[Haskell-cafe] Re: newbie optimization question

[Peter Hercek]

Don Stewart wrote:
> > perfect :: [Int]
> > perfect = [i | i<-[1..10000], i == sum (divisors i)]
>
> This should be a little faster , as sum will fuse,
>
>     perfect :: [Int]
>     perfect = [i | i<-[1..10000], i == sum' (divisors i)]
>         where sum' = foldr (+) 0

sum' did not help. Times are about the same with Int type.

Peter.

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