[Haskell-cafe] A challenge
Peter Verswyvelen
bugfact at gmail.com
Wed Apr 8 12:33:29 EDT 2009
I don't think scanl can work here, since the monadic action has to be
applied to the result of previous one and will have a side effect, so if you
build a list like
[return i, return i >>= f, return i >>= f >>= f, ...]
the first action will do nothing, the second action will have a single side
effect, but the third one will have 3 side effects instead of 2, because it
operates on the side-effect performed by the second one.
This seems to work (a combination of manual state monad and foldM, I could
also have used a state monad transformer I guess)
iterateM n f i = foldM collectEffect (i,[]) (replicate n f) >>= return .
reverse . snd
where
collectEffect (x,rs) f = f x >>= \y -> return (y,y:rs)
Ugly test:
var = unsafePerformIO $ newIORef 0
inc i = do
x <- readIORef var
let y = x+i
writeIORef var y
return y
results in
*Main> iterateM 10 inc 1
[1,2,4,8,16,32,64,128,256,512]
*Main> iterateM 10 inc 1
[513,1026,2052,4104,8208,16416,32832,65664,131328,262656]
but maybe this is not what you're looking for?
On Wed, Apr 8, 2009 at 5:30 PM, Thomas Davie <tom.davie at gmail.com> wrote:
>
> On 8 Apr 2009, at 17:20, Jonathan Cast wrote:
>
> On Wed, 2009-04-08 at 16:57 +0200, Thomas Davie wrote:
>>
>>> We have two possible definitions of an "iterateM" function:
>>>
>>> iterateM 0 _ _ = return []
>>> iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i)
>>>
>>> iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f
>>>
>>> The former uses primitive recursion, and I get the feeling it should
>>> be better written without it. The latter is quadratic time – it
>>> builds up a list of monadic actions, and then runs them each in turn.
>>>
>>
>> It's also quadratic in invocations of f, no? If your monad's (>>=)
>> doesn't object to being left-associated (which is *not* the case for
>> free monads), then I think
>>
>> iterateM n f i = foldl (>>=) (return i) $ replicate n f
>>
>
> But this isn't the same function – it only gives back the final result, not
> the intermediaries.
>
> Bob_______________________________________________
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