java - 控制并发线程数

问题描述

问题解答

如果你了解信号量的实现机制，那么这道题目也是一个意思。

public class Test { private final Integer maxCounter = 3; private Integer current = 0; public void call1() {//在这里补充代码synchronized (this) { try {while (current.equals(maxCounter)) { // 请求 到达上限 wait();} } catch (InterruptedException ex) { } current++; notifyAll();}call2(current);synchronized (this) { try {while (current == 0) { wait();} } catch (InterruptedException ex) { } current--; notifyAll();} } private void call2(Integer current) {System.out.println(Thread.currentThread().getName() + ': I’m called ' + current);// 下面的休眠 2 秒钟用于测试try { Thread.sleep(2000);} catch (InterruptedException ex) { ex.printStackTrace(System.err);} } static class TestThread implements Runnable {private Test t;public TestThread(Test t) { this.t = t;}@Overridepublic void run() { t.call1();} } public static void main(String[] args) {Test t1 = new Test();TestThread tt = new TestThread(t1);for (int i = 0; i < 10; i++) { Thread t = new Thread(tt, 'Thread-' + i); t.start();} }}

运行这段代码，你可以发现每 2 秒内，最多只有 3 （maxCounter）个线程在运行。

用CountDownLatch。。。