Question

当我调用send（）方法时，我收到了ConcurrentModificationException。我只访问数据，没有修改，所以我不确定问题是什么。有什么建议吗？

public class chatServer{

public static LinkedList<serverListener> threadList = new LinkedList<serverListener>();
public static ListIterator<serverListener> li = threadList.listIterator();

public static void main (String [] args)throws IOException{
    ServerSocket incoming = new ServerSocket(58667); //create new ServerSocket
    Socket servSock;
    BufferedReader buffRead;
    while(true){
        servSock = incoming.accept(); //accept incoming connections from clients
        serverListener c1 = new serverListener(servSock); //create a new chat listener object, passes in servSock socket
        c1.start();
        threadList.addLast(c1);
    }//while(running)
    //incoming.close();
}//main

public static synchronized void send(String m) throws IOException{
    while(li.hasNext()){
        li.next().getOS().writeBytes(m);
    }//while
}//send

public static synchronized void removeClient(DataOutputStream d){
    while(li.hasNext()){
        if (li.next().getOS()==d){
            li.remove();
        }//if
    }//while
}//removeClient
}//chatServer

public class serverListener extends Thread{
private Socket sock; //socket used to communicate (set from passed in value)
private static String incoming; //string to store incoming message
private BufferedReader dataIn; //BufferedReader to read input message
private static DataOutputStream dataOut; //data stream to send message

//chatListener constructor, accepts socket as parameter 
public serverListener(Socket s) throws IOException{
    sock = s; //sets sock to parameter value s
    dataIn = new BufferedReader(new InputStreamReader(sock.getInputStream())); //creates input stream reader from socket sock
    dataOut = new DataOutputStream(sock.getOutputStream()); //creates a new output stream from sock
}//listen constructor

//server listener thread, receives messages from connected client, runs until EXITEXIT is received
public void run(){
    do{
        try{
            setMessage(dataIn.readLine());} //read incoming message, store value in incoming string
        catch(IOException e){ //catches error where a message is unable to be read
        }

        if(getMessage().equals("EXITEXIT")){ //if incoming message is EXITEXIT, set running to false, will not print value
            chatServer.removeClient(dataOut);
            try {
                chatServer.send("Client has left");
            } catch (IOException e) {
            }
        } else
            try {
                chatServer.send(getMessage());
            } catch (IOException e) {
            } //send message to synchronized print method
    }while(true); //loop until isRunning is false
}//run


public static DataOutputStream getOS(){
    return dataOut;
}

private static void setMessage(String m){
    incoming = m;
}
private static String getMessage(){
    return incoming;
}
}//public chatListener

在客户端发送消息并调用send方法之前，一切正常。

更新我最终使用for循环：

public static synchronized void send(String m, DataOutputStream d) throws Exception{
    for (int i=0; i<threadList.size(); i++){
        try{
            if(threadList.get(i).getOS()!=d)
                threadList.get(i).getOS().writeBytes(m+'\n');
        }catch(Exception e){}
    }//for
}//send

Answer 1

简而言之，因为你修改了一个列表而没有使用它的迭代器。

在这里，您创建一个列表并获取其迭代器：

public static LinkedList<serverListener> threadList = new LinkedList<serverListener>();
public static ListIterator<serverListener> li = threadList.listIterator();

在这里，您在结构上修改了列表而不使用其迭代器：

while(true){
    servSock = incoming.accept(); //accept incoming connections from clients
    serverListener c1 = new serverListener(servSock); //create a new chat listener object, passes in servSock socket
    c1.start();
    threadList.addLast(c1);
}//while(running)

因此，下次您致电li.next()时，您将获得ConcurrentModificationException

来自LinkedList javadoc（强调我的）：

此类的iterator和listIterator方法返回的迭代器是 fail-fast ：如果在创建迭代器后的任何时候对列表进行结构修改，无论如何除非通过Iterator自己的remove或add方法，迭代器将抛出ConcurrentModificationException 。因此，面对并发修改，迭代器会快速而干净地失败，而不是在未来不确定的时间冒着任意的，非确定性行为的风险。

获取ConcurrentModificationException聊天程序

1 个答案: