4 Dec 07:25
Re: IVars
Conal Elliott <conal <at> conal.net>
2007-12-04 06:25:17 GMT
2007-12-04 06:25:17 GMT
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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