Java：多线程通过同步方法同时访问变量来递减静态变量

Question

public class Person extends Thread {

    private static int count = 10;

    public void decrement() {
        synchronized(Person.class) {
            count--;
        }

    }

    public int getCount() {
        return count;
    }

    public void run(){
        while( count > 0){
            this.decrement();
            System.out.print(this.getCount() + ",");
        }
    }
}

public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {

        Person p1 = new Person();
        Person p2 = new Person();
        Person p3 = new Person();
        Person p4 = new Person();
        Person p5 = new Person();

        p1.start();
        p2.start();
        p3.start();
        p4.start();
        p5.start();
    }
}

8,8,7,6,5,4,3,2,1,0

8,7,5,3,1,0,6,8,4,0

Answer 1

原始代码中的内容如下：

1. 线程1减少计数（计数->9)
2. 线程两次递减计数（计数->8)
3. 线程三次递减计数（计数->7)
4. 线程四次递减计数（计数->6)
5. 线程1输出计数(count:6)
6. 线程两个输出计数（计数：6）
7. 线程三个输出计数（计数：6）
8. 线程四个输出计数（计数：6）

因为您锁定的是减量，而不是减量和输出在一起，所以它看起来是多次减量。

        this.decrement();
        System.out.print(this.getCount() + ",");

public class Person extends Thread {

    private static int count = 10;

    public int decrement() {
        synchronized(Person.class) {
            count = count - 1;
            return count;
        }

    }

    public int getCount() {
        synchronized(Person.class) {
            return count;
        }
    }

    public void run(){
        while( getCount() > 0){
            int count = this.decrement();
            System.out.println(count);
        }
    }
}

import java.util.concurrent.atomic.AtomicInteger;

public class Person extends Thread {

    private static AtomicInteger count = new AtomicInteger(10);

    public int decrement() {
        return count.decrementAndGet();
    }

    public void run(){
        while(count.get() > 0){
            int currentCount = this.decrement();
            System.out.print(currentCount + ",");
        }
    }
}