Question

我正试图与管道InputStream &一道实施一个经过翻新的循环缓冲;但每当我到达Decoder可操作的Mhead.write时,它就每打一次。我认为,在使用单独的胎面时,没有出现僵局的机会。

    private class DecoderTask implements Runnable{

    @Override
    public void run() {
        while(!mStop){
            try {
                    Log.d(TAG,"trying to write");
        mHead.write(decode( 0, 1000));
            mHead.flush();
            Log.d(TAG,"Decoded");
            } catch (DecoderException e) {
                Log.e(TAG,e.toString());
            } catch (IOException e) {
                Log.e(TAG,e.toString());
            }
        }
    }

}
private class WriteTask implements Runnable{

    @Override
    public void run() {
        while(!mStop){
            try {
                                 Log.d(TAG,"trying to read");
                 int read = mTail.read(mByteSlave, 0, mByteSlave.length);
                 mAudioTrack.flush();
                                 mAudioTrack.write(mByteSlave,0,read);
                                 Log.d(TAG,"read");                 
            } catch (IOException e) {
                Log.e(TAG,e.toString());
            }
        }
    }

}


//in some function
mTail = new PipedInputStream();
mHead = new PipedOutputStream(mTail);
mByteSlave = new byte[BUF];
mT1 = new Thread(new DecoderTask(), "Reader");
mT2 = new Thread(new WriteTask(), "Writer");
mT1.start();
mT2.start();
return;

edit:这里是我服务的全部来源:http://pastie.org/1179792“rel=“nofollow”http://pastie.org/1179792。

缩略语:

trying to read
trying to write

Answer 1

该方案没有障碍,只是非常缓慢和效率低下。它使用100%的CPU。问题在于if (mTail. Available() >= mByteSlave.length)——这在大多数情况下会回错,因此,在座右铭上会 bus。如果你能够这样做的话,就这样做了。这个问题随后得到解决。如果你能够这样做,就会变得更加复杂......

还有一个问题:PipedInputStream.read 返回。你们需要使用:

int len = mTail.read(mByteSlave, 0, mByteSlave.length);
mAudioTrack.write(mByteSlave, 0, len);

除此以外,我在此法典中没有发现任何错误。我的完全考验就是这样:

import java.io.*;
public class Test2 {
    PipedOutputStream mHead;
    PipedInputStream mTail;
    byte[] mByteSlave = new byte[1024];
    boolean mStop;
    public static void main(String... ar) throws Exception {
        new Test2().run();
    }
    void run() throws Exception {
        mTail = new PipedInputStream();
        mHead = new PipedOutputStream(mTail);
        Thread mT1 = new Thread(new DecoderTask(), "Reader");
        Thread mT2 = new Thread(new WriteTask(), "Writer");
        mT1.start();
        mT2.start();
    }
    class DecoderTask implements Runnable {
        public void run() {
            while (!mStop) {
                try {
                    mHead.write(new byte[3000]);
                    mHead.flush();
                    System.out.println("decoded 3000");
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
    }
    class WriteTask implements Runnable {
        public void run() {
            while (!mStop) {
                try {
                    int len = mTail.read(mByteSlave, 0, mByteSlave.length);
                    if (len < 0) break; // EOF
                    // mAudioTrack.write(mByteSlave, 0, len);
                    // mAudioTrack.flush();
                    System.out.println("written " + len);
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
    }
}

Answer 2

我曾经历过同样的问题,并在以下几个方面加以解决:。区块,直到所有 by都写到产出流。这可能是缺席的PIPE_SIZE的问题。

www.un.org/Depts/DGACM/index_spanish.htm 毕竟,面积的确很重要;-)

Answer 3

只有排除了现有试验。读本将阻碍任何方面,在没有数据的情况下,你别无选择。

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