutputStream(os);
        dos.writeInt(is.available());

        max = is.available();
        System.out.println("Bytes to send: " + max);
        while ((bytesRead = is.read(buffer)) != -1) {
            startTiming();
            dos.write(buffer, 0, bytesRead);
            stopTiming();
            process += bytesRead;
        }
        os.flush();
        is.close();
        return;
    }
    if ((!closeIn) && closeOut) { // To File from Socket
        action = "Download";
        DataInputStream dis = new DataInputStream(is);
        int bytesToRead = dis.readInt();
        System.out.println("Bytes to read: " + bytesToRead);
        max = bytesToRead;
        int nextBufferSize;
        while ((nextBufferSize = Math.min(BUFFERSIZE, bytesToRead)) > 0) {
            startTiming();
            bytesRead = is.read(buffer, 0, nextBufferSize);
            bytesToRead -= bytesRead;
            process += nextBufferSize;
            os.write(buffer, 0, bytesRead);
            stopTiming();
        }
        os.flush();
        os.close();
        return;
    }
    throw new IllegalArgumentException("The only two boolean combinations are: closeOut == false && closeIn == true AND closeOut == true && closeIn == false");
}

Martijn.

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评论(4

爱的十字路口 2024-08-20 09:06:28

我不确定我是否听从了你的解释。

但是,是的 - 您无法真正控制 BufferedReader 实际读取的量。这种读取器的要点在于,它会根据需要乐观地读取底层资源块以补充其缓冲区。因此,当您第一次调用readLine时,它会发现其内部缓冲区没有足够的空间来满足您的请求,并且继续读取它认为需要的字节数< /strong> 来自底层源,这通常会比您当时要求的要多得多。一旦缓冲区被填充,它就会从缓冲的内容中返回您的行。

因此,一旦将输入流包装在 BufferedReader 中,您应该确保仅通过同一个缓冲读取器读取该流。如果不这样做,您最终将丢失数据(因为一些字节将被消耗并且现在位于 BufferedReader 的缓存中等待服务)。

I'm not sure I've followed your explanation.

However, yes - you have no real control over how much a BufferedReader will actually read. The point of such a reader is that it optimistically reads chunks of the underlying resource as needed to replenish its buffer. So when you first call readLine, it will see that its internal buffer doesn't have enough to serve youthe request, and will go off and read however many bytes it feels like into its buffer from the underlying source, which will generally be much more than you asked for just then. Once the buffer's been populated, it returns your line from the buffered content.

Thus once you wrap an input stream in a BufferedReader, you should be sure to only read that stream through the same buffered reader. If you don't you'll end up losing data (as some bytes will have been consumed and are now sitting in the BufferedReader's cache waiting to be served).

蘑菇王子 2024-08-20 09:06:28

DataInputStream 很可能是您想要使用的。另外,不要使用 available() 方法,因为它通常是无用的。

DataInputStream is most likely what you want to use. Also, don't use the available() method as it is generally useless.

放赐 2024-08-20 09:06:28

BufferedReader 假定它是从底层输入流中唯一读取的内容。

其目的是最大限度地减少从底层流读取的次数(这是昂贵的,因为它们可以委托得相当深)。为此,它保留一个缓冲区,通过在对底层流的单次调用中读取尽可能多的字节来填充该缓冲区。

所以是的,你的诊断是准确的。

A BufferedReader assumes that it is the only one reading from the underlying input stream.

It's purpose is to minimize the number of reads from the underlying stream (which are expensive, as they can delegate quite deeply). To that end, it keeps a buffer, which it fills by reading as many bytes as possible into it in a single call to the underlying stream.

So yes, your diagnosis is accurate.

傲娇萝莉攻 2024-08-20 09:06:28

这里只是一个疯狂的尝试 - 40000 是二进制的 1001110001000000。现在,这里的前 7 位是 1001110,即 78。这意味着,您正在写入 2 个字节的信息,但读取 7 位。

Just a wild stab here - 40000 is 1001110001000000 in binary. Now, the first seven bits here are 1001110 which is 78. Meaning, you're writing 2 bytes of information but reading seven bits.

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