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(); } } }
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