conal at conal.net
Tue Dec 4 01:25:17 EST 2007
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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