On Mon, 30 Dec 2002, Cesar Augusto Acosta Minoli wrote:
> Hello! I'm Working with Lists in Haskell, I�m a Beginner in Functional
> Programming and I would like to know if there is�a way to write a more
> efficient function that return the length of a list, I wrote this one:
>
> long������� ::� [a]->Int
> long p���� =� longitud p 0
> ������������������ where
> ������������������ longitud []������ s=s
> ������������������ longitud (x:xs) s=longitud xs (s+1)
>
> but I think that it have a lineal grow O(n).
Yes, it's O(n), but you can't do any better for calculating the length of
a list. Your second parameter seems to be an accumulator which is the sort
of thing you'd make explicit in an imperative approach but can often be
eliminated in functional code - e.g.,
long [] = 0
long (x:xs) = 1 + long xs
A decent optimizing compiler will probably turn that code into something
that uses an accumulator. This code probably isn't any more efficient than
yours, it's just shorter.
-- Mark
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