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

[Paul Higham]

Yes, I believe that the point of the exercise was only to grok the  
semantics and applicability of foldr, it is a rather unnatural way to  
express init.  But you prompt an interesting question: does it make  
sense to apply foldr to an infinite list?  Or rather how is it  
possible to lazily evaluate foldr?

::paul

On Dec 4, 2010, at 2:48 PM, Daniel Fischer wrote:

> 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.
>
>>
>> Note that snoc is defined at the top of the same page as the exercise
>> in Simon's book.
>>
>> ::paul
>
> Cheers,
> Daniel
>
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