套接字中的“无止境”AudioInputStream

Question

我在从套接字创建AudioInputStream时遇到了问题。以下是重要的部分：

public class SoundStream extends Thread {
    private int port;
    private String IP;
    private Socket socket;

    private SoundObject soundObject;

    private OpenAL openAL;
    private Source source;

    private boolean run = true;

    public SoundStream(int port, String IP, SoundObject soundObject) {
        this.soundObject = soundObject;
        this.port = port;
        this.IP = IP;
    }

    public void run() {
        try {
            this.socket = new Socket(this.IP, this.port);
            this.openAL = new OpenAL();
        } catch (Exception e) {
            e.printStackTrace();
        }
        this.mainCycleMethod();
    }

    private void mainCycleMethod() {
        while (run) {
            this.soundObject.blockAndWait();
            switch (this.soundObject.getAndResetEvent()) {
                case 0:
                    this.run = false;
                    this.close();
                    break;
                case 1:
                    this.setPitch();
                    break;
                case 2:
                    this.closeSource();
                    this.play();
                    break;
                case 3:
                    this.pause(true);
                    break;
                case 4:
                    this.pause(false);
                    break;
            }
        }
    }

    private BufferedInputStream getInputStream() throws Exception {
        return new BufferedInputStream(socket.getInputStream());
    }

    private void setPitch() {
        if(this.source != null) {
            try {
                this.source.setPitch(this.soundObject.getPitch());
            } catch (ALException e) {
                e.printStackTrace();
            }
        }
    }

    private void play() {
        try {
            AudioInputStream audioInputStream = new AudioInputStream(this.getInputStream(), this.soundObject.getAudioFormat(), AudioSystem.NOT_SPECIFIED);
//            AudioInputStream audioInputStream_tmp = AudioSystem.getAudioInputStream(this.getInputStream());
//            AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(this.soundObject.getAudioFormat(), audioInputStream_tmp);
            this.source = openAL.createSource(audioInputStream);
            this.source.setGain(1f);
            this.source.play();
        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }

    private void close() {
        this.closeSource();
        this.openAL.close();
        try {
            this.socket.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private void closeSource() {
        if(this.source!=null) {
            this.source.close();
        }
    }

    private void pause(boolean pause) {
        if(this.source != null) {
            try {
                if (pause) {
                    this.source.pause();
                } else {
                    this.source.play();
                }
            } catch (ALException ex) {
                ex.printStackTrace();
            }
        }
    }
}


public class SoundObject extends AbstractEventObject {
    public AudioFormat getAudioFormat() {
        boolean signed = false;
        //true,false
        boolean bigEndian = false;
        //true,false
        return new AudioFormat(this.frequency, this.bits, this.channels, signed, bigEndian);
    }
.
.
.
.
}

这段代码将UnsupportedAudioFileException抛到这一行：

AudioInputStream audioInputStream_tmp = AudioSystem.getAudioInputStream(this.getInputStream());

但是，当我使用此代码时：

AudioInputStream audioInputStream = new AudioInputStream(this.getInputStream(), this.soundObject.getAudioFormat(), 100000);

它播放声音，但只在它将这100000个样本帧加载到音频输入流之后。在播放完所有的100000帧之后。

我想，如果我能够在第一次AudioFormat初始化过程中将AudioInputStream作为参数直接传递给它，我就会解决这个问题，但这似乎是不可能的。我正在从服务器接收音频格式规范。

我认为一个可能的解决方案是创建一个dataline，我可以将它作为参数传递给AudioInputStream构造函数。但是，我不知道如何从套接字直接获取数据到dataline。我知道一个使用无限循环的解决方案，在这个解决方案中，它读取数据并将它们写入dataline。但这似乎是浪费。是否有更直接的方法？

我希望使用java库解决问题是可能的，因为我需要改变速度，我希望我不必自己去做。

谢谢

Answer 1

我终于解决了这个问题。事实证明，java内置了流支持，但它不在GitHub的文档中，所以我一开始没有注意到。源类中有一个createOutputStream方法，它返回OutputStream。您可以直接将字节写入OutputStream。

这是我的代码：

在这个片段中，我初始化了OpenAL：

public void run() {
    try {
        this.socket = new Socket(this.IP, this.port);
        this.openAL = new OpenAL();
    } catch (Exception ex) {
        Log.severe(ex.toString());
    }
    this.mainCycleMethod();
}

下面是在InputStream可用时调用的play方法：

private void play() {
    try {
        this.source = openAL.createSource();
        this.outputWriter = new OutputWriter(this.socket.getInputStream(), this.source, this.soundObject.getAudioFormat());
        this.source.setGain(1f);
        this.outputWriter.start();
    } catch (Exception ex) {
        Log.severe(ex.toString());
    }
}

您必须使用没有参数的createSource方法，它返回Source的新实例。不要调用源上的play方法，它由SourceOutputStream类处理，createOutputStream方法返回该类实例。手动调用play方法没有什么问题，但是当缓冲区为空时，我的经验很糟糕。基本上，当您开始将数据流到OpenAL时，它不会在稍后开始播放。

下面是我的OutputWriter代码，它负责将字节从InputStream传递给OutputStream：

package cz.speechtech.sound;

import org.urish.openal.ALException;
import org.urish.openal.Source;

import javax.sound.sampled.AudioFormat;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;

/**
 * Created by honza on 16.12.15.
 */
public class OutputWriter extends Thread {
    private InputStream inputStream;
    private OutputStream outputStream;

    private int STREAMING_BUFFER_SIZE = 24000;
    private int NUMBER_OF_BUFFERS = 4;

    private boolean run = true;

    public OutputWriter(InputStream inputStream, Source source, AudioFormat audioFormat) {
        this.inputStream = inputStream;
        try {
            this.outputStream = source.createOutputStream(audioFormat, this.NUMBER_OF_BUFFERS, 1024);
        } catch (ALException e) {
            e.printStackTrace();
        }
    }

    public void run() {
        byte[] buffer = new byte[this.STREAMING_BUFFER_SIZE];
        int i;
        try {
            Thread.sleep(1000); // Might cause problems
            while (this.run) {
                i = this.inputStream.read(buffer);
                if (i == -1) break;
                outputStream.write(buffer, 0, i);
            }
        } catch (IOException e) {
            e.printStackTrace();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }

    public synchronized void stopRunning() {
        this.run = false;
        try {
            this.outputStream.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

}

祝您今天愉快。