跨線程同步

Question

以下代碼應確保跨所有線程同步對「同步」的訪問。

根據a輸出，並非總是如此，請注意Thread-3和Thread-4如何讀取相同的同步值。

我在代碼中丟失了什麼嗎？

[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 

這裏的代碼：

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

你不斷改變其上的所有線程都應該是同步的同步對象。所以，實際上，它們根本不同步。讓每個鎖都應該是最後的同步變量，然後你會看到你的代碼不再編譯。

解決方案：在另一個最終對象上同步，或者使用AtomicInteger並更改其值，或者在this上同步（即使該方法同步）。

Answer 2

Integer是不可變的類，當您在執行synC++時，您正在爲Sync分配一個新的引用，並且您的其他線程可能會持有對較舊同步的引用，因此會引發多線程問題。嘗試定義一個說INTEGER的簡單MUTEX，如：

private final Integer MUTEX = new Integer(1); 

並用它代替同步同步。

Answer 3

你應該上Object

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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