注意:如果重要的话,我使用的是3.1。此外,我已经在codereview上问过这个问题,但到目前为止还没有回应。
我需要一个操作符来允许一个布尔流作为另一个流的门(当门流为真时,让值通过,当门流为假时,丢弃它们)。我通常会使用Switch,但如果源流是冷的,它会不断地重新创建它,这是我不想要的。
我还想自己清理一下,这样如果源或门都完成了,结果就会完成。
public static IObservable<T> When<T>(this IObservable<T> source, IObservable<bool> gate)
{
var s = source.Publish().RefCount();
var g = gate.Publish().RefCount();
var sourceCompleted = s.TakeLast(1).DefaultIfEmpty().Select(_ => Unit.Default);
var gateCompleted = g.TakeLast(1).DefaultIfEmpty().Select(_ => Unit.Default);
var anyCompleted = Observable.Amb(sourceCompleted, gateCompleted);
var flag = false;
g.TakeUntil(anyCompleted).Subscribe(value => flag = value);
return s.Where(_ => flag).TakeUntil(anyCompleted);
}除了总体的冗长之外,我不喜欢我订阅gate,即使结果从来没有订阅过(在这种情况下,这个操作符应该是一个no-op)。有没有办法摆脱那些订阅者?
我也尝试过这个实现,但当涉及到自我清理时,它甚至更糟糕:
return Observable.Create<T>(
o =>
{
var flag = false;
gate.Subscribe(value => flag = value);
return source.Subscribe(
value =>
{
if (flag) o.OnNext(value);
});
});以下是我用来检查实现的测试:
[TestMethod]
public void TestMethod1()
{
var output = new List<int>();
var source = new Subject<int>();
var gate = new Subject<bool>();
var result = source.When(gate);
result.Subscribe(output.Add, () => output.Add(-1));
// the gate starts with false, so the source events are ignored
source.OnNext(1);
source.OnNext(2);
source.OnNext(3);
CollectionAssert.AreEqual(new int[0], output);
// setting the gate to true will let the source events pass
gate.OnNext(true);
source.OnNext(4);
CollectionAssert.AreEqual(new[] { 4 }, output);
source.OnNext(5);
CollectionAssert.AreEqual(new[] { 4, 5 }, output);
// setting the gate to false stops source events from propagating again
gate.OnNext(false);
source.OnNext(6);
source.OnNext(7);
CollectionAssert.AreEqual(new[] { 4, 5 }, output);
// completing the source also completes the result
source.OnCompleted();
CollectionAssert.AreEqual(new[] { 4, 5, -1 }, output);
}
[TestMethod]
public void TestMethod2()
{
// completing the gate also completes the result
var output = new List<int>();
var source = new Subject<int>();
var gate = new Subject<bool>();
var result = source.When(gate);
result.Subscribe(output.Add, () => output.Add(-1));
gate.OnCompleted();
CollectionAssert.AreEqual(new[] { -1 }, output);
}发布于 2018-06-04 20:58:56
这是可行的:
public static IObservable<T> When<T>(this IObservable<T> source, IObservable<bool> gate)
{
return
source.Publish(ss => gate.Publish(gs =>
gs
.Select(g => g ? ss : ss.IgnoreElements())
.Switch()
.TakeUntil(Observable.Amb(
ss.Select(s => true).Materialize().LastAsync(),
gs.Materialize().LastAsync()))));
}这两个测试都通过了。
发布于 2018-06-03 21:41:08
你和Observable.Create的关系是正确的。您应该从可观察对象的两个订阅中调用onError和onCompleted,以便正确完成或在需要时出错。另外,通过在Create委托中返回两个IDisposable,如果您打算在source或gate完成之前处理When订阅,则可以确保两个订阅都被正确清理。
public static IObservable<T> When<T>(this IObservable<T> source, IObservable<bool> gate)
{
return Observable.Create<T>(
o =>
{
var flag = false;
var gs = gate.Subscribe(
value => flag = value,
e => o.OnError(e),
() => o.OnCompleted());
var ss = source.Subscribe(
value =>
{
if (flag) o.OnNext(value);
},
e => o.OnError(e),
() => o.OnCompleted());
return new CompositeDisposable(gs, ss);
});
}一个更短,但更难阅读的版本,只使用Rx运算符。对于冷可观察对象,它可能需要源的发布/引用计数。
public static IObservable<T> When<T>(this IObservable<T> source, IObservable<bool> gate)
{
return gate
.Select(g => g ? source : source.IgnoreElements())
.Switch()
.TakeUntil(source.Materialize()
.Where(s => s.Kind == NotificationKind.OnCompleted));
}https://stackoverflow.com/questions/50666864
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