What would Squeak be like without non-local returns
siguctua at gmail.com
Wed Nov 7 07:29:09 UTC 2007
On 07/11/2007, Andreas Raab <andreas.raab at gmx.de> wrote:
> Igor Stasenko wrote:
> > Let suppose that Mr.Satan will rewrite an example above as following:
> > object eventual ifTrue: [100 atRandom seconds asDelay wait. ^3].
> > 100 atRandom seconds asDelay wait.
> > ^4
> > When, in real case, a random delay could be replaced by any code which
> > needs a time to evaluate.
> > What is the answer then? :)
> Same answer (though strictly speaking you may get an error that you
> can't *wait* in an eventual system).
This doesn't change anything, you may put:
(1 to: 1000000 atRandom) do: [1+1]
instead of delay.
>The reason being that <object> will
> only get an eventual ref to the block so it will only be able to send
> other eventual messages to it. Put differently, ifTrue: would need to be
> implemented as:
> True>>ifTrue: aBlock
> "I am true, so evaluate aBlock"
> aBlock isBlock "if it's an immediate block..."
> ifTrue:[aBlock value] "... evaluate it directly ..."
> ifFalse:[aBlock eventual value] "... otherwise eventual"
> (I'm ignoring the issues of meta-circularity for ifTrue: here; the above
> is purely conceptual) The main point is that because the evaluation of
> the block would be eventual it would *necessarily* evaluate after the
> immediate delay and the entire message has completed *regardless* of how
> long or how short that delay is. If that were different it would violate
> one of the fundamental axioms of event-loop concurrency (the fact that
> only one message is ever executed at the time) and wouldn't count as
> message passing concurrency anymore.
Hmm. An example above was to show that when running in parallel there
is equal chance to get to ^3 or ^4 first.
And in previous message you described behavior, when running code gets
to ^4 first. But what if it gets to ^3 first?
And if there no code running in parallel, then there's nothing to argue about.
> - Andreas
Igor Stasenko AKA sig.
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