Java多线程：tryLock()方法

tryLock(long time, TimeUnit unit) 的作用在给定等待时长内锁没有被另外的线程持有，并且当前线程也没有被中断,则获得该锁，通过该方法可以实现锁对象的限时等待。

package com.wkcto.lock.reentrant;

import java.util.concurrent.TimeUnit;
import java.util.concurrent.locks.ReentrantLock;

/**
 *tryLock(long time, TimeUnit unit) 的基本使用
 */
public class Test07 {
    static class TimeLock implements Runnable{
        private static ReentrantLock lock = new ReentrantLock();    //定义锁对象

        @Override
        public void run() {
            try {
                if ( lock.tryLock(3, TimeUnit.SECONDS) ){       //获得锁返回true
                    System.out.println(Thread.currentThread().getName() + "获得锁,执行耗时任务");
//                    Thread.sleep(4000);         //假设Thread-0线程先持有锁,完成任务需要4秒钟,Thread-1线程尝试获得锁,Thread-1线程在3秒内还没有获得锁的话,Thread-1线程会放弃
                    Thread.sleep(2000);          //假设Thread-0线程先持有锁,完成任务需要2秒钟,Thread-1线程尝试获得锁,Thread-1线程会一直尝试,在它约定尝试的3秒内可以获得锁对象
                }else {         //没有获得锁
                    System.out.println(Thread.currentThread().getName() + "没有获得锁");
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                if (lock.isHeldByCurrentThread()){
                    lock.unlock();
                }
            }
        }
    }

    public static void main(String[] args) {
        TimeLock timeLock = new TimeLock();

        Thread t1 = new Thread(timeLock);
        Thread t2 = new Thread(timeLock);
        t1.start();
        t2.start();
    }
}

tryLock()仅在调用时锁定未被其他线程持有的锁，如果调用方法时，锁对象对其他线程持有，则放弃，调用方法尝试获得没，如果该锁没有被其他线程占用则返回true表示锁定成功; 如果锁被其他线程占用则返回false,不等待。

package com.wkcto.lock.reentrant;

import java.util.concurrent.locks.ReentrantLock;

/**
 *tryLock()
 *  当锁对象没有被其他线程持有的情况下才会获得该锁定
 */
public class Test08 {
    static class Service{
        private ReentrantLock lock = new ReentrantLock();
        public void serviceMethod(){
            try {
                if (lock.tryLock()){
                    System.out.println(Thread.currentThread().getName() + "获得锁定");
                    Thread.sleep(3000);     //模拟执行任务的时长
                }else {
                    System.out.println(Thread.currentThread().getName() + "没有获得锁定");
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                if (lock.isHeldByCurrentThread()){
                    lock.unlock();
                }
            }
        }
    }

    public static void main(String[] args) throws InterruptedException {
        Service service = new Service();
        Runnable r = new Runnable() {
            @Override
            public void run() {
                service.serviceMethod();
            }
        };

        Thread t1 = new Thread(r);
        t1.start();
        Thread.sleep(50);       //睡眠50毫秒,确保t1线程锁定
        Thread t2 = new Thread(r);
        t2.start();
    }
}

package com.wkcto.lock.reentrant;

import java.util.Random;
import java.util.concurrent.locks.ReentrantLock;

/**
 * 使用tryLock()可以避免死锁
 */
public class Test09 {
    static class  IntLock implements Runnable{
        private static ReentrantLock lock1 = new ReentrantLock();
        private static ReentrantLock lock2 = new ReentrantLock();
        private int lockNum;        //用于控制锁的顺序

        public IntLock(int lockNum) {
            this.lockNum = lockNum;
        }

        @Override
        public void run() {
            if ( lockNum % 2 == 0 ){    //偶数先锁1,再锁2
                while (true){
                    try {
                        if (lock1.tryLock()){
                            System.out.println(Thread.currentThread().getName() + "获得锁1, 还想获得锁2");
                            Thread.sleep(new Random().nextInt(100));

                            try {
                                if (lock2.tryLock()){
                                    System.out.println(Thread.currentThread().getName() + "同时获得锁1与锁2 ----完成任务了");
                                    return;         //结束run()方法执行,即当前线程结束
                                }
                            } finally {
                                if (lock2.isHeldByCurrentThread()){
                                    lock2.unlock();
                                }
                            }
                        }
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    } finally {
                        if (lock1.isHeldByCurrentThread()){
                            lock1.unlock();
                        }
                    }
                }
            }else {     //奇数就先锁2,再锁1
                while (true){
                    try {
                        if (lock2.tryLock()){
                            System.out.println(Thread.currentThread().getName() + "获得锁2, 还想获得锁1");
                            Thread.sleep(new Random().nextInt(100));

                            try {
                                if (lock1.tryLock()){
                                    System.out.println(Thread.currentThread().getName() + "同时获得锁1与锁2 ----完成任务了");
                                    return;         //结束run()方法执行,即当前线程结束
                                }
                            } finally {
                                if (lock1.isHeldByCurrentThread()){
                                    lock1.unlock();
                                }
                            }
                        }
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    } finally {
                        if (lock2.isHeldByCurrentThread()){
                            lock2.unlock();
                        }
                    }
                }
            }
        }
    }
    public static void main(String[] args) {
        IntLock intLock1 = new IntLock(11);
        IntLock intLock2 = new IntLock(22);
        Thread t1 = new Thread(intLock1);
        Thread t2 = new Thread(intLock2);
        t1.start();
        t2.start();
        //运行后,使用tryLock()尝试获得锁,不会傻傻的等待,通过循环不停的再次尝试,如果等待的时间足够长,线程总是会获得想要的资源
    }
}