This might be slightly related. When I was assisting a Haskell lab course, I encountered solutions like the following:
removeRoot :: BSTree a -> BSTree a removeRoot (Node x Empty Empty) = Empty removeRoot (Node x left Empty) = left removeRoot (Node x Empty right) = right removeRoot (Node x left right) = undefined {- not needed for discussion -}
The first removeRoot case is unnecessary. Students new to Haskell (or
maybe new to recursion in general?) seem to consider more base cases
than needed.
-Raynor
On 6/8/09, Martijn van Steenbergen
Hello,
While grading a Haskell student's work I ran into an example of a program not being lazy enough. Since it's such a basic and nice example I thought I'd share it with you:
One part of the assignment was to define append :: [a] -> [a] -> [a], another to define cycle2 :: [a] -> [a]. This was her (the majority of the students in this course is female!) solution:
append :: [a] -> [a] -> [a] append [] ys = ys append xs [] = xs append (x:xs) ys = x : (append xs ys)
cycle2 :: [a] -> [a] cycle2 [] = error "empty list" cycle2 xs = append xs (cycle2 xs)
This definition of append works fine on any non-bottom input (empty, finite, infinite), but due to the unnecessary second append case, cycle2 never produces any result.
Martijn. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe