Callable

类比 Runnable,都可以创建线程，但是区别是有返回值
需要搭配Future FutureTask

//无返回值
//不能抛出异常
@FunctionalInterface
public interface Runnable {
    public abstract void run();
}
@FunctionalInterface
public interface Callable<V> {
    V call() throws Exception;
}

使用示例

FutureTask task = new FutureTask(new Callable() {
    @Override
    public Object call() throws Exception {
        System.out.println("通过Callable方式执行任务");
        Thread.sleep(3000);
        return "返回任务结果";
    }
});
new Thread(task).start();

Future

接受Callable的任务，对这个返回的任务的操作

boolean cancel (boolean mayInterruptIfRunning) 取消任务，是立即取消还是等执行完取消
boolean isCancelled () 判断是否已取消
boolean isDone () 判断是否已完成
V get () 获取任务，阻塞式的获取结果
V get (long timeout, TimeUnit unit) ，同上，制定了阻塞的时间限制

FutureTask与Future，

实现关系

FutureTask既可以当做Runnable执行，也可以当做Future接受Callable的返回结果

使用示例

public class FutureTaskDemo {
    public static void main(String[] args) throws ExecutionException, InterruptedException {
        Task task = new Task();
        //构建futureTask
        FutureTask<Integer> futureTask = new FutureTask<>(task);
        //作为Runnable入参
        new Thread(futureTask).start();
        System.out.println("task运行结果："+futureTask.get());
    }
    static class Task implements Callable<Integer> {
        @Override
        public Integer call() throws Exception {
            System.out.println("子线程正在计算");
            int sum = 0;
            for (int i = 0; i < 100; i++) {
                sum += i;
            }
            return sum;
        }
    }
}

使用案例：促销活动中的尚宁信息查询

public class FutureTaskDemo2 {
    public static void main(String[] args) throws ExecutionException, InterruptedException {
        FutureTask<String> ft1 = new FutureTask<>(new T1Task());
        FutureTask<String> ft2 = new FutureTask<>(new T2Task());
        FutureTask<String> ft3 = new FutureTask<>(new T3Task());
        FutureTask<String> ft4 = new FutureTask<>(new T4Task());
        FutureTask<String> ft5 = new FutureTask<>(new T5Task());
        //构建线程池
        ExecutorService executorService = Executors.newFixedThreadPool(5);
        executorService.submit(ft1);
        executorService.submit(ft2);
        executorService.submit(ft3);
        executorService.submit(ft4);
        executorService.submit(ft5);
        //获取执行结果
        System.out.println(ft1.get());
        System.out.println(ft2.get());
        System.out.println(ft3.get());
        System.out.println(ft4.get());
        System.out.println(ft5.get());
        executorService.shutdown();
    }
    static class T1Task implements Callable<String> {
        @Override
        public String call() throws Exception {
            System.out.println("T1:查询商品基本信息...");
            TimeUnit.MILLISECONDS.sleep(50);
            return "商品基本信息查询成功";
        }
    }
    static class T2Task implements Callable<String> {
        @Override
        public String call() throws Exception {
            System.out.println("T2:查询商品价格...");
            TimeUnit.MILLISECONDS.sleep(50);
            return "商品价格查询成功";
        }
    }
    static class T3Task implements Callable<String> {
        @Override
        public String call() throws Exception {
            System.out.println("T3:查询商品库存...");
            TimeUnit.MILLISECONDS.sleep(50);
            return "商品库存查询成功";
        }
    }
    static class T4Task implements Callable<String> {
        @Override
        public String call() throws Exception {
            System.out.println("T4:查询商品图片...");
            TimeUnit.MILLISECONDS.sleep(50);
            return "商品图片查询成功";
        }
    }
    static class T5Task implements Callable<String> {
        @Override
        public String call() throws Exception {
            System.out.println("T5:查询商品销售状态...");
            TimeUnit.MILLISECONDS.sleep(50);
            return "商品销售状态查询成功";
        }
    }
}

Future注意事项

建议使用get加timeout
思考：使用Callable 和Future 产生新的线程了吗？单独使用时并没有
局限性：

并发执行多任务时：get是阻塞等待获取的
无法对多个任务进行链式调用，下面有方法可以实现
无法组合任务，下面的也可以实现
无异常处理，
CompletionService

主要功能就是：一边生成任务，一边获取任务，让两个事分开执行，任务直接不会相互阻塞，不再依赖任务顺序

CompletionService原理

内部通过阻塞队列+FutureTask，先完成的可以先获取，
内部队列保存执行完成的Future，
take, poll获取一个已经执行完成的Future，再get结果

使用案

向不同平台询价，再保存

//    创建线程池 
ExecutorService    executor = Executors.newFixedThreadPool(3); 
//    异步向电商S1询价 
Future<Integer>    f1 = executor.submit(()->getPriceByS1()); 
//    异步向电商S2询价 
Future<Integer>    f2=    executor.submit(()->getPriceByS2());             
//    获取电商S1报价并异步保存 
executor.execute(()->save(f1.get()));        
//    获取电商S2报价并异步保存 
executor.execute(()->save(f2.get())
//问题，就是f1会阻挡f2

//创建线程池
ExecutorService executor = Executors.newFixedThreadPool(10);
//创建CompletionService
CompletionService<Integer> cs = new ExecutorCompletionService<>(executor);
//异步向电商S1询价
cs.submit(() -> getPriceByS1());
//异步向电商S2询价
cs.submit(() -> getPriceByS2());
//异步向电商S3询价
cs.submit(() -> getPriceByS3());
//将询价结果异步保存到数据库
for (int i = 0; i < 3; i++) {
    Integer r = cs.take().get();
    executor.execute(() -> save(r));
}

实现类似 Dubbo 的 Forking Cluster场景
Dubbo 中有一种叫做 Forking 的集群模式，这种集群模式下，支持并行地调用多个服务实例，只要有一个成功就返回结果。

// 创建线程池
ExecutorService executor = Executors.newFixedThreadPool(3);
// 创建CompletionService
CompletionService<Integer> cs = new ExecutorCompletionService<>(executor);
// 用于保存Future对象
List<Future<Integer>> futures = new ArrayList<>(3);
//提交异步任务，并保存future到futures 
futures.add(cs.submit(()->geocoderByS1()));
futures.add(cs.submit(()->geocoderByS2()));
futures.add(cs.submit(()->geocoderByS3()));
// 获取最快返回的任务执行结果
Integer r = 0;
try {
  // 只要有一个成功返回，则break
  for (int i = 0; i < 3; ++i) {
    r = cs.take().get();
    //简单地通过判空来检查是否成功返回
    if (r != null) {
      break;
    }
  }
} finally {
  //取消所有任务
  for(Future<Integer> f : futures)
    f.cancel(true);
}
// 返回结果

应用场景

批量提交异步任务时，有效各个
让异步的执行结果有序化，先执行完的先入阻塞队列
线程池隔离，各个业务不干扰
CompletableFuture，重要
让业务逻辑处理并行啊，聚合，依赖的，扩展，增强，

对Future扩展，增强，还有任务的编排能力
CompletionStage: 默认使用时ForkJoinPool.commonPool线程池

重要API

依赖场景

thenApply（），前面执行完，执行后面
thenCompose()，链接有依赖关系的任务，

描述and聚合关系

thenCombine: 任务合并
thenAccepetBoth: 两个如未能执行完成交给后面的任务，无返回
runAfterBoth,:l两个任务完成，执行下一个操作

描述or的关系

applyToEither:两个任务谁快，执行谁
acceptEither: 谁快消耗那个的结果
runAfterEither:：谁先完成，就下一步

并行

anyOf
allOf

创建异步操作
CompletableFuture 提供了四个静态方法来创建一个异步操作：


//
public static CompletableFuture<Void> runAsync(Runnable runnable)
//不指定就用ForkJoinPool.commonPool() 指定就用executor
//强烈建议你要根据不同的业务类型创建不同的线程池，以避免互相干扰
public static CompletableFuture<Void> runAsync(Runnable runnable, Executor executor)
  //Supplier 接口的 get() 方法是有返回值的（会阻塞）
 public static <U> CompletableFuture<U> supplyAsync(Supplier<U> supplier)
public static <U> CompletableFuture<U> supplyAsync(Supplier<U> supplier, Executor executor)

runAsync&supplyAsync

Runnable runnable = () -> System.out.println("执行无返回结果的异步任务");
CompletableFuture.runAsync(runnable);
CompletableFuture<String> future = CompletableFuture.supplyAsync(() -> {
    System.out.println("执行有返回值的异步任务");
    try {
        Thread.sleep(5000);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    return "Hello World";
});
String result = future.get();
System.out.println(result);

获取结果
join&get,都是获取异步的结果，抛的异常不一样

whenComplete&exceptionally

public CompletableFuture<T> whenComplete(BiConsumer<? super T,? super Throwable> action)
public CompletableFuture<T> whenCompleteAsync(BiConsumer<? super T,? super Throwable> action)
public CompletableFuture<T> whenCompleteAsync(BiConsumer<? super T,? super Throwable> action, Executor executor)

方法不以Async结尾，意味着Action使用相同的线程执行，而Async可能会使用其它的线程去执行(如果使用相同的线程池，也可能会被同一个线程选中执行)。
这几个方法都会返回CompletableFuture，当Action执行完毕后它的结果返回原始的CompletableFuture的计算结果或者返回异常

whenComplete&exceptionally

CompletableFuture<String> future = CompletableFuture.supplyAsync(() -> {
    try {
        TimeUnit.SECONDS.sleep(1);
    } catch (InterruptedException e) {
    }
    if (new Random().nextInt(10) % 2 == 0) {
        int i = 12 / 0;
    }
    System.out.println("执行结束！");
    return "test";
});
future.whenComplete(new BiConsumer<String, Throwable>() {
    @Override
    public void accept(String t, Throwable action) {
        System.out.println(t+" 执行完成！");
    }
});
//失败之后的处理
future.exceptionally(new Function<Throwable, String>() {
    @Override
    public String apply(Throwable t) {
        System.out.println("执行失败：" + t.getMessage());
        return "异常xxxx";
    }
}

thenApply，上个任务的结果作为入参

public <U> CompletableFuture<U> thenApply(Function<? super T,? extends U> fn)
public <U> CompletableFuture<U> thenApplyAsync(Function<? super T,? extends U> fn)

CompletableFuture<Integer> future = CompletableFuture.supplyAsync(() -> {
    int result = 100;
    System.out.println("一阶段：" + result);
    return result;
}).thenApply(number -> {
    // number是上任务的结果
    int result = number * 3;
    System.out.println("二阶段：" + result);
    return result;
});

thenCompose 返回的函数实现计算步骤的记过看实例d

public <U> CompletableFuture<U> thenCompose(Function<? super T, ? extends CompletionStage<U>> fn);
public <U> CompletableFuture<U> thenComposeAsync(Function<? super T, ? extends CompletionStage<U>> fn) ;

CompletableFuture<Integer> future = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(30);
                System.out.println("第一阶段：" + number);
                return number;
            }
        })
        .thenCompose(new Function<Integer, CompletionStage<Integer>>() {
            @Override
            //入参param 技术是上个任务返回的number
            public CompletionStage<Integer> apply(Integer param) {
                return CompletableFuture.supplyAsync(new Supplier<Integer>() {
                    @Override
                    public Integer get() {
                        int number = param * 2;
                        System.out.println("第二阶段：" + number);
                        return number;
                    }
                });
            }
        });
}
System.out.println("最终结果: " + future.get());

thenApply 和 thenCompose的区别,有点类似

thenApply 转换的是泛型中的类型，返回的是同一个CompletableFuture；
thenCompose 将内部的 CompletableFuture 调用展开来并使用上一个
CompletableFutre 调用的结果在下一步的 CompletableFuture 调用中进行运算，是生
成一个新的CompletableFuture。

CompletableFuture<String> future = CompletableFuture.supplyAsync(() -> "Hello");
CompletableFuture<String> result1 = future.thenApply(param -> param + " World");
CompletableFuture<String> result2 = future
        .thenCompose(param -> CompletableFuture.supplyAsync(() -> param + " World"));
System.out.println(result1.get());
System.out.println(result2.get());

下面的是结果消费，只是执行action，不返回CompletableFuture

thenAccept系列：对单个结果进行消费
thenAcceptBoth系列：对两个结果进行消费
thenRun系列：不关心结果，只对结果执行Action
thenAccept，没有返回值了
```java public CompletionStage thenAccept(Consumer<? super T> action); public CompletionStage thenAcceptAsync(Consumer<? super T> action);

```java
CompletableFuture<Void> future = CompletableFuture
        .supplyAsync(() -> {
            int number = new Random().nextInt(10);
            System.out.println("第一阶段：" + number);
            return number;
        }).thenAccept(number ->
                System.out.println("第二阶段：" + number * 5));
System.out.println("最终结果：" + future.get());
//fuure 泛型是void

thenAcceptBoth,前面的两个都执行玩，执行action

public <U> CompletionStage<Void> thenAcceptBoth(CompletionStage<? extends U> other,BiConsumer<? super T, ? super U> action);
public <U> CompletionStage<Void> thenAcceptBothAsync(CompletionStage<? extends U> other,BiConsumer<? super T, ? super U> action);

CompletableFuture<Integer> futrue1 = CompletableFuture.supplyAsync(new Supplier<Integer>() {
    @Override
    public Integer get() {
        int number = new Random().nextInt(3) + 1;
        try {
            TimeUnit.SECONDS.sleep(number);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("第一阶段：" + number);
        return number;
    }
});
CompletableFuture<Integer> future2 = CompletableFuture.supplyAsync(new Supplier<Integer>() {
    @Override
    public Integer get() {
        int number = new Random().nextInt(3) + 1;
        try {
            TimeUnit.SECONDS.sleep(number);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("第二阶段：" + number);
        return number;
    }
});
//就时连个future 结束时在处理
futrue1.thenAcceptBoth(future2, new BiConsumer<Integer, Integer>() {
    @Override
    public void accept(Integer x, Integer y) {
        System.out.println("最终结果：" + (x + y));
    }
}).join();

thenRun，再执行一段另一个逻辑，与上一个CompletabelFuture无关了

public CompletionStage<Void> thenRun(Runnable action);
public CompletionStage<Void> thenRunAsync(Runnable action);

CompletableFuture<Void> future = CompletableFuture.supplyAsync(() -> {
    int number = new Random().nextInt(10);
    System.out.println("第一阶段：" + number);
    return number;
}).thenRun(() ->
        System.out.println("thenRun 执行"));
 System.out.println("最终结果：" + future.get());

thenCombine 合并连个任务的结果，下一步处理

public <U,V> CompletionStage<V> thenCombine(CompletionStage<? extends U> other,BiFunction<? super T,? super U,? extends V> fn);
public <U,V> CompletionStage<V> thenCombineAsync(CompletionStage<? extends U> other,BiFunction<? super T,? super U,? extends V> fn);

CompletableFuture<Integer> future1 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(10);
                System.out.println("第一阶段：" + number);
                return number;
            }
        });
CompletableFuture<Integer> future2 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(10);
                System.out.println("第二阶段：" + number);
                return number;
            }
        });
//跟thenApply没啥区别
CompletableFuture<Integer> result = future1
        .thenCombine(future2, new BiFunction<Integer, Integer, Integer>() {
            @Override
            public Integer apply(Integer x, Integer y) {
                return x + y;
            }
        });
System.out.println("最终结果：" + result.get());

applyToEither

CompletableFuture<Integer> future1 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(10);
                System.out.println("第一阶段start：" + number);
                try {
                    TimeUnit.SECONDS.sleep(number);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("第一阶段end：" + number);
                return number;
            }
        });
CompletableFuture<Integer> future2 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(10);
                System.out.println("第二阶段start：" + number);
                try {
                    TimeUnit.SECONDS.sleep(number);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("第二阶段end：" + number);
                return number;
            }
        });
就是获取运行比较快的那个任务的结果
future1.applyToEither(future2, new Function<Integer, Integer>() {
    @Override
    public Integer apply(Integer number) {
        System.out.println("最快结果：" + number);
        return number * 2;
    }
}).join();

acceptEither，获取执行最快的任务的结果，并有下一步的处理

CompletableFuture<Integer> future1 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(10) + 1;
                try {
                    TimeUnit.SECONDS.sleep(number);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("第一阶段：" + number);
                return number;
            }
        });
CompletableFuture<Integer> future2 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(10) + 1;
                try {
                    TimeUnit.SECONDS.sleep(number);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("第二阶段：" + number);
                return number;
            }
        });
future1.acceptEither(future2, new Consumer<Integer>() {
    @Override
    public void accept(Integer number) {
        System.out.println("最快结果：" + number);
    }
}).join();

runAfterEither，两个任务，有一个完成，就执行下一步

CompletableFuture<Integer> future1 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(5);
                try {
                    TimeUnit.SECONDS.sleep(number);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("第一阶段：" + number);
                return number;
            }
        });
CompletableFuture<Integer> future2 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
            @Override
            public Integer get() {
                int number = new Random().nextInt(5);
                try {
                    TimeUnit.SECONDS.sleep(number);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("第二阶段：" + number);
                return number;
            }
        });
future1.runAfterEither(future2, new Runnable() {
    @Override
    public void run() {
        System.out.println("已经有一个任务完成了");
    }
}).join();

runAfterBoth 两个都完成菜执行下一步

CompletableFuture<Integer> future1 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
    @Override
    public Integer get() {
        try {
            TimeUnit.SECONDS.sleep(1);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("第一阶段：1");
        return 1;
    }
});
CompletableFuture<Integer> future2 = CompletableFuture
        .supplyAsync(new Supplier<Integer>() {
    @Override
    public Integer get() {
        try {
            TimeUnit.SECONDS.sleep(2);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("第二阶段：2");
        return 2;
    }
});
future1.runAfterBoth(future2, new Runnable() {
    @Override
    public void run() {
        System.out.println("上面两个任务都执行完成了。");
    }
}).get();

anyOf 多个completableFuture ,有一个完成就返回这个的Future结果

Random random = new Random();
CompletableFuture<String> future1 = CompletableFuture
        .supplyAsync(() -> {
            try {
                TimeUnit.SECONDS.sleep(random.nextInt(5));
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            return "hello";
        });
CompletableFuture<String> future2 = CompletableFuture
        .supplyAsync(() -> {
            try {
                TimeUnit.SECONDS.sleep(random.nextInt(1));
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            return "world";
        });
CompletableFuture<Object> result = CompletableFuture.anyOf(future1, future2);
System.out.println(result.get());

allOf
allOf方法用来实现多 CompletableFuture 的同时返回。

CompletableFuture<String> future1 = CompletableFuture
        .supplyAsync(() -> {
            try {
                TimeUnit.SECONDS.sleep(2);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("future1完成！");
            return "future1完成！";
        });
CompletableFuture<String> future2 = CompletableFuture
        .supplyAsync(() -> {
            System.out.println("future2完成！");
            return "future2完成！";
        });
CompletableFuture<Void> combindFuture = CompletableFuture
        .allOf(future1, future2);
try {
    combindFuture.get();
} catch (InterruptedException e) {
    e.printStackTrace();
} catch (ExecutionException e) {
    e.printStackTrace();
}
System.out.println("future1: " + future1.isDone() + "，future2: " +
future2.isDone());

CompletableFuture常用方法总结

使用案例：实现最优的“烧水泡茶”程序

public class FutureTaskDemo3{
    public static void main(String[] args) throws ExecutionException, InterruptedException {
        // 创建任务T2的FutureTask
        FutureTask<String> ft2 = new FutureTask<>(new T2Task());
        // 创建任务T1的FutureTask
        FutureTask<String> ft1 = new FutureTask<>(new T1Task(ft2));
        // 线程T1执行任务ft1
        Thread T1 = new Thread(ft1);
        T1.start();
        // 线程T2执行任务ft2
        Thread T2 = new Thread(ft2);
        T2.start();
        // 等待线程T1执行结果
        System.out.println(ft1.get());
    }
}
// T1Task需要执行的任务：
// 洗水壶、烧开水、泡茶
class T1Task implements Callable<String> {
    FutureTask<String> ft2;
    // T1任务需要T2任务的FutureTask
    T1Task(FutureTask<String> ft2){
        this.ft2 = ft2;
    }
    @Override
    public String call() throws Exception {
        System.out.println("T1:洗水壶...");
        TimeUnit.SECONDS.sleep(1);
        System.out.println("T1:烧开水...");
        TimeUnit.SECONDS.sleep(15);
        // 获取T2线程的茶叶
        String tf = ft2.get();
        System.out.println("T1:拿到茶叶:"+tf);
        System.out.println("T1:泡茶...");
        return "上茶:" + tf;
    }
}
// T2Task需要执行的任务:
// 洗茶壶、洗茶杯、拿茶叶
class T2Task implements Callable<String> {
    @Override
    public String call() throws Exception {
        System.out.println("T2:洗茶壶...");
        TimeUnit.SECONDS.sleep(1);
        System.out.println("T2:洗茶杯...");
        TimeUnit.SECONDS.sleep(2);
        System.out.println("T2:拿茶叶...");
        TimeUnit.SECONDS.sleep(1);
        return "龙井";
    }
}

public class CompletableFutureDemo2 {
    public static void main(String[] args) {
        //任务1：洗水壶->烧开水
        CompletableFuture<Void> f1 = CompletableFuture
                .runAsync(() -> {
                    System.out.println("T1:洗水壶...");
                    sleep(1, TimeUnit.SECONDS);
                    System.out.println("T1:烧开水...");
                    sleep(15, TimeUnit.SECONDS);
                });
        //任务2：洗茶壶->洗茶杯->拿茶叶
        CompletableFuture<String> f2 = CompletableFuture
                .supplyAsync(() -> {
                    System.out.println("T2:洗茶壶...");
                    sleep(1, TimeUnit.SECONDS);
                    System.out.println("T2:洗茶杯...");
                    sleep(2, TimeUnit.SECONDS);
                    System.out.println("T2:拿茶叶...");
                    sleep(1, TimeUnit.SECONDS);
                    return "龙井";
                });
        //任务3：任务1和任务2完成后执行：泡茶
        CompletableFuture<String> f3 = f1.thenCombine(f2, (__, tf) -> {
                    System.out.println("T1:拿到茶叶:" + tf);
                    System.out.println("T1:泡茶...");
                    return "上茶:" + tf;
                });
        //等待任务3执行结果
        System.out.println(f3.join());
    }
    static void sleep(int t, TimeUnit u){
        try {
            u.sleep(t);
        } catch (InterruptedException e) {
        }
    }
}

Future

Callable

Future

FutureTask与Future，

使用示例

使用案例：促销活动中的尚宁信息查询

Future注意事项

CompletionService

CompletionService原理

使用案

应用场景

CompletableFuture，重要

重要API

whenComplete&exceptionally

thenApply，上个任务的结果作为入参

thenCompose 返回的函数实现计算步骤的记过看实例d

thenAccept，没有返回值了

thenAcceptBoth,前面的两个都执行玩，执行action

thenRun，再执行一段另一个逻辑，与上一个CompletabelFuture无关了

thenCombine 合并连个任务的结果，下一步处理

applyToEither

acceptEither，获取执行最快的任务的结果，并有下一步的处理

runAfterEither，两个任务，有一个完成，就执行下一步

runAfterBoth 两个都完成菜执行下一步

anyOf 多个completableFuture ,有一个完成就返回这个的Future结果