使用ThreadPoolExecutor线程池实现并发操作并返回结果

在日常业务开发中,有时对一些没有关联的业务操作,如查询多个结果,使用串行调用并不是一个理想的处理方式,可以优化为使用线程池做并发调用,这样在一定程度上能提高性能,如下测试demo方法,使用TimeUnit.SECONDS.sleep(xxx)模拟业务处理时长:

public static String getName () throws Exception {
        TimeUnit.SECONDS.sleep(5);

        return "阿三";
    }

    public static int getAge () throws Exception {
        TimeUnit.SECONDS.sleep(15);

        return 18;
    }

    public static String getBirthday () throws Exception {
        TimeUnit.SECONDS.sleep(8);

        return "1001";
    }

1.串行调用耗时测试

  // 1、串行测试
  StopWatch watch = new StopWatch();
  watch.start("串行");
  String name = getName();
  int age = getAge();
  String birthday = getBirthday();
  watch.stop();
  System.out.println(name + "-" + age + "-" + birthday + ", 串行 耗时:" + watch.getTotalTimeMillis());

2.并行耗时测试

  2.1 使用ThreadPoolExecutor的submit方法,参数为Callable类型的可以有返回值

public <T> Future<T> submit(Callable<T> task) 
    /**
     * 使用线程池
     */
    static ThreadPoolExecutor poolExecutor = new ThreadPoolExecutor(10, 20, 2000L,
            TimeUnit.SECONDS, new LinkedBlockingQueue<>());

    public static void main(String[] args) throws Exception {

        
        // 2.并行测试1
        StopWatch watch2 = new StopWatch();
        watch2.start("并行");
        // 此方式get方法阻塞了后面的服务,用法不对
        String submit = poolExecutor.submit(() -> getName()).get();
        Integer submit1 = poolExecutor.submit(() -> getAge()).get();
        String submit2 = poolExecutor.submit(() -> getBirthday()).get();
        watch2.stop();
        System.out.println(submit + "-" + submit1 + "-" + submit2  + ", 并行1 耗时:" + watch2.getTotalTimeMillis());

        // 3.并行测试2
        StopWatch watch3 = new StopWatch();
        watch3.start("并行2");
        Future<String> future = poolExecutor.submit(() -> getName());
        Future<Integer> future1 = poolExecutor.submit(() -> getAge());
        Future<String> future2 = poolExecutor.submit(() -> getBirthday());
        String submit3 = future.get();
        Integer submit4 = future1.get();
        String submit5 = future2.get();
        watch3.stop();
        System.out.println(submit3 + "-" + submit4 + "-" + submit5 + ", 并行3 耗时:" + watch3.getTotalTimeMillis());

    }

  2.2 对返回值为统一类型的可以使用invokeAll方法:

public <T> List<Future<T>> invokeAll(Collection<? extends Callable<T>> tasks)
        // 4.并行测试3
        StopWatch watch4 = new StopWatch();
        watch4.start("并行3");
        Callable<String> taskName = () -> getName();
        Callable<String> taskBirthday = () -> getBirthday();
        List<Callable<String>> taskList = Lists.newArrayList();
        taskList.add(taskName);
        taskList.add(taskBirthday);
        List<Future<String>> futures = poolExecutor.invokeAll(taskList);

        StringBuilder sb = new StringBuilder();
        for (Future<String> resultFuture : futures) {
            sb.append(resultFuture.get()).append("-");
        }
        watch4.stop();
        System.out.println(sb.toString() + ", 并行3 耗时:" + watch4.getTotalTimeMillis());    

3.测试结果

可以看出最后两种方式耗时和最大耗时服务的用时差不多


Github源码参照

posted @ 2020-06-03 13:22  傻不拉几猫  阅读(217)  评论(0编辑  收藏