Le Thu, 3 Dec 2009 13:20:31 +0100,
David Virebayre
It doesn't work this way : Strings are just lists of Chars. Comparison is made recursively, Char by Char. You can have a look at the source to make sure :
instance (Eq a) => Eq [a] where [] == [] = True (x:xs) == (y:ys) = x == y && xs == ys _xs == _ys = False
Hello Thank you David and Bulat for your answers. I don't see the proof you see. Because GHC could store two sames objects juste once and by the definition of == on lists it could deduce that "forall x; List x => x==x". GHC have all informations to do this optimization job, because haskell functions definitions are mathematics definitions. Bulat says that this optimization is not done, so I will do it by my hands (ho my poor lazy hands). -- Emmanuel Chantréau