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

On Sun, Dec 5, 2010 at 12:37 PM, Hein Hundal wrote:

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From: Daniel Fischer Subject: Re: [Haskell-beginners] defining 'init' in terms of 'foldr' On Saturday 04 December 2010 23:20:51, Paul Higham wrote:

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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