ReentrantLock tryLock(timeout,timeUnit) 超时未返回

Question

以下代码中，tryLock(timeout,timeUnit) 方法在超过2s后并未输出 “T2 got no lock end” 而是在T1线程结束后才有输出，但是如果以debug 模式运行或者替换代码中的Math.random()为其他耗时操作比如"TimeUnit.SECONDS.sleep(1)"时却可以在2s后输出 "T2 got no lock end",请问为什么会这样？

public class MyService2 {
public ReentrantLock lock = new ReentrantLock();

public void waitMethod() {
    try {
        System.out.println(System.currentTimeMillis() + " " + Thread.currentThread().getName() + " enter ");
        boolean b = lock.tryLock(2, TimeUnit.SECONDS);
        if (b) {
            System.out.println(System.currentTimeMillis() + " lock begin:" + Thread.currentThread().getName());
            for (int i = 0; i < Integer.MAX_VALUE / 10; i++) {
                Math.random();
            }
            System.out.println(System.currentTimeMillis() + " lock end " + Thread.currentThread().getName());
            return;
        }
        System.out.println(System.currentTimeMillis() + " " + Thread.currentThread().getName() + " got no lock end ");
    } catch (InterruptedException e) {
        e.printStackTrace();
    } finally {
        if (lock.isHeldByCurrentThread()) {
            lock.unlock();
        }
    }
}

public static void main(String[] args) throws InterruptedException {
    MyService2 myService2 = new MyService2();
    Runnable runnable = myService2::waitMethod;
    Thread thread1 = new Thread(runnable);
    thread1.setName("T1");
    thread1.start();
    TimeUnit.MILLISECONDS.sleep(10);
    Thread thread2 = new Thread(runnable);
    thread2.setName("T2");
    thread2.start();
}

Answer 1

这个太像虚拟机优化的现象了

for (int i = 0; i < Integer.MAX_VALUE / 10; i++) {
                Math.random();
            }

这段代码应该是没有运行的,如果你换成

for (int i = 0; i < 100000; i++) {
                System.out.println(Math.random());
            }

结果应该会像你想的那样,这样会对外界产生影响