Java-为什么这个程序不会抛出并发的修改异常 [英] Java- Why this program not throwing concurrent Modification exception

问题描述

我试图通过访问HashMap实例变量引用来诱发并发修改异常，但是这个程序不会抛出错误。非常感谢您能帮助我理解。

I am trying to induce a concurrent modification exception by accessing a HashMap instance variable reference, but this program is not throwing the error. Appreciate if you could help me to understand.

package Threads;
import java.util.HashMap;

public class ProducerConsumer {

    private HashMap<String, String> sharedMap = new HashMap<String,String>();

    public void putMethod(){

        for (int count=0; count<100; count++)
        {
        System.out.println("Adding into sharedMap:"+count);
        sharedMap.put(Integer.toString(count),Integer.toString(count));
        }
    }

    public static void main(String [] args) throws InterruptedException{

    final ProducerConsumer pc1=new ProducerConsumer();
    Thread t1= new Thread( new Runnable(){

        @Override
        public void run() {
            pc1.putMethod();
        }

    });

    Thread t2= new Thread( new Runnable(){

        @Override
        public void run() {
            pc1.putMethod();
            }   
    });
    t1.start();
    t2.start();
    t1.join();
    t2.join();
    }
}

推荐答案

add（）方法不检测并发修改，因此不会抛出 ConcurrentModificationException （这是匿名已经说的）。

The add() method does not detect concurrent modifications and therefore will not throw a ConcurrentModificationException (that's what anonymous already said).

但是，并发访问HashMap可能很危险。 请在另一篇文章中阅读更多信息。

However, concurrent access to a HashMap can be dangerous, though. Read more about this in another post.

如果您从 HashMap 中并行读取，您可以强制执行 ConcurrentModificationException ：

You can enforce a ConcurrentModificationException if you read from the HashMap in parallel:

...
public void putMethod() { ... }
public void iterateMethod() {
    sharedMap.keySet().stream().forEach((s) -> {
        System.out.println("Read key " + s);
    }
}

public static void main(String[] args) throws InterruptedException {
    ...
    t1.start();
    Thread.sleep(20); // sleep time depends on your computer's speed ;-)
    t2.start();
    ...
}
...

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