java - 信号量不起作用

Question

我正在尝试创建一个小程序来演示信号量的使用。我创建了 2 个线程，运行 Farmer 的两个实例:一个使用字符串“north”作为参数，另一个使用“south”作为参数。它们似乎同时完成，而不是先完成 1 个线程，然后再完成第 2 个线程(如输出所示:

农夫过桥，向北走
农夫过桥，向南
农夫已经过了桥，正在向北行驶
农夫已过桥，正在向南行驶

谁能告诉我我在这里做错了什么？

import java.util.concurrent.Semaphore;
public class Farmer implements Runnable
{
    private String heading;
    private final Semaphore bridge = new Semaphore(1);
    public Farmer(String heading)
    {
        this.heading = heading;
    }

    public void run() 
    {
        if (heading == "north")
        {
            try 
            {
                //Check if the bridge is empty
                bridge.acquire();
                System.out.println("Farmer going over the bridge, heading north");
                Thread.sleep(1000);
                System.out.println("Farmer has crossed the bridge and is now heading north");
                bridge.release();
            } 
            catch (InterruptedException e) 
            {
                e.printStackTrace();
            }
            //Farmer crossed the bridge and "releases" it

        }
        else if (heading == "south")
        {
            try 
            {
                //Check if the bridge is empty
                bridge.acquire();
                System.out.println("Farmer going over the bridge, heading south");
                Thread.sleep(1000);
                //Farmer crossed the bridge and "releases" it
                System.out.println("Farmer has crossed the bridge and is now heading south");
                bridge.release();
            } 
            catch (InterruptedException e) 
            {
                e.printStackTrace();
            }
        }
    }
}

Answer 1

每个 Farmer 都在创建自己的信号量，这意味着它可以独立于其他人获取和释放信号量。

更改此:

private final Semaphore bridge = new Semaphore(1);

对此:

private final static Semaphore bridge = new Semaphore(1);

然后信号量将在所有 Farmer 实例之间共享。