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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