Good point. By fold/unfold transformation you get the following: contains = flip elem [Eureka] = contains xs e = flip elem xs e [Expose data structures] = contains [] e = False contains (x:xs) e = flip elem (x:xs) e [Instantiate] = contains [] e = False contains (x:xs) e = elem e x:[] || flip elem xs e [Unfold flip one step] = contains [] e = False contains (x:xs) e = elem e x:[] || contains xs e [Fold back to original defintion] = contains [] e = False contains (x:xs) = e==x || contains xs e [Substitute] Apparently, the fold/unfold transformation law will always yield an equally or more efficient computation. So this begs the question... contains [] e = False contains (x:xs) = e==x || contains xs e OR contains = flip elem Neil Mitchell wrote:
Hi
contains :: Eq a => [a]->a->Bool contains [] e = False contains (x:xs) e = if x==e then True else contains xs e
contains = flip elem
And even if not using the elem function, the expression:
if x==e then True else contains xs e
can be written as:
x==e || contains xs e
Thanks
Neil _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
-- View this message in context: http://www.nabble.com/Knowledge-tp14423007p14577605.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.