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Java多线程的临界资源问题的解决方法,很多新手对此不是很清楚,为了帮助大家解决这个难题,下面小编将为大家详细讲解,有这方面需求的人可以来学习下,希望你能有所收获。
临界资源问题的原因:某一个线程在对临界资源进行访问时,还没来得及完全修改临界资源的值,临界资源就被其他线程拿去访问,导致多个线程访问同一资源。直观表现为打印结果顺序混乱。
解决方法:加锁
静态方法中用类锁,非静态方法中用对象锁。
1.同步代码段:synchronized(){...}
2.同步方法:使用关键字synchronized修饰的方法
3.使用显式同步锁ReentrantLock
锁池描述的即为锁外等待的状态
方法一:同步代码段:synchronized(){...}
public class SourceConflict { public static void main(String[] args) { //实例化4个售票员,用4个线程模拟4个售票员 Runnable r = () -> { while (TicketCenter.restCount > 0) { synchronized(" ") { if (TicketCenter.restCount <= 0) { return; } System.out.println(Thread.currentThread().getName() + "卖出一张票,剩余" + --TicketCenter.restCount + "张票"); } } }; //用4个线程模拟4个售票员 Thread thread1 = new Thread(r, "thread-1"); Thread thread2 = new Thread(r, "thread-2"); Thread thread3 = new Thread(r, "thread-3"); Thread thread4 = new Thread(r, "thread-4"); //开启线程 thread1.start(); thread2.start(); thread3.start(); thread4.start(); } }//实现四名售票员共同售票,资源共享,非独立//Lambda表达式或匿名内部类内部捕获的局部变量必须显式的声明为 final 或实际效果的的 final 类型,而捕获实例或静态变量是没有限制的class TicketCenter{ public static int restCount = 100; }
方法二:同步方法,即使用关键字synchronized修饰的方法
public class SourceConflict2 { public static void main(String[] args) { //实例化4个售票员,用4个线程模拟4个售票员 Runnable r = () -> { while (TicketCenter.restCount > 0) { sellTicket(); } }; //用4个线程模拟4个售票员 Thread thread1 = new Thread(r, "thread-1"); Thread thread2 = new Thread(r, "thread-2"); Thread thread3 = new Thread(r, "thread-3"); Thread thread4 = new Thread(r, "thread-4"); //开启线程 thread1.start(); thread2.start(); thread3.start(); thread4.start(); } private synchronized static void sellTicket() { if (TicketCenter.restCount <= 0) { return; } System.out.println(Thread.currentThread().getName() + "卖出一张票,剩余" + --TicketCenter.restCount + "张票"); }}class TicketCenter{ public static int restCount = 100; }
方法三:使用显式同步锁ReentrantLock
import java.util.concurrent.locks.ReentrantLock;public class SourceConflict3 { public static void main(String[] args) { //实例化4个售票员,用4个线程模拟4个售票员 //显式锁 ReentrantLock lock = new ReentrantLock(); Runnable r = () -> { while (TicketCenter.restCount > 0) { lock.lock(); if (TicketCenter.restCount <= 0) { return; } System.out.println(Thread.currentThread().getName() + "卖出一张票,剩余" + --TicketCenter.restCount + "张票"); lock.unlock(); } }; //用4个线程模拟4个售票员 Thread thread1 = new Thread(r, "thread-1"); Thread thread2 = new Thread(r, "thread-2"); Thread thread3 = new Thread(r, "thread-3"); Thread thread4 = new Thread(r, "thread-4"); //开启线程 thread1.start(); thread2.start(); thread3.start(); thread4.start(); } }class TicketCenter{ public static int restCount = 100; }
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