java - ServerSocket.accept() 抛出 SocketTimeoutException 并显示 null 消息

Question

我正在尝试从 ServerSocket 对象获取 Socket 的实例。但是，它总是打印

SocketTimeoutException = null

这是我的代码:

try {
    // We listen for new connections
    Log.d("ShairPort", "Service >> In TRY ");
    try {
        servSock = new ServerSocket(port);
    } catch (IOException e) {
        Log.d("ShairPort", "Service >> In TRY # IO Exception = " +e.getMessage());
        servSock = new ServerSocket();
    }catch(Exception e) {
        Log.d("ShairPort", "Service >> In TRY # E xception = " +e.getMessage());
    }

    // DNS Emitter (Bonjour)
    byte[] hwAddr = getHardwareAdress();
    emitter = new BonjourEmitter(name, getStringHardwareAdress(hwAddr), port);

    //Setting Timeout
    servSock.setSoTimeout(10000);


    Log.d("ShairPort", "Service >> stopThread = " +stopThread);

    while (!stopThread) {
        try {
            //********** This is throwing Exception ***************//
            Socket socket = servSock.accept();
            servSock.setReuseAddress(true);
            Log.d("ShairPort", "Service >> got connection from " + socket.toString());

            new RTSPResponder(hwAddr, socket).start();

        } catch(SocketTimeoutException e) {
            e.printStackTrace();
            Log.d("ShairPort", "Service >> SocketTimeoutException = " + e.getMessage());
           //********* here I am getting exception **************//         

        }catch(Exception e) {
            Log.d("ShairPort", "Service >> Exception = " + e.getMessage());
            servSock.close();
        }
    }
} catch (IOException e) {

    Log.d("ShairPort", "Service >> TRY # CATCH IOException = " + e.getMessage());
    throw new RuntimeException(e);

} 

如果我做错了什么，请告诉我，以便我解决此问题。

02-06 18:14:08.300: W/System.err(2245): java.net.SocketTimeoutException
02-06 18:14:08.300: W/System.err(2245):     at java.net.PlainSocketImpl.accept(PlainSocketImpl.java:108)
02-06 18:14:08.300: W/System.err(2245):     at java.net.ServerSocket.implAccept(ServerSocket.java:203)
02-06 18:14:08.310: W/System.err(2245):     at java.net.ServerSocket.accept(ServerSocket.java:128)
02-06 18:14:08.310: W/System.err(2245):     at vavi.apps.shairport.LaunchThread.run(LaunchThread.java:99)
02-06 18:14:08.310: W/System.err(2245): Caused by: libcore.io.ErrnoException: accept failed: EAGAIN (Try again)
02-06 18:14:08.320: W/System.err(2245):     at libcore.io.Posix.accept(Native Method)
02-06 18:14:08.320: W/System.err(2245):     at libcore.io.BlockGuardOs.accept(BlockGuardOs.java:55)
02-06 18:14:08.320: W/System.err(2245):     at java.net.PlainSocketImpl.accept(PlainSocketImpl.java:98)
02-06 18:14:08.320: W/System.err(2245):     ... 3 more

Answer 1

您正在服务器套接字上设置十秒的超时。

如果十秒内没有连接到达，accept() 方法将抛出 SocketTimeoutException。

如果您不希望出现这种行为，请不要设置超时或提高超时。

我不知道为什么当你自己设置超时时你会感到惊讶。

注意，绑定(bind)服务器套接字后，在服务器套接字上调用 setReuseAddress() 是完全没有意义的，更不用说每次接受新连接时了。