On Wednesday 16 March 2011 18:31:00, Yves Parès wrote: > Hello, > > A question recently popped into my mind: does lazy evaluation reduce the > need to "proper" tail-recursion? > I mean, for instance : > > fmap f [] = [] > fmap f (x:xs) = f x : fmap f xs > > Here fmap is not tail-recursive, but thanks to the fact that operator > (:) is lazy, I think that it may still run in constant space/time, am I > right?
Yes, and a tail-recursive map couldn't run in constant space, as far as I can see (time is O(length) for both of course, if the result is compeltely consumed). Tail recursion is good for strict stuff, otherwise the above pattern - I think it's called guarded recursion - is better, have the recursive call as a non-strict field of a constructor. _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
