跨线程同步

Question

以下代码应确保在所有线程之间同步对“ sync”的访问。

根据输出，情况并非总是如此，请注意Thread-3和Thread-4如何读取相同的sync值。

我在代码中缺少什么吗？

[Thread-0] before value of sync is 0
[Thread-0] after value of sync is 1
[Thread-3] before value of sync is 1
[Thread-3] after value of sync is 2
[Thread-4] before value of sync is 1
[Thread-4] after value of sync is 3
[Thread-2] before value of sync is 3
[Thread-2] after value of sync is 4
[Thread-1] before value of sync is 4
[Thread-1] after value of sync is 5

这里的代码：

package com.mypackage.sync;

public class LocalSync implements Runnable {

    private Integer sync = 0;

    public void someMethod() {
        synchronized (sync) {
            System.out.println("[" + Thread.currentThread().getName() + "]" + " before value of sync is " + sync);
            sync++;
            System.out.println("[" + Thread.currentThread().getName() + "]" + " after value of sync is " + sync);
        }
    }

    @Override
    public void run() {
        someMethod();
    }

    public static void main(String[] args) {

        LocalSync localSync = new LocalSync();
        Thread[] threads = new Thread[5];
        for (int i = 0; i < threads.length; i++) {
            threads[i] = new Thread(localSync, "Thread-" + i);
            threads[i].start();
        }

    }
}

Answer 1

您正在不断更改应该在其上同步所有线程的sync对象。 因此，实际上，它们根本不同步。 将您的同步变量定为最终变量，因为每个锁都应该如此，您将看到代码不再编译。

解决方案：在另一个最终对象上同步，或使用AtomicInteger更改其值，或在this对象上同步（即，使方法同步）。

Answer 2

Integer是不可变的类，在执行sync ++时，您正在为Sync分配一个新引用，而其他线程可能会将该引用保存为一个较早的sync，因此会出现多线程问题。 尝试定义一个简单的像INTEGER这样的MUTEX：

private final Integer MUTEX = new Integer(1);

并使用它而不是同步。

Answer 3

您应该在一个Object上同步

private Object synchObj = new Object();
private Integer sync = 0;

public void someMethod() {
    synchronized (synchObj) {
        System.out.println("[" + Thread.currentThread().getName() + "]" + " before value of sync is " + sync);
        sync++;
        System.out.println("[" + Thread.currentThread().getName() + "]" + " after value of sync is " + sync);
    }
}

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