令人困惑的结果测试两个线程使用同步方法递增单个 int

Question

我正在尝试弄清楚多个线程如何对单个数据进行更改。我最近了解了 synchronized 关键字的作用，但某个测试仍然给我带来混乱的结果。

完整代码如下：

public class TestSyncTest{
    public static void main(String[] args){
        TestSync job = new TestSync();
        
        Thread a = new Thread(job);
        Thread b = new Thread(job);
        
        a.start();
        b.start();
    }
}

class TestSync implements Runnable{
    private int balance;
    
    public void run(){
        for(int i = 0; i < 50; i++){
            increment();
            System.out.println("balance is " + balance);
        }
    }
    
    public synchronized void increment(){
        int i = balance;
        
        try{
            Thread.sleep(10);
        } catch(InterruptedException ex){
            ex.printStackTrace();
        }
        
        balance = i + 1;
    }
}

主要是为了演示如何让JVM用一个线程完成对单条数据的一系列操作，而不需要其他线程参与和破坏。 increment() 是 synchronized 使该方法成为一个线程一次处理的单个块。 sleep() 只是为了模拟大量需要时间的代码，此时其他线程可能会跳入。当这段代码运行时，sleep() 是否存在并不重要，因为整个方法是 synchronized。所以输出看起来像这样漂亮、干净和正确：

>java TestSyncTest
balance is 1
balance is 2
balance is 3
balance is 4
balance is 5
balance is 6
. . .
balance is 98
balance is 99
balance is 100

balance 完全符合预期。当 synchronized 被删除时，这是输出：

>java TestSyncTest
balance is 1
balance is 1
balance is 2
balance is 2
balance is 3
balance is 4
balance is 4
balance is 5
balance is 5
balance is 6
balance is 6
. . .
balance is 48
balance is 48
balance is 49
balance is 49
balance is 50
balance is 50
balance is 51

这也符合预期，因为另一个线程不断跳入当前线程的敏感增量过程的中间，导致一致错误。我运行遇到的问题是 increment() 方法如下所示：

public synchronized void increment(){
    int i = balance;
    balance = i + 1;
}

这里的方法是synchronized，但是里面的第二个语句或多或少会立即跟在第一个后面，这是一个完整输出的例子：

>java TestSyncTest
balance is 2
balance is 3
balance is 1
balance is 5
balance is 6
balance is 7
balance is 8
balance is 9
balance is 10
balance is 11
balance is 12
balance is 13
balance is 14
balance is 15
balance is 16
balance is 17
balance is 18
balance is 19
balance is 20
balance is 21
balance is 22
balance is 23
balance is 24
balance is 25
balance is 26
balance is 27
balance is 28
balance is 29
balance is 30
balance is 31
balance is 32
balance is 33
balance is 34
balance is 35
balance is 36
balance is 37
balance is 38
balance is 39
balance is 40
balance is 41
balance is 42
balance is 43
balance is 44
balance is 45
balance is 46
balance is 47
balance is 48
balance is 49
balance is 50
balance is 51
balance is 52
balance is 53
balance is 4
balance is 54
balance is 55
balance is 56
balance is 57
balance is 58
balance is 59
balance is 60
balance is 61
balance is 62
balance is 63
balance is 64
balance is 65
balance is 66
balance is 67
balance is 68
balance is 69
balance is 70
balance is 71
balance is 72
balance is 73
balance is 74
balance is 75
balance is 76
balance is 77
balance is 78
balance is 79
balance is 80
balance is 81
balance is 82
balance is 83
balance is 84
balance is 85
balance is 86
balance is 87
balance is 88
balance is 89
balance is 90
balance is 91
balance is 92
balance is 93
balance is 94
balance is 95
balance is 96
balance is 97
balance is 98
balance is 99
balance is 100

它以 2 开头，1 出现在 4 的位置后面，4 出现在 53 和 54 之间。其中一些数字总是会混淆，但最终结果总是 100，它本身并不总是输出中的最后一个。

请帮助我了解这里发生了什么。

Answer 1

即使方法 increment() 是 synchronized 你仍然有一个 race-condition on:

public void run(){
    for(int i = 0; i < 50; i++){
        increment();
        System.out.println("balance is " + balance); // <-- here 
    }
}

在读取变量balance期间System.out.println。一个线程可能正在修改方法 increment() 中的该变量，同时另一个线程正在读取该变量。

尽管如此，race-condition 并未反映在您的输出中，因为 System.out.println 方法内部 synchronizes.

在您的第一个代码片段中，由于 sleep 调用，线程按顺序输出结果。当您删除该调用时，无法保证 System.out.println 产生的输出将被排序。

令人困惑的结果测试两个线程使用同步方法递增单个 int

confusing results testing two threads incrementing a single int with synchronized method

java

parallel-processing

multithreading

synchronization

thread-safety