blocks|key|4641803|text|这取决于实现方式。这不是必须的，但它很有可能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4641804|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It depends on the implementation. It doesn't have to, but it very well could.

blocks|key|1554529|text|我为此创建了一些快速测试代码，似乎IDisposable将保留一个引用并将IObservable保存在内存中。或者至少可以，就像Jason所说的那样。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1554530|Subject<int>+subject+=+new+Subject<int>();
WeakReference+wr+=+new+WeakReference(subject);

IDisposable+disposable+=+subject.Subscribe(i+=>+Console.WriteLine("Next:+"+%2B+i));

subject.OnNext(5);
subject.OnNext(10);
subject+=+null;

GC.Collect();
Console.WriteLine(wr.IsAlive+?+"Subject+is+still+alive."+:+"Subject+is+not+alive.");|code-block|syntax|javascript|1554531|当按照上面的方式编写时，弱引用仍然是活动的。如果我注释掉可处理的赋值，那么subject就会被GC‘。|1554532|/*IDisposable+disposable+=+*/+subject.Subscribe(i+=>+Console.WriteLine("Next:+"+%2B+i));|1554533|entityMap^0|H|B|11|B|0|0|11|7|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

I created some quick test code for this, and it seems that the <code>IDisposable</code> will keep a reference and hold the <code>IObservable</code> in memory. Or at least very well could, as Jason says.

<pre><code>Subject&lt;int&gt; subject = new Subject&lt;int&gt;();
WeakReference wr = new WeakReference(subject);

IDisposable disposable = subject.Subscribe(i =&gt; Console.WriteLine("Next: " + i));

subject.OnNext(5);
subject.OnNext(10);
subject = null;

GC.Collect();
Console.WriteLine(wr.IsAlive ? "Subject is still alive." : "Subject is not alive.");
</code></pre>

When written as above, the weak reference will still be alive. If I comment out the disposable assignment, then <code>subject</code> gets GC'ed.

<pre><code>/*IDisposable disposable = */ subject.Subscribe(i =&gt; Console.WriteLine("Next: " + i));
</code></pre>

blocks|key|604859|text|这里有一个测试，它似乎表明可观察物是GC‘’ed的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|604860|var+xs+=+Observable.Timer(TimeSpan.FromSeconds(1.0));
var+disposable+=+xs.Subscribe(x+=>+Console.WriteLine(x));

var+wr+=+new+WeakReference(xs);
xs+=+null;

Thread.Sleep(2000);

Console.WriteLine(wr.IsAlive);+//+True

GC.Collect();

Console.WriteLine(wr.IsAlive);+//+False|code-block|syntax|javascript|604861|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a test that seems to show that the observable is GC'ed:

<pre><code>var xs = Observable.Timer(TimeSpan.FromSeconds(1.0));
var disposable = xs.Subscribe(x =&gt; Console.WriteLine(x));

var wr = new WeakReference(xs);
xs = null;

Thread.Sleep(2000);

Console.WriteLine(wr.IsAlive); // True

GC.Collect();

Console.WriteLine(wr.IsAlive); // False
</code></pre>

Does an <code>IDisposable</code> from <code>IObservable.Subscribe()</code> hold a reference to the <code>IObservable</code>?

If the <code>IDisposable</code> is rooted and can't be garbage collected, does it hold a reference to the subscription and the <code>IObservable</code>?

In other words, if the intended lifetime of the <code>IObservable</code> is shorter than the lifetime of the returned <code>IDisposable</code>, can the <code>IObservable</code> be GC'ed?

Does an IDisposable from IObservable.Subscribe() hold a reference to the source?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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来自IObservable.Subscribe()的IDisposable是否包含对IObservable的引用如果IDisposable是根的并且不能被垃圾收集，它是否包含对订阅和IObservable的引用换句话说，如果IObservable的预期生存期比返回的IDisposable的生存期短，那么IObserva...

问来自IObservable.Subscribe()的IDisposable是否包含对源的引用？
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问来自IObservable.Subscribe()的IDisposable是否包含对源的引用？EN