剑指JUC原理-13.线程池（中）:https://developer.aliyun.com/article/1413642

scheduleAtFixedRate 例子

以固定的速率去执行任务

ScheduledExecutorService pool = Executors.newScheduledThreadPool(1);
log.debug("start...");
pool.scheduleAtFixedRate(() -> {
  log.debug("running...");
}, 1, 1, TimeUnit.SECONDS);
参数：任务对象、延时时间、执行间隔、时间单位

输出

21:45:43.167 c.TestTimer [main] - start... 
21:45:44.215 c.TestTimer [pool-1-thread-1] - running... 
21:45:45.215 c.TestTimer [pool-1-thread-1] - running... 
21:45:46.215 c.TestTimer [pool-1-thread-1] - running... 
21:45:47.215 c.TestTimer [pool-1-thread-1] - running...

scheduleAtFixedRate 例子（任务执行时间超过了间隔时间）：

ScheduledExecutorService pool = Executors.newScheduledThreadPool(1);
log.debug("start...");
pool.scheduleAtFixedRate(() -> {
  log.debug("running...");
  sleep(2);
}, 1, 1, TimeUnit.SECONDS);

输出分析：一开始，延时 1s，接下来，由于任务执行时间 > 间隔时间，间隔被『撑』到了 2s

21:44:30.311 c.TestTimer [main] - start... 
21:44:31.360 c.TestTimer [pool-1-thread-1] - running... 
21:44:33.361 c.TestTimer [pool-1-thread-1] - running... 
21:44:35.362 c.TestTimer [pool-1-thread-1] - running... 
21:44:37.362 c.TestTimer [pool-1-thread-1] - running...

scheduleWithFixedDelay 例子：

ScheduledExecutorService pool = Executors.newScheduledThreadPool(1);
log.debug("start...");
pool.scheduleWithFixedDelay(()-> {
  log.debug("running...");
  sleep(2);
}, 1, 1, TimeUnit.SECONDS);

输出分析：一开始，延时 1s，scheduleWithFixedDelay 的间隔是上一个任务结束 <-> 延时 <-> 下一个任务开始所以间隔都是 3s

21:40:55.078 c.TestTimer [main] - start... 
21:40:56.140 c.TestTimer [pool-1-thread-1] - running... 
21:40:59.143 c.TestTimer [pool-1-thread-1] - running... 
21:41:02.145 c.TestTimer [pool-1-thread-1] - running... 
21:41:05.147 c.TestTimer [pool-1-thread-1] - running...

评价整个线程池表现为：线程数固定，任务数多于线程数时，会放入无界队列排队。任务执行完毕，这些线程也不会被释放。用来执行延迟或反复执行的任务

正确处理执行任务异常

方法1：主动捉异常

ExecutorService pool = Executors.newFixedThreadPool(1);
        pool.submit(() -> {
            try {
                log.debug("task1");
                int i = 1 / 0;
            } catch (Exception e) {
                log.error("error:", e);
            }
        });

输出

21:59:04.558 c.TestTimer [pool-1-thread-1] - task1 
21:59:04.562 c.TestTimer [pool-1-thread-1] - error: 
java.lang.ArithmeticException: / by zero 
 at cn.itcast.n8.TestTimer.lambda$main$0(TestTimer.java:28) 
 at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:511) 
 at java.util.concurrent.FutureTask.run(FutureTask.java:266) 
 at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1149) 
 at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:624) 
 at java.lang.Thread.run(Thread.java:748)

方法2：使用 Future

ExecutorService pool = Executors.newFixedThreadPool(1);
        Future<Boolean> f = pool.submit(() -> {
            log.debug("task1");
            int i = 1 / 0;
            return true;
        });
        log.debug("result:{}", f.get());

输出

21:54:58.208 c.TestTimer [pool-1-thread-1] - task1 
Exception in thread "main" java.util.concurrent.ExecutionException: 
java.lang.ArithmeticException: / by zero 
 at java.util.concurrent.FutureTask.report(FutureTask.java:122) 
 at java.util.concurrent.FutureTask.get(FutureTask.java:192) 
 at cn.itcast.n8.TestTimer.main(TestTimer.java:31) 
Caused by: java.lang.ArithmeticException: / by zero 
 at cn.itcast.n8.TestTimer.lambda$main$0(TestTimer.java:28) 
 at java.util.concurrent.FutureTask.run(FutureTask.java:266) 
 at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1149) 
 at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:624) 
 at java.lang.Thread.run(Thread.java:748)

任务调度线程池模拟定时执行任务

如何让每周四 18:00:00 定时执行任务？

// 获得当前时间
        LocalDateTime now = LocalDateTime.now();
        // 获取本周四 18:00:00.000
        LocalDateTime thursday =
                    now.with(DayOfWeek.THURSDAY).withHour(18).withMinute(0).withSecond(0).withNano(0);
        // 如果当前时间已经超过 本周四 18:00:00.000， 那么找下周四 18:00:00.000
        if(now.compareTo(thursday) >= 0) {
            thursday = thursday.plusWeeks(1);
        }
        // 计算时间差，即延时执行时间
        long initialDelay = Duration.between(now, thursday).toMillis();
        // 计算间隔时间，即 1 周的毫秒值
        long oneWeek = 7 * 24 * 3600 * 1000;
        ScheduledExecutorService executor = Executors.newScheduledThreadPool(2);
        System.out.println("开始时间：" + new Date());
        executor.scheduleAtFixedRate(() -> {
            System.out.println("执行时间：" + new Date());
        }, initialDelay, oneWeek, TimeUnit.MILLISECONDS);

Tomcat线程池

Tomcat 在哪里用到了线程池呢？

LimitLatch 用来限流，可以控制最大连接个数，类似 J.U.C 中的 Semaphore 后面再讲
Acceptor 只负责【接收新的 socket 连接】
Poller 只负责监听 socket channel 是否有【可读的 I/O 事件】
一旦可读，封装一个任务对象（socketProcessor），提交给 Executor 线程池处理
Executor 线程池中的工作线程最终负责【处理请求】，也是tomcat实现高并发的保障

Tomcat 线程池扩展了 ThreadPoolExecutor，行为稍有不同

如果总线程数达到 maximumPoolSize，这时不会立刻抛 RejectedExecutionException 异常，而是再次尝试将任务放入队列，如果还失败，才抛出 RejectedExecutionException 异常。

tomcat 7 部分源码（后续版本在该模块改动并不大）

public void execute(Runnable command, long timeout, TimeUnit unit) {
        submittedCount.incrementAndGet();
        try {
            super.execute(command);
        } catch (RejectedExecutionException rx) {
            if (super.getQueue() instanceof TaskQueue) {
                final TaskQueue queue = (TaskQueue)super.getQueue();
                try {
                    if (!queue.force(command, timeout, unit)) {
                        submittedCount.decrementAndGet();
                        throw new RejectedExecutionException("Queue capacity is full.");
                    }
                } catch (InterruptedException x) {
                    submittedCount.decrementAndGet();
                    Thread.interrupted();
                    throw new RejectedExecutionException(x);
                }
            } else {
                submittedCount.decrementAndGet();
                throw rx;
            }
        }
    }

一旦超过了最大核心数了，父类的实现就会抛执行异常。

子类使用 try catch捕获到

首先拿到线程池中的阻塞队列，尝试将这个任务再次加入到队列中，如果这次成功了就不抛异常了。

TaskQueue.java

public boolean force(Runnable o, long timeout, TimeUnit unit) throws InterruptedException {
        if ( parent.isShutdown() )
            throw new RejectedExecutionException(
                    "Executor not running, can't force a command into the queue"
            );
        return super.offer(o,timeout,unit); //forces the item onto the queue, to be used if the task 
        is rejected
    }

Fork/Join

概念

Fork/Join 是 JDK 1.7 加入的新的线程池实现，它体现的是一种分治思想，适用于能够进行任务拆分的 cpu 密集型运算

所谓的任务拆分，是将一个大任务拆分为算法上相同的小任务，直至不能拆分可以直接求解。跟递归相关的一些计算，如归并排序、斐波那契数列、都可以用分治思想进行求解

Fork/Join 在分治的基础上加入了多线程，可以把每个任务的分解和合并交给不同的线程来完成，进一步提升了运算效率

Fork/Join 默认会创建与 cpu 核心数大小相同的线程池

使用

提交给 Fork/Join 线程池的任务需要继承 RecursiveTask（有返回值）或 RecursiveAction（没有返回值），例如下面定义了一个对 1~n 之间的整数求和的任务

class AddTask1 extends RecursiveTask<Integer> {
    int n;
    public AddTask1(int n) {
        this.n = n;
    }
    @Override
    public String toString() {
        return "{" + n + '}';
    }
    @Override
    protected Integer compute() {
        // 如果 n 已经为 1，可以求得结果了
        if (n == 1) {
            log.debug("join() {}", n);
            return n;
        }
        // 将任务进行拆分(fork)
        AddTask1 t1 = new AddTask1(n - 1);
        t1.fork();
        log.debug("fork() {} + {}", n, t1);
        // 合并(join)结果
        int result = n + t1.join();
        log.debug("join() {} + {} = {}", n, t1, result);
        return result;
    }
}

然后提交给 ForkJoinPool 来执行

public static void main(String[] args) {
 ForkJoinPool pool = new ForkJoinPool(4);
 System.out.println(pool.invoke(new AddTask1(5)));
}

结果

[ForkJoinPool-1-worker-0] - fork() 2 + {1} 
[ForkJoinPool-1-worker-1] - fork() 5 + {4} 
[ForkJoinPool-1-worker-0] - join() 1 
[ForkJoinPool-1-worker-0] - join() 2 + {1} = 3 
[ForkJoinPool-1-worker-2] - fork() 4 + {3} 
[ForkJoinPool-1-worker-3] - fork() 3 + {2} 
[ForkJoinPool-1-worker-3] - join() 3 + {2} = 6 
[ForkJoinPool-1-worker-2] - join() 4 + {3} = 10 
[ForkJoinPool-1-worker-1] - join() 5 + {4} = 15 
15

用图来表示

改进

class AddTask3 extends RecursiveTask<Integer> {
    int begin;
    int end;
    public AddTask3(int begin, int end) {
        this.begin = begin;
        this.end = end;
    }
    @Override
    public String toString() {
        return "{" + begin + "," + end + '}';
    }
    @Override
    protected Integer compute() {
        // 5, 5
        if (begin == end) {
            log.debug("join() {}", begin);
            return begin;
        }
        // 4, 5
        if (end - begin == 1) {
            log.debug("join() {} + {} = {}", begin, end, end + begin);
            return end + begin;
        }
        // 1 5
        int mid = (end + begin) / 2; // 3
        AddTask3 t1 = new AddTask3(begin, mid); // 1,3
        t1.fork();
        AddTask3 t2 = new AddTask3(mid + 1, end); // 4,5
        t2.fork();
        log.debug("fork() {} + {} = ?", t1, t2);
        int result = t1.join() + t2.join();
        log.debug("join() {} + {} = {}", t1, t2, result);
        return result;
    }
}

然后提交给 ForkJoinPool 来执行

public static void main(String[] args) {
 ForkJoinPool pool = new ForkJoinPool(4);
 System.out.println(pool.invoke(new AddTask3(1, 5)));
}

结果

[ForkJoinPool-1-worker-0] - join() 1 + 2 = 3 
[ForkJoinPool-1-worker-3] - join() 4 + 5 = 9 
[ForkJoinPool-1-worker-0] - join() 3 
[ForkJoinPool-1-worker-1] - fork() {1,3} + {4,5} = ? 
[ForkJoinPool-1-worker-2] - fork() {1,2} + {3,3} = ? 
[ForkJoinPool-1-worker-2] - join() {1,2} + {3,3} = 6 
[ForkJoinPool-1-worker-1] - join() {1,3} + {4,5} = 15 
15