Re: [Oorexx-devel] Multithreading question

[Rick M. <obj...@gm...>]

On Fri, Jul 1, 2011 at 10:46 AM, Rony G. Flatscher
<Ron...@wu...> wrote:
>
> On 01.07.2011 16:32, Rick McGuire wrote:
>
> Guarded state nests.  So invoking an method that is unguarded does not
> remove any guard protections of any other methods that might be on the
> call stack.  It would be absolutely wrong for the interpreter to do
> so, as the method in question has requested exclusive access to the
> object.  Having the guarded state come and go based on any subsequent
> method invocations would violate that contract.
>
>
> Thank you very much for this clarification!
>
> Maybe two last questions in this context:
>
> if executing a SetGuardOn() multiple times in the same native method and
> invocation, followed by a single SetGuardOff(), would that single
> SetGuardOff() remove the guard lock? Or would one need to invoke
> SetGuardOff() as many times as SetGuardOn()?

The lock created by SetGuardOn() is specific to that method
invocation, so it does not nest.  That should work ok (but a bit
strange, in my opinion).


>
> Would it be o.k. for a native method to do a SetGuardOff() at the entry into
> the native method and a SetGuardOn() right before returning from the native
> method in order to reliably define a defined state at the beginning of it?

Not sure I understand what you're trying to gain there.  Calling
SetGuardOn() right before returning is pointless since the method
termination will immediately release the lock.  I don't see how this
does anything to reliably define anything.  Once you call
SetGuardOff() at the beginning, then all bets are off regarding
reliable state.

Rick

>
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>
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