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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