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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