[PHP知识库]多个线程同步执行，多个线程依次执行，多个线程交错执行

1. 同步执行使用CountDownLatch作为计数工具;

2. 依次执行使用volatile关键字保证可见性；

3. 交错执行使用Semaphore作为信号量控制；

同步执行案例如下图：

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.Semaphore;

public class ThreadSafeDemo {

    public static void main(String[] args) {
        CountDownLatch countDownLatch = new CountDownLatch(1);
        //生成3个线程打印时间
        for(int i =0;i<3;i++){
            new Thread(()->{
                try{
                    countDownLatch.await();
                    System.out.println(System.currentTimeMillis());
                    Thread.sleep(100);
                }catch(Exception e){
                    e.printStackTrace();
                }
            }).start();
        }

        System.out.println(System.currentTimeMillis());
        countDownLatch.countDown();
    }


}

依次执行代码案例如下图：

public class OrderThreadDemo {

    static volatile int ticket = 1;

    public static void main(String[] args) {

        Thread t1 = new Thread(()->{
            while(true){
                if(ticket == 1){
                    try{
                        Thread.sleep(100);
                        for(int i =0;i<10;i++){
                            System.out.println("A"+i);
                        }
                    }catch (Exception e){
                        e.printStackTrace();
                    }
                    ticket = 2;
                    return;
                }
            }

        });

        Thread t2 = new Thread(()->{
            while(true){
                if(ticket == 2){
                    try{
                        Thread.sleep(100);
                        for(int i =0;i<10;i++){
                            System.out.println("B"+i);
                        }
                    }catch (Exception e){
                        e.printStackTrace();
                    }
                    ticket = 3;
                    return;
                }
            }

        });

        Thread t3 = new Thread(()->{
            while(true){
                if(ticket == 3){
                    try{
                        Thread.sleep(100);
                        for(int i =0;i<10;i++){
                            System.out.println("C"+i);
                        }
                    }catch (Exception e){
                        e.printStackTrace();
                    }
                    ticket = 1;
                    return;
                }
            }

        });
        t1.start();
        t2.start();
        t3.start();
    }
}

交错执行代码案例如下图：

import java.util.concurrent.Semaphore;

public class OrderThreadSempDemo {

    private static Semaphore s1 = new Semaphore(1);
    private static Semaphore s2 = new Semaphore(1);
    private static Semaphore s3 = new Semaphore(1);

    public static void main(String[] args) {

        try{
            //获取许可
            s1.acquire();
            s2.acquire();
        }catch(Exception e){
            e.printStackTrace();
        }

        new Thread(()->{
            while(true){
                try{
                    s1.acquire();
                }catch (Exception e){
                    e.printStackTrace();
                }
                try{
                    Thread.sleep(500);
                }catch (Exception e){
                    e.printStackTrace();
                }
                System.out.println("A");
                s2.release();
            }
        }).start();

        new Thread(()->{
            while(true){
                try{
                    s2.acquire();
                }catch (Exception e){
                    e.printStackTrace();
                }
                try{
                    Thread.sleep(500);
                }catch (Exception e){
                    e.printStackTrace();
                }
                System.out.println("B");
                s3.release();
            }
        }).start();

        new Thread(()->{
            while(true){
                try{
                    s3.acquire();
                }catch (Exception e){
                    e.printStackTrace();
                }
                try{
                    Thread.sleep(500);
                }catch (Exception e){
                    e.printStackTrace();
                }
                System.out.println("C");
                s1.release();
            }
        }).start();
    }
}