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should? or you want to say "actually do that so the optimization does is not done"?
The correct amswer is not the sould you mention, but True (IMHO). So the optimization can be done anyway.
2009/12/3 Neil Brown
<nccb2@kent.ac.uk>
Emmanuel CHANTREAU wrote:
Le Thu, 3 Dec 2009 13:20:31 +0100,
David Virebayre <dav.vire+haskell@gmail.com> a écrit :
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.
p :: String
p = error "Haha!"
q :: String
q = repeat 'a'
pEqualsP :: Bool
pEqualsP = p == p
qEqualsQ :: Bool
qEqualsQ = q == q
By your rule, pEqualsP and qEqualsQ should be True. In fact, the correct answer is that pEqualsP should produce an error and qEqualsQ should never terminate. Since Strings can contain such errors and infinite lists, you can't know for certain that an object equals itself without checking its entire length, which is what the original definition for equals did anyway. There may be strict data structures for which your optimisation might be applicable, though.
Neil.