信号量公平性参数不遵循先进先出

Question

我正在开发一个简单的信号量程序，其中我正在初始化一个计数为4的信号量，并启动6个线程。在run方法中，我获取信号量锁，在每个线程完成后，我释放锁。

这是我的代码：

import java.util.concurrent.Semaphore;

public class SemaphoreTest {
    
    static Semaphore semaphore = new Semaphore(4, true);
    
    static class MyThread extends Thread{
        
        String  name = "";
        
        public MyThread(String name){
            this.name = name;
            
        }
        
        public void run(){
            
            System.out.println(name+" going to acquire lock...");
            System.out.println("Available Permits = "+semaphore.availablePermits());
            
            try {
                semaphore.acquire();
                System.out.println(name+" got permit.");
                
                try{
                    for(int i=1;i<=1;i++){
                        System.out.println(name+" is performing operation "+i+". Available Semaphore permits are : "+semaphore.availablePermits());
                        Thread.sleep(1000);
                    }
                        
                }finally{
                    System.out.println(name+" Releasing lock...");
                    semaphore.release();
                    System.out.println("Available permits after releasing "+"name"+" = "+semaphore.availablePermits());
                }
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            
            
        }
    }
    
    public static void main(String[] args){
        Thread t1 = new MyThread("A");
        t1.start();
        
        Thread t2 = new MyThread("B");
        t2.start();
        
        Thread t3 = new MyThread("C");
        t3.start();
        
        Thread t4 = new MyThread("D");
        t4.start();
        
        Thread t5 = new MyThread("E");
        t5.start();
        
        Thread t6 = new MyThread("F");
        t6.start();
    }

}

结果如下：

A going to acquire lock...
Available Permits = 4
C going to acquire lock...
A got permit.
A is performing operation 1. Available Semaphore permits are : 3
B going to acquire lock...
Available Permits = 3
B got permit.
F going to acquire lock...
E going to acquire lock...
Available Permits = 2
Available Permits = 3
D going to acquire lock...
Available Permits = 0
C got permit.
C is performing operation 1. Available Semaphore permits are : 0
E got permit.
E is performing operation 1. Available Semaphore permits are : 0
Available Permits = 2
B is performing operation 1. Available Semaphore permits are : 2
A Releasing lock...
E Releasing lock...
Available permits after releasing name = 2
D got permit.
D is performing operation 1. Available Semaphore permits are : 1
B Releasing lock...
C Releasing lock...
Available permits after releasing name = 1
F got permit.
F is performing operation 1. Available Semaphore permits are : 2
Available permits after releasing name = 2
Available permits after releasing name = 2
D Releasing lock...
F Releasing lock...
Available permits after releasing name = 3
Available permits after releasing name = 4

现在，从java文档开始：

java.util.concurrent.信号量。信号量（int许可，布尔公平）

使用给定的许可数量和给定的公平性设置创建信号量。

参数：
允许可用许可的初始数量。此值可能为负值，在这种情况下，必须在授予任何获取之前发布。
如果此信号量将保证在争用情况下先进先出地授予许可，则为true，否则为false

构造函数信号量（int-allows，boolean-fair）保证先进先出。但就这个程序的输出而言，它是不一样的。锁的获取如下：

A-

锁按如下方式释放：

A-

请建议是否符合预期？还是我遗漏了什么？




Answer 1

这里有一个关于线程计时的误解：假设线程一输出消息就会获得锁，但事实上，没有理由不在这两者之间暂停线程。

C: System.out.println(...);
C: thread gets put on hold
A: System.out.println(...);
A: aquires lock
B: System.out.println(...);
B: aquires lock
C: resumes
C: aquires lock

Answer 2

匿名用户

释放许可证的顺序只是run方法花费的时间的结果，与公平无关。

这里的FIFO意味着如果两个线程调用信号量。acquire（）并且没有可用的许可证，当有许可证可用时，首先调用它的线程将是第一个接收许可证的线程。

在您的示例中，A、B、C、E获取许可证是因为它们首先调用acquire，而D和F必须等待许可证可用。然后，似乎D在F之前调用了acquire，因此获得了第一个可用的许可证。