# [Haskell-cafe] Optimization with Strings ?

Alberto G. Corona agocorona at gmail.com
Thu Dec 3 14:21:14 EST 2009

" In fact, the correct answer is that pEqualsP should produce an error and
qEqualsQ should never terminate"

¿¿???
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 at kent.ac.uk>

> Emmanuel CHANTREAU wrote:
>
>> Le Thu, 3 Dec 2009 13:20:31 +0100,
>> 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
>>
>>
>> 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.
>>
>>
> Besides any other reasons, Haskell has the error function, and infinite
> lists.  Consider:
>
> 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.
>
> _______________________________________________