在我的项目中,我使用 Java 中 vert.x 的 Future
实现。到目前为止一切都很顺利。但是,目前我在按顺序对对象列表执行操作时遇到问题。问题在于 Java 中的 reduce
方法分别“减少”和“组合”结果。这导致所有操作同时启动。正如在 easy
方法中所见,实现顺序执行是可能的。
private Future<Void> action(String object) {
System.out.println("started " + object);
Future<Void> f = Future.future();
vertx.setTimer(1000, res -> {
System.out.println("doing " + object);
f.complete();
});
return f;
}
private void easy() {
action("one")
.compose(ignore -> action("two"))
.compose(ignore -> action("three"))
.setHandler(ignore -> System.out.println("completed"));
}
private void list() {
List<String> l = new ArrayList<>();
l.add("one");
l.add("two");
l.add("three");
Future<Void> f = Future.future();
l.stream().reduce(f,
(f1, s) -> action(s),
(f1, f2) -> f2.compose(ignore -> f1)
).setHandler(res -> {
System.out.println("completed");
});
}
执行easy时的输出:
started one
doing one
started two
doing two
started three
doing three
completed
执行列表时的输出:
started one
started two
started three
doing one
doing two
doing three
completed
Javascript 中的相同代码片段也有效,因为reduce 函数在一步中完成减少和合并:
function action(object) {
return new Promise((resolve, reject) => {
console.log("started " + object)
window.setTimeout(() => {
console.log("doing " + object);
resolve()
}, 1000);
});
}
function easy() {
action("one")
.then(() => action("two"))
.then(() => action("three"))
.then(() => console.log("completed"));
}
function list() {
l = ["one", "two", "three"]
l.reduce((p, s) => p.then(() => action(s)), Promise.resolve())
.then(() => console.log("completed"));
}
// easy()
list()
easy
和list
的输出与Java代码的easy方法相同。我正在寻找的是一种修复Java中的reduce方法的方法,或者是实现相同结果的替代方法。
最佳答案
好的。我找到了 FoldLeft 方法的实现 here现在顺序执行工作正常...
private void list() {
List<String> l = new ArrayList<>();
l.add("one");
l.add("two");
l.add("three");
Future<Void> f = Future.succeededFuture();
foldLeft(l.iterator(), f,
(f1, s) -> f1.compose(ignore -> action(s)))
.setHandler(res -> {
System.out.println("completed");
});
}
private static <A, B> B foldLeft(Iterator<A> iterator, B identity, BiFunction<B, A, B> bf) {
B result = identity;
while(iterator.hasNext()) {
A next = iterator.next();
result = bf.apply(result, next);
}
return result;
}
关于java - 在 vert.x 中按顺序减少 Future,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51513678/