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MemoryStream.Close() or MemoryStream.Dispose()

Which one do I call?

Is it necessary to call both?

Will the other throw an exception if I have already called one of them?

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ashwnacharya Avatar asked Nov 25 '10 07:11

ashwnacharya


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1 Answers

Close() and Dispose(), when called on a MemoryStream, only serve to do two things:

  • Mark the object disposed so that future accidental usage of the object will throw an exception.
  • Possibly1 release references to managed objects, which can make the GC's job a bit easier depending on the GC implementation. (On today's GC algorithms it makes no real difference, so this is a point for an academic discussion and has no significant real-world impact.)

MemoryStream does not have any unmanaged resources to dispose, so you don't technically have to dispose of it. The effect of not disposing a MemoryStream is roughly the same thing as dropping a reference to a byte[] -- the GC will clean both up the same way.

Which one do I call? Is it necessary to call both?

The Dispose() method of streams delegate directly to the Close() method2, so both do exactly the same thing.

Will the other throw an exception if I have already called one of them?

The documentation for IDisposable.Dispose() specifically states it is safe to call Dispose() multiple times, on any object3. (If that is not true for a particular class then that class implements the IDisposable interface in a way that violates its contract, and this would be a bug.)

All that to say: it really doesn't make a huge difference whether you dispose a MemoryStream or not. The only real reason it has Close/Dispose methods is because it inherits from Stream, which requires those methods as part of its contract to support streams that do have unmanaged resources (such as file or socket descriptors).


1Mono's implementation does not release the byte[] reference. I don't know if the Microsoft implementation does.

2"This method calls Close, which then calls Stream.Dispose(Boolean)."

3"If an object's Dispose method is called more than once, the object must ignore all calls after the first one."

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cdhowie Avatar answered Oct 21 '22 20:10

cdhowie