[Haskell-beginners] defining 'init' in terms of 'foldr'
Paul Higham
polygame at mac.com
Sat Dec 4 23:20:51 CET 2010
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])
Note that snoc is defined at the top of the same page as the exercise
in Simon's book.
::paul
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