[Haskell-cafe] Lazy Evaluation in Monads
Artyom Kazak
artyom.kazak at gmail.com
Tue May 31 22:20:30 CEST 2011
Suppose iRecurse looks like this:
iRecurse = do
x <- launchMissiles
r <- iRecurse
return 1
As x is never needed, launchMissiles will never execute. It obviously is
not what is needed.
But in Haskell, standart file input|output is often lazy. It's a
combination of buffering and special tricks, not the usual rule.
Scott Lawrence <bytbox at gmail.com> писал(а) в своём письме Tue, 31 May 2011
22:49:02 +0300:
> I was under the impression that operations performed in monads (in this
> case, the IO monad) were lazy. (Certainly, every time I make the
> opposite assumption, my code fails :P .) Which doesn't explain why the
> following code fails to terminate:
>
> iRecurse :: (Num a) => IO a
> iRecurse = do
> recurse <- iRecurse
> return 1
>
> main = (putStrLn . show) =<< iRecurse
>
> Any pointers to a good explanation of when the IO monad is lazy?
>
>
> === The long story ===
>
> I wrote a function unfold with type signature (([a] -> a) -> [a]), for
> generating a list in which each element can be calculated from all of
> the previous elements.
>
> unfold :: ([a] -> a) -> [a]
> unfold f = unfold1 f []
>
> unfold1 :: ([a] -> a) -> [a] -> [a]
> unfold1 f l = f l : unfold1 f (f l : l)
>
> Now I'm attempting to do the same thing, except where f returns a monad.
> (My use case is when f randomly selects the next element, i.e. text
> generation from markov chains.) So I want
>
> unfoldM1 :: (Monad m) => ([a] -> m a) -> [a] -> m [a]
>
> My instinct, then, would be to do something like:
>
> unfoldM1 f l = do
> next <- f l
> rest <- unfoldM1 f (next : l)
> return (next : rest)
>
> But that, like iRecurse above, doesn't work.
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