4 Dec
2007
4 Dec
'07
6:25 a.m.
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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