我有一个Java应用程序是Voip。我正在使用一个套接字通过线程同时发送和接收信息。代码如下所示。您可以同时写入套接字输入和输出流吗?
Socket clientSocket = sockList.accept();
OutputStream outSock = clientSocket.getOutputStream();
InputStream inSock = clientSocket.getInputStream();
new Thread(new Capture(outSock)).start();
new Thread(new PlayAudio(inSock)).start();
outSock.close();
clientSocket.close();
我发现的问题是,当我写入输出流时,它会在第一次写入时阻塞。我发送的字节数不多。贝娄是我写的代码。
private class Capture implements Runnable{
private OutputStream out;
public Capture(OutputStream out){
this.out = out;
}
@Override
public void run() {
try{
int numBytesRead;
TargetDataLine outLine = getMic();
outLine.open();
outLine.start();
byte[] data = new byte[outLine.getBufferSize()/5];
byte[] test = {0x1,0x1,0x1};
while(true) {
//numBytesRead = outLine.read(data, 0, data.length);
//System.out.println(numBytesRead);
out.write(test, 0, test.length);
out.flush();
/*if(numBytesRead > 0){
out.write(data, 0, data.length);
System.out.println("C");
}*/
}
}catch(Exception ex){}
}
}
另一个线程读取的音频码是...
private class PlayAudio implements Runnable{
private InputStream in;
public PlayAudio(InputStream in){
this.in = in;
}
@Override
public void run() {
int write;
try{
SourceDataLine inLine = getSpeaker();
inLine.open();
inLine.start();
byte[] data = new byte[inLine.getBufferSize()];
byte[] test = new byte[3];
while(true){
System.out.println(1);
//write = in.read(data, 0, data.length);
in.read(test, 0 , test.length);
System.out.println(2);
/*if(write > 0){
inLine.write(data, 0, write);
System.out.println(3);
System.out.println(write);
}*/
}
} catch(Exception ex){}
}
}
我评论的实际代码的很大一部分,因为我只是想获得它的工作。我的写入功能在第一次写入时无限期地阻塞。这可能是我的线程有问题吗?我唯一的想法是,输出和输入流共享我的套接字对象,这可能会导致死锁或什么。请让我知道最新情况。
我认为,由于进出(两个独立的流)是独立的对象,同时对它们进行操作应该没有关系。所以你说我应该仍然担心同步,不管? –