Juc09_CompletableFuture异步编排（三）

        CompletableFuture.supplyAsync(() -> {
            return 1;
        }).thenApply(f -> {
            return f+2;
        }).thenApply(f -> {
            return f+3;
        }).thenAccept(r -> System.out.println(r));
        // 任务A执行完执行B,并且B不需要A的结果
        System.out.println(CompletableFuture.supplyAsync(() -> "resultA").thenRun(() -> {}).join());
        // 任务A执行完成执行B,B需要A的结果,但是任务B无返回值
        System.out.println(CompletableFuture.supplyAsync(() -> "resultA").thenAccept(resultA -> {}).join());
        // 任务A执行完成执行B,B需要A的结果,同时任务B有返回值
        System.out.println(CompletableFuture.supplyAsync(() -> "resultA").thenApply(resultA -> resultA + " resultB").join());

        //这个方法表示的是,谁快就用谁的结果,类似于我们在打跑得快,或者麻将谁赢了就返回给谁
        //public <U> CompletableFuture<U> applyToEither(CompletionStage<? extends T> other, Function<? super T, U> fn);
        //下面这个在第一个中停留1s,在第二种停留2s,返回的结果是1
        System.out.println(CompletableFuture.supplyAsync(() -> {
            //暂停几秒钟线程
            try { TimeUnit.SECONDS.sleep(1);  } catch (InterruptedException e) {e.printStackTrace();}
            return 1;
        }).applyToEither(CompletableFuture.supplyAsync(() -> {
            try { TimeUnit.SECONDS.sleep(2);  } catch (InterruptedException e) {e.printStackTrace();}
            return 2;
        }), r -> {
            return r;
        }).join());
        //暂停几秒钟线程
        try { TimeUnit.SECONDS.sleep(3); } catch (InterruptedException e) { e.printStackTrace(); }

        //public <U,V> CompletableFuture<V> thenCombine
        //(CompletionStage<? extends U> other,BiFunction<? super T,? super U,? extends V> fn)
        //两个CompletionStage任务都完成后,最终把两个任务的结果一起交给thenCombine来处理
        //先完成的先等着,等待其他分支任务
        System.out.println(CompletableFuture.supplyAsync(() -> {
            return 10;
        }).thenCombine(CompletableFuture.supplyAsync(() -> {
            return 20;
        }), (r1, r2) -> {
            return r1 + r2;
        }).thenCombine(CompletableFuture.supplyAsync(() -> {
            return 30;
        }), (r3, r4) -> {
            return r3 + r4;
        }).join());
        System.out.println(CompletableFuture.supplyAsync(() -> {
            return 10;
        }).thenCombine(CompletableFuture.supplyAsync(() -> {
            return 20;
        }), (r1, r2) -> {
            return r1 + r2;
        }).join());