java 多线程:线程安全问题synchronized关键字解决

背景:

多个线程同时修改一个变量时,有概率导致两次修改其中某些次被覆盖。

例如:如下案例一个变量值为3,三个线程同时对其-1,如果按顺序执行,每次减完的结果应该是2,1,0。但实际运行中有可能变为0,0,0 ;0 1 1 等情况

/**
 * @ClassName VarNotSafe
 * @projectName: object1
 * @author: Zhangmingda
 * @description: XXX
 * date: 2021/4/20.
 */
public class VarNotSafe {
    public static int num = 3;

    public static void main(String[] args) {
        Runnable r = () -> {
            num--;
            System.out.println(Thread.currentThread().getName() + "结果==>:" + num);
        };
        Thread t1 = new Thread(r,"t1");
        Thread t2 = new Thread(r,"t2");
        Thread t3 = new Thread(r,"t3");
        t1.start();
        t2.start();
        t3.start();
    }
}

 

000实例原因:三个线程分别都做了num--环节就把执行权让出给下一个线程了。结果三个线程都把前半部分做完才执行System.out.println。导致000

变量被同时使用测试代码2:

例如数字10000, 10个线程每个线程循环1000次减去1,最终结果为零。但是并发执行覆盖可能导致未减为0

/**
 * @ClassName VarNotSafe2
 * @projectName: object1
 * @author: Zhangmingda
 * @description: XXX
 * date: 2021/4/20.
 */
public class VarNotSafe2 {
    private static int num = 10000;
    private static class MinusNumRannable implements Runnable{
        @Override
        public void run() {
            for(int i=0; i<1000; i++){
                num--;
            }
            System.out.println(Thread.currentThread().getName() + "计算结果:" + num);
        }
    }

    public static void main(String[] args) {
        Runnable runnable = new MinusNumRannable();
        for(int i=0; i<10; i++){
            new Thread(runnable).start();
        }
        try {
            Thread.sleep(3000);
            System.out.println("最终结果:" + num);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}
VarNotSafe2

  

 

 

synchronized关键字解决

/**
 * @ClassName VarNotSafe
 * @projectName: object1
 * @author: Zhangmingda
 * @description: XXX
 * date: 2021/4/20.
 */
public class VarSafeSynchronized {
    public static int num = 3;

    public static class SafeThread implements Runnable {

        @Override
        public synchronized void run() {
            num--;
            System.out.println(Thread.currentThread().getName() + "结果==>:" + num);
        }
    }
    public static void main(String[] args) {
        Runnable safeR = new SafeThread();
        Thread t1 = new Thread(safeR,"t1");
        Thread t2 = new Thread(safeR,"t2");
        Thread t3 = new Thread(safeR,"t3");
        t1.start();
        t2.start();
        t3.start();
    }
}

 

posted on 2021-04-20 11:26  zhangmingda  阅读(290)  评论(0编辑  收藏  举报

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