[Haskell-cafe] A challenge
Peter Verswyvelen
bugfact at gmail.com
Wed Apr 8 12:38:24 EDT 2009
Oh, I could have written it in more point free style (with arguments
reversed) as
iterateM n i = fmap (reverse . snd) .
foldM collectEffect (i,[]) .
replicate n
where
collectEffect (x,rs) f = f x >>= \y -> return (y,y:rs)
and I'm sure collectEffect could also be improved, but I'm still in
newbieeee land
On Wed, Apr 8, 2009 at 6:33 PM, Peter Verswyvelen <bugfact at gmail.com> wrote:
> 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_______________________________________________
>>
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
>> http://www.haskell.org/mailman/listinfo/haskell-cafe
>>
>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: http://www.haskell.org/pipermail/haskell-cafe/attachments/20090408/32114a46/attachment.htm
More information about the Haskell-Cafe
mailing list