[Haskell-beginners] defining 'init' in terms of 'foldr'

[Chaddaï Fouché]

On Sun, Dec 5, 2010 at 12:37 PM, Hein Hundal <hundalhh at yahoo.com> wrote:
>> From: Daniel Fischer <daniel.is.fischer at googlemail.com>
>> Subject: Re: [Haskell-beginners] defining 'init' in terms
>> of 'foldr'
>> On Saturday 04 December 2010 23:20:51, Paul Higham wrote:
>> > Not sure if this thread is still active but I also
>> > struggled with this
>> > same exercise.  I offer the following solution as
>> > a thing to shoot at:
>> >
>> > myInit :: [a] -> [a]
>> > myInit ys = foldr snoc [] $ (\(x:xs) -> xs) $ foldr snoc [] ys
>> >         where snoc = (\x xs -> xs ++ [x])
>>
>> init === reverse . tail . reverse
>> only holds for finite lists, for infinite lists xs, reverse
>> xs = _|_, but
>> init xs = xs.
>> Also, it's inefficient, but that's not the point of the
>> exercise.
>
> Here is my incomplete beginner solution:
>
> --Start Code
>
> import Data.Maybe
> myinit v = fromJust $ foldr f Nothing v
>   where
>      f a Nothing = Just []
>      f a (Just v) = Just (a:v)
>
> --End Code
>
> This worked for the finite lists that I tried, but it did not work for infinite lists.  I was surprised that foldr works with infinite lists.  If I set
>
> v = take 5 $ foldr (\x xs -> (x*x):xs) [] [1..]
>
>
> then the value of v is [1,2,9,16].  Is there another way to create the init function with foldr that works for infinite lists?

Keeping your idea but being lazier about it :

> myInit v = fromMaybe (error "myInit : empty list") $ foldr f Nothing v
>   where
>      f x xs = Just (case xs of Nothing -> []; Just ys -> x:ys)

-- 
Jedaï