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

[Hein Hundal]

> 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?

Hein