java测试两个线程使用同步方法增加单个int的结果令人困惑

Question

我试图弄清楚多个线程如何对单个数据段进行更改。我最近了解了synchronized关键字的作用，但某个测试仍然给出了令人困惑的结果

以下是完整的代码：

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;
    }
}

基本上，它是为了演示如何让JVM用一个线程在单个数据块上完成一系列操作，而不让其他线程参与进来并破坏事情increment()是synchronized使该方法成为一个线程一次处理的单个块sleep()只是为了模拟大量代码，这些代码需要时间，否则其他线程可能会跳入其中。当此代码运行时，sleep()是否存在并不重要，因为整个方法都是synchronized。因此，输出看起来很好、干净且正确，如下所示：

>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与预期的一样完美。删除synchronized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

这也是意料之中的，因为另一个线程不断跳入当前线程敏感增量过程的中间，导致一致性错误。我遇到的问题是increment()方法如下所示：

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

这里，方法是synchronized，但内部的第二条语句或多或少会立即跟随第一条语句，这是整个输出的一个示例：

>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

它从2开始，1在后面出现，而不是4，4在53到54之间出现。这些数字中总会有一些混淆，但最终结果总是100，这本身并不总是输出中的最后一个

请帮我了解这里发生了什么

Answer 1

即使方法increment()是synchronized，您仍然有一个竞争条件：

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

在读取变量期间，在^{中平衡。一个线程可能正在方法increment（）中修改该变量，而另一个线程正在读取该变量

尽管如此，竞争条件不会反映在输出上，因为System.out.println方法在内部synchronizes

在第一个代码段中，由于sleep调用，线程按顺序输出结果。删除该调用后，无法保证System.out.println生成的输出将被排序