问题:
假设有个全局变量var初始化为0
MyThread线程循环+1
MyThread2线程检测到var大于10时退出循环
问题来了,我们发现MyThread一直没有退出循环
也就是说线程没有及时刷新内存
解决方法:给全局变量添加 volatile关键字
Java提供了volatile来保证可见性。当一个变量被volatile修饰后,表示着线程本地内存无效,当一个线程修改共享变量后他会立即被更新到主内存中,当其他线程读取共享变量时,它会直接从主内存中读取。
abstract class ThreadMemo implements Runnable { private Thread t; protected String threadName; // public static int var = 0; volatile public static int var = 0; // 只需要添加volatile关键字就可以了 public ThreadMemo(String name) { threadName = name; System.out.println("Creating " + threadName); } abstract public void run(); public void start() { System.out.println("Starting " + threadName); if (t == null) { t = new Thread(this, threadName); t.start(); } } } class MyThread extends ThreadMemo { MyThread(String name) { super(name); } public void run() { var = super.var; while(true) { var = var + 1; // sleep(50); // time.sleep(50); try { Thread.sleep(50); } catch (InterruptedException e) { System.out.println("Thread " + threadName + " interrupted."); } // System.out.println("mythread 1 var: " + var); } } } class MyThread2 extends ThreadMemo { MyThread2(String name) { super(name); } public void run() { var = super.var; while(true) { if(var < 10) { // try { // Thread.sleep(1); // } catch (InterruptedException e) { // System.out.println("Thread " + threadName + " interrupted."); // } // Object o = new Object(); continue; } System.out.println("mythread2 var = " + var); break; } } } public class Test2 { // public static int var = 0; public static void main(String[] args) { ThreadMemo t1 = new MyThread("Thread-1"); ThreadMemo t2 = new MyThread2("Thread-2"); Thread a = new Thread(t1); Thread b = new Thread(t2); a.start(); b.start(); } }
然后我们发现,给读线程的循环中添加一些语句,循环就能退出了
V2EX上发现相同的问题 Java 多线程并发,线程什么时候会刷新 "工作内存"
据说是遇到IO阻塞才会刷新内存