9

Ok so what I am trying to do here is to have a method "running" a process for a given amount of "time" this all seams to work to a degree but it keeps giveing these eceptions. This is the first execption it gives

     Exception in thread "main" java.util.ConcurrentModificationException

and then later on in the exicutio it gives this

    at java.util.LinkedList$ListItr.checkForComodification(LinkedList.java:761)
at java.util.LinkedList$ListItr.next(LinkedList.java:696)
at parta.PartA.runQueueOne(PartA.java:273)

I dont know what I am doing wrong here am i supposed make this concurrent or something? If so how? I thought that linked list were synchronous by nature? maybe that where im getting this messed up.

Well any way here is the method that im usings:

public static void runQueueOne(LinkedList<MyProcess> q1, LinkedList<MyProcess> q2, LinkedList<MyProcess> q3, LinkedList<MyProcess> q4, int ct) {
    System.out.println("Running Level One Queue");


    for (MyProcess p : q1) {
        if (p.name.equalsIgnoreCase(q1.getFirst().name)) {
            //add 3 millsedonds to the service time
            q1.getFirst().serviceTimeTotal += 3;
            System.out.println(q1.getFirst().name + " is running");

        } else {
            //add 3 millseconds to wait time fr the un busy one
            p.waitTimeTotal += 3;
        }
    }

    for (MyProcess p : q2) {
        p.waitTimeTotal += 3;
    }
    for (MyProcess p : q3) {
        p.waitTimeTotal += 3;
    }
    for (MyProcess p : q4) {
        p.waitTimeTotal += 3;
    }

    //calculate all the priority
    for (MyProcess p : q1) {
        p.calculatePriority();
        switch (p.priority) {
            case 1:
                break;
            case 2:
                q1.remove(p);
                q2.add(p);
                break;
            case 3:
                q1.remove(p);
                q3.add(p);
                break;
            case 4:
                q1.remove(p);
                q4.add(p);
                break;
        }

    }
    ct += 3;
}

and here is where I call it in the main method

while (!allProcessDone) {
    //arrival queue
    for (MyProcess p : al) {
        addToQueue(qOne, p, currentTime);

        //cheack to see if all the processes are done
        if (p1.isDone == true &
                p2.isDone == true &
                p3.isDone == true &
                p4.isDone == true &
                p5.isDone == true &
                p6.isDone == true &
                p7.isDone == true &
                p8.isDone == true &
                p9.isDone == true &
                p10.isDone == true) {
            //end the loop
            allProcessDone = true;
            System.out.println("All proccess have been completed");
            break;
        }


        switch (robin) {
            case 1:
                runQueueOne(qOne, qTwo, qThree, qFour, currentTime);
                robin = 2;
                break;
            case 2:
                runQueueTwo(qOne, qTwo, qThree, qFour, currentTime);
                robin = 3;
                break;
            case 3:
                runQueueThree(qOne, qTwo, qThree, qFour, currentTime);
                robin = 4;
                break;
            case 4:
                runQueueFour(qOne, qTwo, qThree, qFour, currentTime);
                robin = 1;
                break;

        }
    }
4

2 に答える 2

13

コレクションに同時にアクセスして変更していますが、これは for-each ループから直接行うことはできません。

Iteratorこの問題を解決するには、 を使用する必要があります。

LinkedList<MyProcess> q1 = new LinkedList<MyProcess>();

Iterator<MyProcess> iterator = q1.iterator();

while (iterator.hasNext()){
     MyProcess mp = iterator.next();

     if (mp.name.equals("xyz")){
         iterator.remove();    // You can do the modification here.
     }
 }
于 2012-10-22T14:37:54.830 に答える
3

ConcurrentModificationExceptionは、forループを使用してリストを反復処理しているときに、リストから要素を削除しようとすると発生します。

私はあなたのエラーがこれらの行から来ていると推測しています：

for (MyProcess p : q1) {
    p.calculatePriority();

    switch (p.priority) {
        case 1:
            break;
        case 2:
            q1.remove(p);
            q2.add(p);
            break;
        case 3:
            q1.remove(p);
            q3.add(p);
            break;
        case 4:
            q1.remove(p);
            q4.add(p);
            break;
    }
}

エラーを修正するには、iterator.remove（）メソッドを使用します

于 2012-10-22T14:33:17.043 に答える