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 <[EMAIL PROTECTED]> wrote: > You can make them from MVars. > > On Dec 2, 2007 8:03 PM, Conal Elliott <[EMAIL PROTECTED]> wrote: > > > what became of (assign-once) IVars? afaict, they were in concurrent > > haskell and now aren't. > > > > _______________________________________________ > > Haskell mailing list > > Haskell@haskell.org > > http://www.haskell.org/mailman/listinfo/haskell > > > > >
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