Re: [Haskell-cafe] Re: [Haskell] IVars

Thanks, Luke. I'd been unconsciously assuming that the IVar would get written to (if ever) by a thread other than the one doing the reading. (Even then, there could be a deadlock.) - Conal On Dec 9, 2007 9:37 AM, Luke Palmer wrote:

On Dec 9, 2007 5:09 PM, Conal Elliott wrote:

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(moving to haskell-cafe)

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readIVar' :: IVar a -> a readIVar' = unsafePerformIO . readIVar

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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

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- Conal

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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

On Dec 8, 2007 11:12 AM, Marc A. Ziegert wrote: libraries, maybe as "unsafeReadIVar" or "unsafeReadMVar".

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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) _|_.

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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:

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Conal Elliott wrote:

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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.

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