CompletableFuture:等待第一个正常返回？
CompletableFuture:等待第一个正常返回？
提问于 2015-11-25 17:47:30
我有一些CompletableFuture，我想并行运行它们，等待第一个返回的，通常是。
我知道我可以使用CompletableFuture.anyOf来等待第一个返回，但这通常会返回或我想忽略异常。
List<CompletableFuture<?>> futures = names.stream().map(
(String name) ->
CompletableFuture.supplyAsync(
() ->
// this calling may throw exceptions.
new Task(name).run()
)
).collect(Collectors.toList());
//FIXME Can not ignore exceptionally returned takes.
Future any = CompletableFuture.anyOf(futures.toArray(new CompletableFuture<?>[]{}));
try {
logger.info(any.get().toString());
} catch (Exception e) {
e.printStackTrace();
}回答 4
Stack Overflow用户
发布于 2015-12-09 03:03:13
您可以使用以下辅助方法：
public static <T>
CompletableFuture<T> anyOf(List<? extends CompletionStage<? extends T>> l) {
CompletableFuture<T> f=new CompletableFuture<>();
Consumer<T> complete=f::complete;
l.forEach(s -> s.thenAccept(complete));
return f;
}您可以像这样使用它，以演示它将忽略先前的异常，但返回第一个提供的值：
List<CompletableFuture<String>> futures = Arrays.asList(
CompletableFuture.supplyAsync(
() -> { throw new RuntimeException("failing immediately"); }
),
CompletableFuture.supplyAsync(
() -> { LockSupport.parkNanos(TimeUnit.SECONDS.toNanos(5));
return "with 5s delay";
}),
CompletableFuture.supplyAsync(
() -> { LockSupport.parkNanos(TimeUnit.SECONDS.toNanos(10));
return "with 10s delay";
})
);
CompletableFuture<String> c = anyOf(futures);
logger.info(c.join());这种解决方案的一个缺点是，如果所有期货都异常完成，它将永远不会完成。一个解决方案，如果有成功的计算，将提供第一个值，但如果根本没有成功的计算，则会异常失败，这是一个更复杂的解决方案：
public static <T>
CompletableFuture<T> anyOf(List<? extends CompletionStage<? extends T>> l) {
CompletableFuture<T> f=new CompletableFuture<>();
Consumer<T> complete=f::complete;
CompletableFuture.allOf(
l.stream().map(s -> s.thenAccept(complete)).toArray(CompletableFuture<?>[]::new)
).exceptionally(ex -> { f.completeExceptionally(ex); return null; });
return f;
}它利用了这样一个事实，即allOf的异常处理程序仅在所有期货都已完成(异常或未完成)后才会被调用，并且未来只能完成一次(抛开obtrude…之类的特殊事情不谈)。当执行异常处理程序时，任何使用结果来完成未来的尝试都已经完成，因此，只有在之前没有成功完成的情况下，才会尝试异常地完成它。
它可以与第一个解决方案完全相同的方式使用，并且只有在所有计算失败时才会表现出不同的行为，例如：
List<CompletableFuture<String>> futures = Arrays.asList(
CompletableFuture.supplyAsync(
() -> { throw new RuntimeException("failing immediately"); }
),
CompletableFuture.supplyAsync(
// delayed to demonstrate that the solution will wait for all completions
// to ensure it doesn't miss a possible successful computation
() -> { LockSupport.parkNanos(TimeUnit.SECONDS.toNanos(5));
throw new RuntimeException("failing later"); }
)
);
CompletableFuture<String> c = anyOf(futures);
try { logger.info(c.join()); }
catch(CompletionException ex) { logger.severe(ex.toString()); }上面的例子使用了一个延迟，演示了当没有成功时，解决方案将等待所有的完成，而this example on ideone将演示稍后的成功将如何将结果转化为成功。请注意，由于Ideones缓存结果，您可能不会注意到延迟。
请注意，在所有期货都失败的情况下，不能保证将报告哪些异常。由于它在错误的情况下等待所有的完成，所以任何一个都可以到达最终结果。
Stack Overflow用户
发布于 2015-12-09 03:01:03
嗯，这是一种框架应该支持的方法。首先，我以为CompletionStage.applyToEither会做类似的事情，但事实证明并非如此。所以我想出了这个解决方案：
public static <U> CompletionStage<U> firstCompleted(Collection<CompletionStage<U>> stages) {
final int count = stages.size();
if (count <= 0) {
throw new IllegalArgumentException("stages must not be empty");
}
final AtomicInteger settled = new AtomicInteger();
final CompletableFuture<U> future = new CompletableFuture<U>();
BiConsumer<U, Throwable> consumer = (val, exc) -> {
if (exc == null) {
future.complete(val);
} else {
if (settled.incrementAndGet() >= count) {
// Complete with the last exception. You can aggregate all the exceptions if you wish.
future.completeExceptionally(exc);
}
}
};
for (CompletionStage<U> item : stages) {
item.whenComplete(consumer);
}
return future;
}要查看它的实际效果，这里有一些用法：
import java.util.ArrayList;
import java.util.Collection;
import java.util.List;
import java.util.concurrent.CompletableFuture;
import java.util.concurrent.CompletionStage;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.Executors;
import java.util.concurrent.ScheduledExecutorService;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.atomic.AtomicInteger;
import java.util.function.BiConsumer;
public class Main {
public static <U> CompletionStage<U> firstCompleted(Collection<CompletionStage<U>> stages) {
final int count = stages.size();
if (count <= 0) {
throw new IllegalArgumentException("stages must not be empty");
}
final AtomicInteger settled = new AtomicInteger();
final CompletableFuture<U> future = new CompletableFuture<U>();
BiConsumer<U, Throwable> consumer = (val, exc) -> {
if (exc == null) {
future.complete(val);
} else {
if (settled.incrementAndGet() >= count) {
// Complete with the last exception. You can aggregate all the exceptions if you wish.
future.completeExceptionally(exc);
}
}
};
for (CompletionStage<U> item : stages) {
item.whenComplete(consumer);
}
return future;
}
private static final ScheduledExecutorService worker = Executors.newSingleThreadScheduledExecutor();
public static <U> CompletionStage<U> delayed(final U value, long delay) {
CompletableFuture<U> future = new CompletableFuture<U>();
worker.schedule(() -> {
future.complete(value);
}, delay, TimeUnit.MILLISECONDS);
return future;
}
public static <U> CompletionStage<U> delayedExceptionally(final Throwable value, long delay) {
CompletableFuture<U> future = new CompletableFuture<U>();
worker.schedule(() -> {
future.completeExceptionally(value);
}, delay, TimeUnit.MILLISECONDS);
return future;
}
public static void main(String[] args) throws InterruptedException, ExecutionException {
System.out.println("Started...");
/*
// Looks like applyToEither doesn't work as expected
CompletableFuture<Integer> a = CompletableFuture.completedFuture(99);
CompletableFuture<Integer> b = Main.<Integer>completedExceptionally(new Exception("Exc")).toCompletableFuture();
System.out.println(b.applyToEither(a, x -> x).get()); // throws Exc
*/
try {
List<CompletionStage<Integer>> futures = new ArrayList<>();
futures.add(Main.<Integer>delayedExceptionally(new Exception("Exception #1"), 100));
futures.add(Main.<Integer>delayedExceptionally(new Exception("Exception #2"), 200));
futures.add(delayed(1, 1000));
futures.add(Main.<Integer>delayedExceptionally(new Exception("Exception #4"), 400));
futures.add(delayed(2, 500));
futures.add(Main.<Integer>delayedExceptionally(new Exception("Exception #5"), 600));
Integer value = firstCompleted(futures).toCompletableFuture().get();
System.out.println("Completed normally: " + value);
} catch (Exception ex) {
System.out.println("Completed exceptionally");
ex.printStackTrace();
}
try {
List<CompletionStage<Integer>> futures = new ArrayList<>();
futures.add(Main.<Integer>delayedExceptionally(new Exception("Exception B#1"), 400));
futures.add(Main.<Integer>delayedExceptionally(new Exception("Exception B#2"), 200));
Integer value = firstCompleted(futures).toCompletableFuture().get();
System.out.println("Completed normally: " + value);
} catch (Exception ex) {
System.out.println("Completed exceptionally");
ex.printStackTrace();
}
System.out.println("End...");
}
}Stack Overflow用户
发布于 2021-08-05 15:02:29
对上面的代码做了一些修改，允许测试第一个结果是否是预期的。
public class MyTask implements Callable<String> {
@Override
public String call() throws Exception {
int randomNum = ThreadLocalRandom.current().nextInt(5, 20 + 1);
for (int i = 0; i < randomNum; i++) {
TimeUnit.SECONDS.sleep(1);
}
return "MyTest" + randomNum;
}
}
public class CompletableFutureUtils {
private static <T> T resolve(FutureTask<T> futureTask) {
try {
futureTask.run();
return futureTask.get();
} catch (Exception e) {
throw new RuntimeException(e);
}
}
private static <V> boolean predicate(Predicate<V> predicate, V v) {
try {
return predicate.test(v);
} catch (Exception e) {
return false;
}
}
public static <T> void cancel(List<FutureTask<T>> futureTasks) {
if (futureTasks != null && futureTasks.isEmpty() == false) {
futureTasks.stream().filter(f -> f.isDone() == false).forEach(f -> f.cancel(true));
}
}
public static <V> CompletableFuture<V> supplyAsync(List<FutureTask<V>> futureTasks, Predicate<V> predicate) {
return supplyAsync(futureTasks, predicate, null);
}
public static <V> CompletableFuture<V> supplyAsync(List<FutureTask<V>> futureTasks, Predicate<V> predicate,
Executor executor) {
final int count = futureTasks.size();
final AtomicInteger settled = new AtomicInteger();
final CompletableFuture<V> result = new CompletableFuture<V>();
final BiConsumer<V, Throwable> action = (value, ex) -> {
settled.incrementAndGet();
if (result.isDone() == false) {
if (ex == null) {
if (predicate(predicate, value)) {
result.complete(value);
cancel(futureTasks);
} else if (settled.get() >= count) {
result.complete(null);
}
} else if (settled.get() >= count) {
result.completeExceptionally(ex);
}
}
};
for (FutureTask<V> futureTask : futureTasks) {
if (executor != null) {
CompletableFuture.supplyAsync(() -> resolve(futureTask), executor).whenCompleteAsync(action, executor);
} else {
CompletableFuture.supplyAsync(() -> resolve(futureTask)).whenCompleteAsync(action);
}
}
return result;
}
}
public class DemoApplication {
public static void main(String[] args) {
List<FutureTask<String>> tasks = new ArrayList<FutureTask<String>>();
for (int i = 0; i < 2; i++) {
FutureTask<String> task = new FutureTask<String>(new MyTask());
tasks.add(task);
}
Predicate<String> test = (s) -> true;
CompletableFuture<String> result = CompletableFutureUtils.supplyAsync(tasks, test);
try {
String s = result.get(20, TimeUnit.SECONDS);
System.out.println("result=" + s);
} catch (Exception e) {
e.printStackTrace();
CompletableFutureUtils.cancel(tasks);
}
}
}调用CompletableFutureUtils.cancel(tasks);非常重要，因为当超时发生时，它将取消后台任务。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/33913193
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