不可变对象上的同步(在 java 中)

Synchronization on immutable objects (in java)(不可变对象上的同步(在 java 中))

本文介绍了不可变对象上的同步(在 java 中)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

Code snippet - 1

class RequestObject implements Runnable
{
    private static Integer nRequests = 0;

    @Override
    public void run()
    {       
        synchronized (nRequests)
        {
            nRequests++;
        }
    }
}

Code snippet - 2

public class Racer implements Runnable
{
    public static Boolean won = false;    

    @Override
    public void run()
    {
        synchronized (won)
        {
            if (!won)
            won = true;
        }
    }        
}

I was having a race condition with the first code snippet. I understood that this was because I was obtaining a lock on an immutable object(of type Integer).

I have written a second code snippet which is again impervious to 'Boolean' being immutable. But this works(no race condition is displayed in an output run). If I have understood the solution to my previous question properly the below is is one possible way in which things can go wrong

  1. Thread 1 gets a lock on an object(say A) pointed by won
  2. Thread 2 tries to now get a lock on the object pointed to by won and gets in the wait queue for A
  3. Thread 1 goes into the synchronized block, verifies that A is false and creates a new object(say B) by saying won = true(A thinks it won the race).
  4. 'won' now points to B. Thread 1 releases the lock on object A(no longer pointed to by won)
  5. Now, thread-2 which was in the wait queue of object A gets woken up and gets a lock on object A which is still false(immutably so). It now goes into the synchronized block and assumes that it has also won, which is not correct.

Why is the second code snippet working fine all the time??

解决方案

    synchronized (won)
    {
        if (!won)
        won = true;
    }

Here you have a transient race condition which you don't notice because it goes away after the first execution of the run method. After that the won variable constantly points to the same instance of Boolean representing true, which thus serves properly as a mutex lock.

This is not to say that you should ever write such code in real projects. All lock objects should be assigned to final variables to make sure they never change.

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本文标题为:不可变对象上的同步(在 java 中)

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