[Haskell] IVars

[Lennart Augustsson]

That argument doesn't totally fly since non-termination isn't considered an
effect in Haskell.  Bottom doesn't commute with a IO operations normally
either.
But not having readIVar return 'IO a' does make me a little quesy. :)

  -- Lennart

On Dec 4, 2007 8:25 AM, Simon Peyton-Jones <simonpj at microsoft.com> wrote:

>  But since the read may block, it matters **when** you perform it.  For
> example if you print "Hello" and then read the IVar, you'll block after
> printing; but if you read the IVar and then print, the print won't come
> out.  If the operation was pure (no IO) then you'd have a lot less control
> over when it happened.
>
>
>
> Simon
>
>
>
> *From:* haskell-bounces at haskell.org [mailto:haskell-bounces at haskell.org] *On
> Behalf Of *Lennart Augustsson
> *Sent:* 04 December 2007 08:19
> *To:* Conal Elliott
> *Cc:* haskell at haskell.org
> *Subject:* Re: [Haskell] IVars
>
>
>
> Good question.  That must be a matter of taste, because as you say the
> read will always produce the same result.  But it sill is a bit of a strange
> operation.
>
>   -- Lennart
>
> On Dec 4, 2007 6:25 AM, Conal Elliott < conal at conal.net> wrote:
>
> Oh.  Simple enough.  Thanks.
>
> Another question:  why the IO in readIVar :: IVar a -> IO a, instead of
> just readIVar :: IVar a -> a?  After all, won't readIVar iv yield the same
> result (eventually) every time it's called?
>
>
>
> On Dec 3, 2007 12:29 AM, Lennart Augustsson <lennart at augustsson.net>
> wrote:
>
> You can make them from MVars.
>
> On Dec 2, 2007 8:03 PM, Conal Elliott <conal at conal.net> wrote:
>
>  what became of (assign-once) IVars?  afaict, they were in concurrent
> haskell and now aren't.
>
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