java8学习：CompletableFuture方法使用

下面是自己总结的一部分的api的使用,当然下面有些方法后面加上Async就会变为异步的

/**
 * 变换:参数为Function接口
 * thenApply:将上一个结果当做这次thenApply的参数传入,加以处理后,返回一个值
 * 可以传入T返回T也可以传入T返回U
 */
public void thenApply() {
    CompletableFuture<String> future = CompletableFuture.supplyAsync(()-> 1).thenApply(i -> String.valueOf(i));
    String i = future.join();
    System.out.println(i); //1
}

/**
 * 消费
 * 参数为Consumer接口
 * thenAccept:将上一个的结果当做这次的参数传入,但是无返回值
 */
public void thenAccept() {
    CompletableFuture<Void> future = CompletableFuture.supplyAsync(() -> 1).thenAccept((integer -> System.out.println(integer)));
    future.join();  //1
}

public void thenCompose() {
    CompletableFuture<Integer> future = CompletableFuture.supplyAsync(() -> 1) //提供第一个CompletableFuture结果值
            .thenCompose((i) ->    //将上一个结果值传入下一个CompletableFuture
                    CompletableFuture.supplyAsync(() -> i + 2));  //对传入的结果值进行处理
    Integer join = future.join();
    System.out.println(join);
}

/**
 * 结合两个结果得到一个最终值
 * 逻辑处理需要的是BiFunction接口
 * 需要两个CompletableFuture结果,并且需要有处理两个结果的逻辑
 * 两个CompletableFuture返回的结果不需要一致
 */
public void thenCombine() {
    CompletableFuture<String> future = CompletableFuture.supplyAsync(() -> "Hello ")   //第一个CompletableFuture返回结果
            .thenCombine(CompletableFuture.supplyAsync(() -> 1),          //第二个CompletableFuture返回结果
                    (future1, future2) -> future1 + future2);      //处理两个结果的逻辑
    String join = future.join();
    System.out.println(join); //Hello World
}

/**
 * 跟上面的差不多,只不过这里处理逻辑需要的是一个BiConsumer接口,无返回值而已
 */
public void thenAcceptBoth() {
    CompletableFuture<Void> both = CompletableFuture.supplyAsync(() -> "Hello ")   //第一个CompletableFuture返回结果
            .thenAcceptBoth(CompletableFuture.supplyAsync(() -> "World"), //第二个CompletableFuture返回结果
                    (future1, future2) -> System.out.println(future1 + future2));//处理两个结果的逻辑
    both.join(); //Hello World
}

/**
 * 两个CompletableFuture一起执行,哪个先执行完就用哪个CompletableFuture的结果
 * 处理逻辑需要的参数是一个Function接口
 * 下面整体代码的逻辑是返回1的睡眠一秒,返回2的睡眠0.3秒,所以返回2的肯定比1要快,所以结果为2,然后应用到处理逻辑上
 * 最终结果就是为 2 done
 * 两个CompletableFuture返回的参数需要一致
 * 这适用于两种渠道完成同一个事情，就可以调用这个方法，找一个最快的结果进行处理，最终有返回值。
 */
public void applyToEither() {
    CompletableFuture<String> apply = CompletableFuture.supplyAsync(() -> {
        try {
            Thread.sleep(1000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        return "1";
    }).applyToEither(
            CompletableFuture.supplyAsync(() -> {
                try {
                    Thread.sleep(300);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                return "2";
            }),
            (future) -> future + " done"      //对于首先完成的CompletableFuture任务之后执行的逻辑
    );
    String join = apply.join();
    System.out.println(join);//2 done
}

/**
 * 与上面使用方法一致,只是多加了一个选项
 * 结果为3 three
 */
public void applyToEithers() {
    CompletableFuture<String> apply = CompletableFuture.supplyAsync(() -> {
        try {
            Thread.sleep(1000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        return "1";
    }).applyToEither(
            CompletableFuture.supplyAsync(() -> {
                try {
                    Thread.sleep(800);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                return "2";
            }),(future) -> future + " done"      //对于首先完成的CompletableFuture任务之后执行的逻辑
    ).applyToEither(
            CompletableFuture.supplyAsync(() -> {
                try {
                    Thread.sleep(300);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                return "3";
            }),(future -> future + " three")
    );
    String join = apply.join();
    System.out.println(join);//2 done
}

/**
 * 对于出错的处理过程进行补偿,即如果future出错,会运行exceptionally里的Function函数
 * 当然没有错就不会运行后面指定的函数
 */
@Test
public void exceptionally() {
    CompletableFuture<String> exceptionally = CompletableFuture.supplyAsync(() -> {
        int i = 1 / 0 ;
        return "1";
    }).exceptionally(ex -> {
        ex.printStackTrace();
        return "error";
    });
    String join = exceptionally.join();
    System.out.println(join);
}

/**
 * 输出
 * error
 * java.util.concurrent.CompletionException: java.lang.ArithmeticException: / by zero
 *     at java.util.concurrent.CompletableFuture.encodeThrowable(CompletableFuture.java:273)
 *     at java.util.concurrent.CompletableFuture.completeThrowable(CompletableFuture.java:280)
 *     at java.util.concurrent.CompletableFuture$AsyncSupply.run(CompletableFuture.java:1592)
 *     at java.util.concurrent.CompletableFuture$AsyncSupply.exec(CompletableFuture.java:1582)
 *     at java.util.concurrent.ForkJoinTask.doExec(ForkJoinTask.java:289)
 *     at java.util.concurrent.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:1056)
 *     at java.util.concurrent.ForkJoinPool.runWorker(ForkJoinPool.java:1692)
 *     at java.util.concurrent.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:157)
 * Caused by: java.lang.ArithmeticException: / by zero
 *     at com.qidai.demotest.MyTestx.lambda$exceptionally$18(MyTestx.java:125)
 *     at java.util.concurrent.CompletableFuture$AsyncSupply.run(CompletableFuture.java:1590)
 *     ... 5 more
 */

/**
 * 与上面的thenAccept相比较,这个方法并不需要输入参数,相同点是都是无返回值
 */
public void thenRun() {
    CompletableFuture<Void> runnable = CompletableFuture.supplyAsync(() -> "1").thenRun(() -> System.out.println("runnable"));
    runnable.join();
}