Question

今晚在一个手上的项目上工作了......突然间，JVM开始轰动我。我已经能够创建一个可以重现崩溃的简单程序，并将其提交给Sun，但我想知道这里是否有人可以看一下这个程序，并告诉我我是不是在做一些愚蠢的事情（即有一种简单的方法可以避免引发崩溃。）

设置就像这样。我在同一台机器上有两个进程A和B.两者都接受ServerSocketChannel上的连接。一旦A连接到B，B有时会想要打开另一个连接回A（实际的应用程序是Peer 2 Peer风格的网络，他们都希望充当客户端和服务器）。一旦B尝试将SocketChannel.Open（...）返回到A，VM就会崩溃。

已经在Vista和Windows 7上进行了测试，并且使用了不同的线程设置（在真实应用中，SocketChannel.Open（..）发生在特定于未来连接的单独线程上）。

以下代码重现了该行为。为了触发它，你只需在5秒内启动两个进程：

  java SocketAcceptor RESPONDER 25002 25003
  java SocketAcceptor INITIATOR 25003 25002

我尝试将连接“回到另一个进程”的方式有什么明显的愚蠢之处吗？谢谢你的帮助

import java.io.IOException;
import java.net.InetSocketAddress;

import java.nio.channels.ServerSocketChannel;
import java.nio.channels.SocketChannel;


public class SocketAcceptor implements Runnable{

    private final Thread internalThread;
    private final ServerSocketChannel ssc;
private final boolean responder;
private final int remotePort;


public SocketAcceptor(int port, boolean responder, int remotePort) throws IOException{
    this.responder = responder;
    this.remotePort = remotePort;

    ssc = ServerSocketChannel.open();
    ssc.socket().bind(new InetSocketAddress("localhost", port));

    internalThread = new Thread(this);
    internalThread.start();
}


@Override
public void run() {

    while (true){

        try{
            //Wait for a new connection 
            SocketChannel sc = ssc.accept();

            if (responder){
                //If we are a responder, make a connection back immediately
                InetSocketAddress addr = new InetSocketAddress(sc.socket().getInetAddress(), remotePort);
                SocketChannel.open(addr);
            }

        }catch (Exception e){
            e.printStackTrace();
        }
    }
}


public static void main(String[] args) throws Exception{
    if (args.length != 3){
        System.out.println("Syntax: java Main <RESPONDER/INITIATOR> <ownPort> <remotePort>");
        System.exit(1);
    }

    boolean responder = args[0].equals("RESPONDER");
    int ownPort = Integer.parseInt(args[1]);
    int remotePort = Integer.parseInt(args[2]);

    if (responder){
        SocketAcceptor sa = new SocketAcceptor(ownPort, true, remotePort);  

    }else{
        SocketAcceptor sa = new SocketAcceptor(ownPort, false, 0);
    }

    //Wait for 5 seconds, to give the human a chance to start both processes
    Thread.sleep(5000);


    if(!responder){
        //The initiator starts the show by making a connection to the other process
        InetSocketAddress addr = new InetSocketAddress("localhost", remotePort);
        SocketChannel.open(addr);
    }
}

}

更详细的事件序列演练。

  A starts listening on a ServerSocketChannel port 25002
  B starts listening on a ServerSocketChannel port 25003
  B creates a connection to A on "localhost"/25002 using SocketChannel.open(...)
  A accepts the connection
  A determines the InetAddress of the incomming connection from B = 127.0.0.1
  A tries to open a connection to 127.0.0.1/25003 using SocketChannel.open(..), but instead crashes

最后，来自崩溃转储的错误文本的顶部

#
# A fatal error has been detected by the Java Runtime Environment:
#
#  EXCEPTION_ACCESS_VIOLATION (0xc0000005) at pc=0x6d9011fd, pid=6988, tid=7576
#
# JRE version: 6.0_16-b01
# Java VM: Java HotSpot(TM) Client VM (14.2-b01 mixed mode, sharing windows-x86 )
# Problematic frame:
# V  [jvm.dll+0x1011fd]
#
# If you would like to submit a bug report, please visit:
#   http://java.sun.com/webapps/bugreport/crash.jsp
#

---------------  T H R E A D  ---------------

Current thread (0x04421400):  JavaThread "pool-1-thread-1" [_thread_in_vm, id=7576, stack(0x03ef0000,0x03f40000)]

siginfo: ExceptionCode=0xc0000005, reading address 0x00000000

Registers:
EAX=0x00000000, EBX=0x04421400, ECX=0x00000006, EDX=0x6da53028
ESP=0x03f3f714, EBP=0x03f3f730, ESI=0x00000000, EDI=0x2c016740
EIP=0x6d9011fd, EFLAGS=0x00010246

Top of Stack: (sp=0x03f3f714)
0x03f3f714:   2c016740 04421510 03f3f800 04421400
0x03f3f724:   00000000 03f3fb2c 6d9ee000 03f3f768
0x03f3f734:   6d617feb 04421400 00000000 00000000
0x03f3f744:   00000010 03f3f758 04421400 2749dfd0
0x03f3f754:   2749dfd8 01b89518 01baee28 01b89510
0x03f3f764:   01b89518 03f3f7a8 6d6325a0 00000000
0x03f3f774:   03f3f800 000061ab 03f3f788 03f3f7a4
0x03f3f784:   00000000 00000006 00000008 04421400 

Instructions: (pc=0x6d9011fd)
0x6d9011ed:   8d 4d f0 e8 db d7 07 00 8b 75 10 85 f6 8b 45 0c
0x6d9011fd:   8b 10 7c 50 8b 45 14 85 c0 7c 49 8b 7a 08 8d 0c 


Stack: [0x03ef0000,0x03f40000],  sp=0x03f3f714,  free space=317k
Native frames: (J=compiled Java code, j=interpreted, Vv=VM code, C=native code)
V  [jvm.dll+0x1011fd]
C  [net.dll+0x7feb]
C  [nio.dll+0x25a0]
j  sun.nio.ch.Net.connect(Ljava/io/FileDescriptor;Ljava/net/InetAddress;II)I+0
j  sun.nio.ch.SocketChannelImpl.connect(Ljava/net/SocketAddress;)Z+162
j  java.nio.channels.SocketChannel.open(Ljava/net/SocketAddress;)Ljava/nio/channels/SocketChannel;+6
j  PeerClientConnection.connect(Ljava/net/InetSocketAddress;)V+2
j  PeerClientConnection$1.run()V+8
j  java.util.concurrent.Executors$RunnableAdapter.call()Ljava/lang/Object;+4
j  java.util.concurrent.FutureTask$Sync.innerRun()V+30
j  java.util.concurrent.FutureTask.run()V+4
j  java.util.concurrent.ThreadPoolExecutor$Worker.runTask(Ljava/lang/Runnable;)V+59
j  java.util.concurrent.ThreadPoolExecutor$Worker.run()V+28
j  java.lang.Thread.run()V+11
v  ~StubRoutines::call_stub
V  [jvm.dll+0xecf9c]
V  [jvm.dll+0x1741e1]
V  [jvm.dll+0xed167]
V  [jvm.dll+0xed1dd]
V  [jvm.dll+0x116290]
V  [jvm.dll+0x1d0424]
V  [jvm.dll+0x173e5c]
C  [msvcr71.dll+0x9565]
C  [kernel32.dll+0x4d0e9]
C  [ntdll.dll+0x419bb]
C  [ntdll.dll+0x4198e]

Answer 1

想想我找到了解决方法。量变到质变：

if (responder){
  InetSocketAddress addr = new InetSocketAddress(sc.socket().getInetAddress(), remotePort);
  SocketChannel.open(addr);
}

为：

if (responder){
  InetSocketAddress addr = new InetSocketAddress(sc.socket().getInetAddress().getHostAddress(), remotePort);
  SocketChannel.open(addr);
}

因此强迫它将ip作为一个字符串并根据它构建一个新的InetSocketAddress，似乎可以解决问题。对于为什么有任何建议（即，我首先做错了什么）？

JVM在打开“返回”SocketChannel时崩溃

1 个答案: