Question

这个java应用程序将启动三个线程并倒数一个数字。一旦计数器达到零，线程将完成并退出。不幸的是，计数器将绕过零检查。怎么解决？

import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class ThreadTest {
    volatile Long totalMsg = 100L;
    final Lock maxLimitLock = new ReentrantLock();

    public void start() {
        int threadNumber = 3;
        ExecutorService executor = Executors.newFixedThreadPool(threadNumber);
        for (int i = 0; i < threadNumber; i++) {
            executor.submit(new ThreadTest.ConsumerThread(i));
        }      
    }
    class ConsumerThread implements Runnable {
        private int threadNumber;
        public ConsumerThread(int threadNumber) {
            this.threadNumber = threadNumber;
        }
        public void run() {
            while (true) {
                try {
                    maxLimitLock.lock();
                    totalMsg--;
                    System.out.println(totalMsg + " left");
                    if (totalMsg == 0) {
                        System.out.println("finished all of " + totalMsg);
                        break;
                    }
                } finally {
                    maxLimitLock.unlock();
                }
            }
            System.out.println("thread " + this.threadNumber + " finished");
        }
    }
    public static void main(String[] args) throws Exception {
        ThreadTest reader = new ThreadTest();
        reader.start();
    }
}

感谢。

Answer 1

代码中的问题在这里

totalMsg--;
...
if (totalMsg == 0) {

您的代码总是递减计数器，以便只能释放一个线程，后续线程将具有一个计数器＆lt; 0因此它们永远不会被释放，所以你应该用

替换它

if (totalMsg <= 0) {

或

if (totalMsg > 0) {
    totalMsg--;
    System.out.println(totalMsg + " left");
} else {
    System.out.println("finished all of " + totalMsg);
    break;
}

注意：考虑使用AtomicLong作为您的计数器。

Java线程倒计时数字

1 个答案: