如何找到我的 Java 信号量不起作用的原因

Question

我用线程和信号量编写了交通规则代码。

我认为我的代码没问题，但它不起作用。我的意思是它不会锁定我在第一个线程下写的第二个线程。我怎样才能找到我的错误？为什么这不锁定我的第二个线程？

我想看到这个结果： 我将有 6 辆车在交通中。 3 辆车在一条路线上（坐标不同），另外 3 辆车在另一条路线上（坐标不同）。比如A坐标是5，B是10，C是15。

相反，D 是-6，E 是-11，F 是-16。有路口。如果 A、B 和 C 车点正在减少（每两秒它们协调减少 1 减 1）。他们想要达到“upTarget”，例如 3。并且 D、E、F 坐标一一增加以达到 -3。如果 A 在 D 到达他的左目标（到 -3）之前到达上目标（到 3），所有左行将等到上行离开 0,0 坐标

我的意思是如果A达到3，那么A B C线将绿灯熄灭，D，E，F将等待它们直到C坐标为0

主类

package sample;

import java.util.Scanner;
import java.util.concurrent.Semaphore;

public class Main {
    public static int x1,x2,x3,y1,y2,y3,upTarget,leftTarget;

    public static void main(String[] args) throws InterruptedException{

        Scanner obj=new Scanner(System.in);

        System.out.println("Insert coordinant of A");
        x1=obj.nextInt();
        System.out.println("Insert coordinant of B");
        x2=obj.nextInt();
        System.out.println("Insert coordinant of C");
        x3=obj.nextInt();

        System.out.println("Insert coordinant of D");
        y1=obj.nextInt();
        System.out.println("Insert coordinant of E");
        y2=obj.nextInt();
        System.out.println("Insert coordinant of F");
        y3=obj.nextInt();

        System.out.println("Insert up target level");
        upTarget=obj.nextInt();

        System.out.println("Insert left target level");
        leftTarget=obj.nextInt();

        System.out.println("Traffic was start");
        Semaphore sem=new Semaphore(1);


        Thread t1=new MyBackend(sem,x1, x2, x3, upTarget, y1, y2, y3, leftTarget,"A");
        Thread t2=new MyBackend(sem,x1, x2, x3, upTarget, y1, y2, y3, leftTarget,"B");





        t1.start();
        t2.start();
        t1.join();
        t2.join();
    }


}

我的后端代码 => 我的线程类

   import java.util.concurrent.Semaphore;

public class MyBackend extends Thread{
     Semaphore sem;
    private int x1,x2,x3,y1,y2,y3,upTarget,leftTarget;
    private String value;

    public MyBackend(Semaphore sem,int x1,int x2,int x3,int upTarget,int y1,int y2,int y3,int leftTarget,String value){
        super(value);
        this.sem=sem;
        this.x1=x1;
        this.x2=x2;
        this.x3=x3;
        this.y1=y1;
        this.y2=y2;
        this.y3=y3;
        this.upTarget=upTarget;
        this.leftTarget=leftTarget;
        this.value=value;
    }

    @Override
    public void run() {
        System.out.println("yeeeeeeeeeeeeeeeeeeeeeeeeeee"+sem.availablePermits());
        if(this.getName().equals("A")){
            try{

                while(x3>=0){
                    x1=x1-1;
                    x2=x2-1;
                    x3=x3-1;
                    Thread.sleep(2000);
                    System.out.println("A coordinate "+x1);
                    System.out.println("B coordinate "+x2);
                    System.out.println("C coordinate "+x3);

                    if(x1==upTarget){
                        sem.acquire();
                        System.out.println("yeeeeeeeeeeeeeeeeeeeeeeeeeee"+sem.availablePermits());
                        System.out.println("First line was reached!!!");
                    }
                }

            }catch(Exception e){}
            sem.release();
        }
        else {
            try{
                while(y3<=0){
                    y1=y1+1;
                    y2=y2+1;
                    y3=y3+1;
                    Thread.sleep(2000);
                    System.out.println("D coordinate "+y1);
                    System.out.println("E coordinate "+y2);
                    System.out.println("F coordinate "+y3);

                    if(y3==leftTarget){
                        sem.acquire();
                        System.out.println("Second line was reached!!!");
                    }
                }
            }catch(Exception e){}
            sem.release();
        }    }
}

Answer 1

您的代码工作正常，信号量工作正常。请在代码中更改线程 B 的 while 循环以添加更多 system.out.println() 如下所示，并查看当线程 A 获取信号量时，线程 B 实际上已挂起（如预期的那样）。

                while(y3<=0){
                    y1=y1+1;
                    y2=y2+1;
                    y3=y3+1;
                    Thread.sleep(2000);
                    System.out.println("D coordinate "+y1);
                    System.out.println("E coordinate "+y2);
                    System.out.println("F coordinate "+y3);

                    if(y3==leftTarget){
                        System.out.println("Thread B acquiring sem: "+sem.availablePermits() + "If this is 0, D,E and F should hang");
                        sem.acquire();
                        System.out.println("Semaphore acquired, continue thread B for D, E and F"+sem.availablePermits());
                        System.out.println("Second line was reached!!!");
                    }
                }

如果您期待其他内容，请澄清。

更新

根据您的评论，我认为您可以添加另一个 if 条件以检查信号量是否已被其他线程获取，然后更改坐标如下；

先设置一个静态同步变量，显示哪个线程先获取信号量。

import java.util.concurrent.Semaphore;

    public class MyBackend extends Thread{
         Semaphore sem;
        private volatile String acquiredBy = null;
        private int x1,x2,x3,y1,y2,y3,upTarget,leftTarget;
        private String value;

        public MyBackend(Semaphore sem,int x1,int x2,int x3,int upTarget,int y1,int y2,int y3,int leftTarget,String value){
            super(value);
            this.sem=sem;
            this.x1=x1;
            this.x2=x2;
            this.x3=x3;
            this.y1=y1;
            this.y2=y2;
            this.y3=y3;
            this.upTarget=upTarget;
            this.leftTarget=leftTarget;
            this.value=value;
        }

        @Override
        public void run() {
            System.out.println("yeeeeeeeeeeeeeeeeeeeeeeeeeee"+sem.availablePermits());
            if(this.getName().equals("A")){
                try{

                    while(x3>=0){
                        if(acquiredBy != null && !acquiredBy.equals(this.getName())){
                              sem.acquire();
                        }
                        x1=x1-1;
                        x2=x2-1;
                        x3=x3-1;
                        Thread.sleep(2000);
                        System.out.println("A coordinate "+x1);
                        System.out.println("B coordinate "+x2);
                        System.out.println("C coordinate "+x3);

                        if(x1==upTarget){
                            sem.acquire();
                            acquiredBy = this.getName();

                             System.out.println("yeeeeeeeeeeeeeeeeeeeeeeeeeee"+sem.availablePermits());
                            System.out.println("First line was reached!!!");
                        }
                    }

                }catch(Exception e){}
                sem.release();
            }
            else {
                try{
                    while(y3<=0){
                        if(acquiredBy != null &&  !acquiredBy.equals(this.getName())){
                              sem.acquire();
                        }
                        y1=y1+1;
                        y2=y2+1;
                        y3=y3+1;
                        Thread.sleep(2000);
                        System.out.println("D coordinate "+y1);
                        System.out.println("E coordinate "+y2);
                        System.out.println("F coordinate "+y3);

                        if(y3==leftTarget){
                            sem.acquire();
                            acquiredBy = this.getName();
                            System.out.println("Second line was reached!!!");
                        }
                    }
                }catch(Exception e){}
                sem.release();
            }    }
    }

这应该可行！