Re: [Haskell-cafe] A challenge

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 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 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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