我刚刚查看了源代码CompletableFuture.allOf
,发现它基本上创建了一次处理两个阶段的节点的二叉树。我们可以轻松实现类似的逻辑,而无需使用toCompletableFuture()
显式地并一次性处理结果列表的生成:
public static <T> CompletionStage<List<T>> allOf(
Stream<? extends CompletionStage<? extends T>> source) {
return allOf(source.collect(Collectors.toList()));
}
public static <T> CompletionStage<List<T>> allOf(
List<? extends CompletionStage<? extends T>> source) {
int size = source.size();
if(size == 0) return CompletableFuture.completedFuture(Collections.emptyList());
List<T> result = new ArrayList<>(Collections.nCopies(size, null));
return allOf(source, result, 0, size-1).thenApply(x -> result);
}
private static <T> CompletionStage<Void> allOf(
List<? extends CompletionStage<? extends T>> source,
List<T> result, int from, int to) {
if(from < to) {
int mid = (from+to)>>>1;
return allOf(source, result, from, mid)
.thenCombine(allOf(source, result, mid+1, to), (x,y)->x);
}
return source.get(from).thenAccept(t -> result.set(from, t));
}
就是这样。
您可以使用此解决方案来实现问题代码的逻辑,如下所示
public static CompletionStage<List<String>> translateAllAsync(List<String> input) {
return allOf(input.stream().map(s -> translate(s)))
.thenApply(list -> list.stream()
.map(s -> s.toUpperCase())
.collect(Collectors.toList()));
}
虽然使用它会更自然
public static CompletionStage<List<String>> translateAllAsync(List<String> input) {
return allOf(input.stream().map(s -> translate(s).thenApply(String::toUpperCase)));
}
请注意,此解决方案保持顺序,因此无需对测试用例中的结果进行排序:
@Test
public void testTranslate() throws Exception {
List<String> list = translateAllAsync(Arrays.asList("foo", "bar")).toCompletableFuture().get();
assertEquals(list, Arrays.asList("TRANSLATED FOO", "TRANSLATED BAR"));
}