[Haskell] IVars

[Lennart Augustsson]

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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