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CompletableFuture的入门

2023-03-31 11:03:38 时间

runAsync 和 supplyAsync

runAsync接受一个Runable的实现,无返回值

CompletableFuture.runAsync(()->System.out.println("无返回结果的运行"));

supplyAsync接受一个Supplier的实现,有返回值

CompletableFuture<Integer> completableFuture = CompletableFuture.supplyAsync(() -> {
      System.out.println("有返回结果的运行");
      return 1;
  });

获取结果的get和join

都是堵塞,直到返回结果
get方法抛出是经过处理的异常,ExecutionException或**InterruptedException **,需要用户手动捕获

try {
   System.out.println(CompletableFuture.supplyAsync(() -> {
    System.out.println("有返回结果的运行");
    return 1;
  }).get());
} catch (InterruptedException e) {
  e.printStackTrace();
} catch (ExecutionException e) {
  e.printStackTrace();
}

join方法抛出的就不用捕获,是经过包装的**CompletionException **或 CancellationException

        System.out.println(CompletableFuture.supplyAsync(() -> {
            try {
                TimeUnit.MILLISECONDS.sleep(1000);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("有返回结果的运行");
            return 1;
        }).join());

常用方法

获取结果的getjoingetNow

get():一直等待
get(timeout,unit):等待,除非超时
getNow(valueIfAbsent):计算完返回计算的结果,未计算完返回默认的结果

CompletableFuture<Integer> completableFuture = CompletableFuture.supplyAsync(() -> {

            try {
                TimeUnit.SECONDS.sleep(1);
                ;
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            return 1;
        });

        System.out.println("立即获取:"+completableFuture.getNow(9999));
        try {
            TimeUnit.SECONDS.sleep(2);
            System.out.println("doing");
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("等一会获取:"+completableFuture.getNow(9999));

join() 同get()

thenApplyhandle

执行完前面的,前面返回的结果返回,然后传给后面再,执行完后面任务,一步一步来。

CompletableFuture<Integer> completableFuture = CompletableFuture.supplyAsync(() -> {
    System.out.println("step 1");
    return 1;
}).thenApply(a -> {
    System.out.println("step 2");
    return a + 2;
}).thenApply(a -> {
    System.out.println("step 3");
    return a + 3;
});
System.out.println(completableFuture.get());

执行结果:

CompletableFuture<Integer> completableFuture = CompletableFuture.supplyAsync(() -> {
    System.out.println("step 1");
    int a=1/0;
    return 1;
}).handle((a,b) -> {
    System.out.println("step 2");
    if (b!=null) {
        System.out.println(b.getMessage());
        return 0;
    }
    return a + 2;
}).handle((a,b) -> {
    System.out.println("step 3");
    if (b!=null) {
        System.out.println(b.getMessage());
        return 0;
    }
    return a + 3;
});
System.out.println(completableFuture.get());

执行结果:

thenApply和handle的区别:
thenApply执行的时候,有异常的则整个执行链会中断,直接抛出异常。

handle有异常也可以往下一步走,根据带的异常参数可以进一步处理

thenAccept

接收前面任务的返回结果,当前节点处理,并不返回结果。

CompletableFuture.supplyAsync(()->{
    System.out.println("step 1");
    return 10;
}).thenAccept(a->{
    System.out.println("res "+a);
});

applyToEither

在多个任务段同时执行时,哪个任务段用时最少,就返回哪个

CompletableFuture<Integer> completableFuture = CompletableFuture.supplyAsync(() -> {
    System.out.println("step 1");
    try {
        TimeUnit.SECONDS.sleep(1);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    return 1;
}).applyToEither(CompletableFuture.supplyAsync(() -> {
    System.out.println("step 2");
    try {
        TimeUnit.SECONDS.sleep(2);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    return 2;
}), a -> {
    return a;
});
System.out.println(completableFuture.get());

执行结果:

thenCombine

合并多个任务段的返回结果

CompletableFuture<Integer> completableFuture = CompletableFuture.supplyAsync(() -> {
            System.out.println("step 1");
            return IntStream.range(1, 11).sum();
        }).thenCombine(CompletableFuture.supplyAsync(() -> {
            System.out.println("step 2");
            return IntStream.range(11, 21).sum();
        }), (a, b) -> a + b)
        .thenCombine(CompletableFuture.supplyAsync(() -> {
            System.out.println("step 3");
            return IntStream.range(21, 31).sum();
        }), (a, b) -> a + b);
System.out.println(completableFuture.get());