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@haskell.org [mailto:haskell-bounces@haskell.org] On Behalf Of Lennart Augustsson
Sent: 04 December 2007 08:19
To: Conal Elliott
Cc: haskell@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@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@augustsson.net> wrote:

You can make them from MVars.

On Dec 2, 2007 8:03 PM, Conal Elliott <conal@conal.net> wrote:

what became of (assign-once) IVars?  afaict, they were in concurrent haskell and now aren't.

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