java - volatile中i++的原子性问题
怪我咯
怪我咯 2017-04-18 10:52:11
0
3
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/**
 * @create on 17/3/27
 * @description
 */
public class Main {
    static volatile int i = 0;
    public static class PlusTask implements Runnable{
        @Override
        public void run(){
            for(int k=0; k<10000; k++)
                i++;
        }
    }

    public static void main(String[] args) throws InterruptedException{
        Thread[] threads = new Thread[10];
        for(int i=0;i<10;i++){
            threads[i] = new Thread(new PlusTask());
            threads[i].start();
        }

        for(int i=0;i<10;i++){
            threads[i].join();
        }

        System.out.println(i);
    }
}

请教各位大牛 为什么这里的输出总是小于10000? 已经调用了thread.join

怪我咯
怪我咯

走同样的路,发现不同的人生

reply all(3)
Peter_Zhu

You may need this:

public void run() {
            for (int k = 0; k < 10000; k++)
                synchronized (Main.class) {
                    i++;
                }
        }

or define oneAtomicInteger

  • volatile only functions as a multi-thread cache consistency, and does not guarantee that only one thread will write variables at a certain time.

Ty80

volitile does not guarantee atomicity

刘奇

The problem of multi-threaded operation of shared variables. Volitale cannot guarantee the atomicity of composite operations, which means that the latter thread cannot always see the modified value i of the previous thread. The solution is to lock or atomically operate

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