On Dec 9, 2007 5:09 PM, Conal Elliott <[EMAIL PROTECTED]> wrote: > (moving to haskell-cafe) > > > readIVar' :: IVar a -> a > > readIVar' = unsafePerformIO . readIVar > > > so, we do not need readIVar'. it could be a nice addition to the > libraries, maybe as "unsafeReadIVar" or "unsafeReadMVar". > > The same argument applies any to pure function, doesn't it? For instance, a > non-IO version of succ is unnecessary. My question is why make readIVar a > blocking IO action rather than a blocking pure value, considering that it > always returns the same value?
But I don't think it does. If we're single-threaded, before we writeIVar on it, it "returns" bottom, but afterward it returns whatever what was written. It's a little fuzzy, but that doesn't seem referentially transparent. Luke > - Conal > > On Dec 8, 2007 11:12 AM, Marc A. Ziegert <[EMAIL PROTECTED]> wrote: > > many many answers, many guesses... > > let's compare these semantics: > > > > > > readIVar :: IVar a -> IO a > > readIVar' :: IVar a -> a > > readIVar' = unsafePerformIO . readIVar > > > > so, we do not need readIVar'. it could be a nice addition to the > libraries, maybe as "unsafeReadIVar" or "unsafeReadMVar". > > but the other way: > > > > readIVar v = return $ readIVar' v > > > > does not work. with this definition, readIVar itself does not block > anymore. it's like hGetContents. > > and... > > > > readIVar v = return $! readIVar' v > > > > evaluates too much: > > it wont work if the stored value evaluates to 1) undefined or 2) _|_. > > it may even cause a 3) deadlock: > > > > do > > writeIVar v (readIVar' w) > > x<-readIVar v > > writeIVar w "cat" > > return x :: IO String > > > > readIVar should only return the 'reference'(internal pointer) to the read > object without evaluating it. in other words: > > readIVar should wait to receive but not look into the received "box"; it > may contain a nasty undead werecat of some type. (Schrödinger's Law.) > > > > - marc > > > > > > > > > > > > Am Freitag, 7. Dezember 2007 schrieb Paul Johnson: > > > > > > > > > Conal Elliott 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? > > > Because it won't necessarily yield the same result the next time you run > > > it. This is the same reason the stuff in System.Environment returns > > > values in IO. > > > > > > Paul. > > > > > > > > > _______________________________________________ > > > Haskell mailing list > > > [EMAIL PROTECTED] > > > > > http://www.haskell.org/mailman/listinfo/haskell > > > > > > > > > > > > _______________________________________________ > Haskell-Cafe mailing list > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe > > _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
