跨線程同步

Question

以下代碼應確保在所有線程之間同步對“ sync”的訪問。

根據輸出，情況並非總是如此，請注意Thread-3和Thread-4如何讀取相同的sync值。

我在代碼中缺少什么嗎？

[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

這里的代碼：

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

您正在不斷更改應該在其上同步所有線程的sync對象。 因此，實際上，它們根本不同步。 將您的同步變量定為最終變量，因為每個鎖都應該如此，您將看到代碼不再編譯。

解決方案：在另一個最終對象上同步，或使用AtomicInteger更改其值，或在this對象上同步（即，使方法同步）。

Answer 2

Integer是不可變的類，在執行sync ++時，您正在為Sync分配一個新引用，而其他線程可能會將該引用保存為一個較早的sync，因此會出現多線程問題。 嘗試定義一個簡單的像INTEGER這樣的MUTEX：

private final Integer MUTEX = new Integer(1);

並使用它而不是同步。

Answer 3

您應該在一個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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