JAVA多线程
直接上代码去跑发现其中的规律即可
public class ThreadPollUtil { private static int corePoolSize = Runtime.getRuntime().availableProcessors(); /** * corePoolSize 用于指定核心线程数量 * maximumPoolSize 指定最大线程数 * keepAliveTime和TimeUnit指定线程空闲后的最大存活时间 */ public static ThreadPoolExecutor executor = new ThreadPoolExecutor(corePoolSize, corePoolSize+1, 10l, TimeUnit.SECONDS, new LinkedBlockingQueue<Runnable>(1000)); /* 让主线程等待 CountDownLatch 任务计数器 countDownLatch这个类使一个线程等待其他线程各自执行完毕后再执行。 是通过一个计数器来实现的,计数器的初始值是线程的数量。每当一个线程执行完毕后,计数器的值就-1,当计数器的值为0时,表示所有线程都执行完毕,然后在闭锁上等待的线程就可以恢复工作了。 */ public static void main(String[] args) throws InterruptedException { //创建一个三个子线程 初始化计数器 Map<Object, Object> map = new HashMap<>(); CountDownLatch countDownLatch = new CountDownLatch(3); Long m=1000000000L; long time11 = System.currentTimeMillis(); ThreadPollUtil.executor.submit(new Runnable() { @Override public void run() { long time1 = System.currentTimeMillis(); int i=0; for (int i1 = 0; i1 < m; i1++) { i=i1+i; } map.put("1",i); //计数器-1 countDownLatch.countDown(); long time2 = System.currentTimeMillis(); System.out.println("线程一"+(time1 - time2) + "毫秒。"); } }); ThreadPollUtil.executor.submit(new Runnable() { @Override public void run() { long time1 = System.currentTimeMillis(); int i=0; for (int i1 = 0; i1 < m; i1++) { i=i1+i; } map.put("1",i); //计数器-1 countDownLatch.countDown(); long time2 = System.currentTimeMillis(); System.out.println("线程二"+(time1 - time2) + "毫秒。"); } }); ThreadPollUtil.executor.submit(new Runnable() { @Override public void run() { long time1 = System.currentTimeMillis(); int i=0; for (int i1 = 0; i1 < m; i1++) { i=i1+i; } map.put("1",i); //计数器-1 countDownLatch.countDown(); long time2 = System.currentTimeMillis(); System.out.println("线程三"+(time1 - time2) + "毫秒。"); } }); /** * 通过await方法让主线程等待 */ countDownLatch.await(); long time22 = System.currentTimeMillis(); System.out.println("总线程"+(time11 - time22) + "毫秒。"); } }
再做个比较没有使用多线程的自上而下的逻辑
class Xxx{ public static void main(String[] args) throws InterruptedException { long time1 = System.currentTimeMillis(); int i=0; Long m=1000000000L; for (int i1 = 0; i1 < m; i1++) { i=i1+i; } System.out.println("执行一"); for (int i1 = 0; i1 < m; i1++) { i=i1+i; } System.out.println("执行二"); for (int i1 = 0; i1 < m; i1++) { i=i1+i; } System.out.println("执行三"); long time2 = System.currentTimeMillis(); System.out.println("总线程"+(time1 - time2) + "毫秒。"); } }