临界区互斥访问的方法

it2024-04-19  48

0 同步机制遵循准则

空闲让进:临界区空闲时,允许请求进入临界区忙则等待:有线程进入临界区时,其他线程必须等待有限等待:等待不能无限制持续下去让权等待:释放等待线程的 CPU 资源

1 单标志

问题:违背空闲让进(当有一个线程结束,另外一个即使没有结束也无法进入临界区)

public class SingleFlag { public static volatile boolean flag; public static volatile int num = 0; public static void main(String[] args) { new Thread(() -> { while (true) { while (flag) ; System.out.println(Thread.currentThread().getName() + " " + ++num); flag = false; } }).start(); new Thread(() -> { while (true) { while (flag) ; System.out.println(Thread.currentThread().getName() + " " + ++num); flag = true; } }).start(); } }

2 双标志先检查

问题:违背忙则等待(恰好执行完 1 时线程切换,此时 2 判断也能通过,则会存在安全问题)

public class DoubleFlag { public static volatile boolean[] flag={false, true}; public static volatile int num = 0; public static void main(String[] args) { new Thread(() -> { while (true) { while (flag[0]); // 1 flag[1] = true; System.out.println(Thread.currentThread().getName() + " " + ++num); flag[1] = false; } }).start(); new Thread(() -> { while (true) { while (flag[1]); // 2 flag[0] = true; System.out.println(Thread.currentThread().getName() + " " + ++num); flag[0] = false; } }).start(); } }

3 双标志后检查

问题:违背有限等待原则(两个线程无限制谦让,导致活锁)

public class DoubleFlagAfterCheck { public static volatile boolean[] flag = {false, true}; public static volatile int num = 0; public static void main(String[] args) { new Thread(() -> { while (true) { flag[1] = true; while (flag[0]); System.out.println(Thread.currentThread().getName() + " " + ++num); flag[1] = false; } }).start(); new Thread(() -> { while (true) { flag[0] = true; while (flag[1]); System.out.println(Thread.currentThread().getName() + " " + ++num); flag[0] = false; } }).start(); } }

4 Peterson

完美解决并发安全性问题 问题:无法达到让权等待(CPU 一直在死循环判断,占用 CPU 资源)

public class Peterson { public static volatile boolean[] flag = {false, false}; public static volatile int turn = 0; public static volatile int num = 0; public static void main(String[] args) { new Thread(() -> { while (true) { flag[0] = true; turn = 2; while (flag[1] && turn == 2) ; System.out.println(Thread.currentThread().getName() + " " + ++num); flag[0] = false; } }).start(); new Thread(() -> { while (true) { flag[1] = true; turn = 1; while (flag[0] && turn == 1) ; System.out.println(Thread.currentThread().getName() + " " + ++num); flag[1] = false; } }).start(); } }
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