Hi,
Thank you all for you replies.
I think now I understand Haskell better.
Best regards,
Frodo
2011/10/12 Frodo Chao
> Hi,
>
> I came upon this when playing with foldr and filter. Compare the two
> definitions:
>
> testr n = foldr (\x y -> x && y) True [t |
es to ||. Should I mind the
orderings of the parameters (especially the accumulator) in the function
passed to foldr?
Thak you for reading.
Sincerely yours,
Frodo Chao
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