Alexteslin
I am really sorry, but i still can't define the function. The chapter the exercise is in precedes algebraic types or Maybe a types. And is seems that must being easy to define. I answered some exercises on using foldr such as summing for example, but this one i am trying:
myLast :: [Int] -> Int myLast (x:xs) = foldr (some function) x xs.
With summing as i understood foldr goes through the list and sums up by (+) operator and one argument like 0. Like: foldr (+) 0 xs
You can think of foldr f e [a,b,c] as a `f` (b `f` (c `f` e)) so myLast [x,a,b,c] is going to be a `f` (b `f` (c `f` x)) so you need an f so that c `f` x is c (for any c and x) and yet (b `f` c) is c for any c and b -- this is impossible (or I'm asleep). So your skeleton above isn't going to work. You can do something like myLast l = unpick (foldr toLast [] l) where unpick ... toLast ... The above contains a strong hint as to the type of toLast (and hence unpick) -- Jón Fairbairn Jon.Fairbairn@cl.cam.ac.uk