Squeak process implicitly yields?

Masashi Umezawa umejava at mars.dti.ne.jp
Mon Apr 26 17:38:48 UTC 2004


Hi,

Mon, 26 Apr 2004 18:01:56 +0200
"Andreas Raab" <andreas.raab at gmx.de> wrote:
> > In my understanding, Smalltalk processes are non-preemptive if their
> > priorities are the same.
> [...]
> > Are there implicit yields in Squeak?
> 
> Yes. While it is true that Squeak's processes are non-preemptive if they run
> at the same priority their behavior is somewhat out of the ordinary when
> they get preempted by a higher priority process. Rather than being added as
> the first process in the priority group (which would give you the expected
> behavior) it is added as the last process thereby implicitly yielding to the
> next process in this priority group.
> 

Aha-. Now I see the results. Thanks!

So, is it intentional? 
I'm not sure what is the merit. Wouldn't it make the squeak code
non-portable?

Cheers,
---
[:masashi | ^umezawa]




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