2 Mar
2011
2 Mar
'11
2:11 p.m.
Hi,
my questions are: 1. is this substitution ok? 2. is []a = [a]? 3. is []b = [b]?
if 2. and 3. are valid then the following is obtained: fmap::(a->b)->[a]->[b] which is the same as map type and therefore: fmap = map. QED.
Can you please answer questions 2 and 3 above?
Yes, they are equivalent. Actually both forms are valid Haskell: a :: [Int] a = [1] a' :: [] Int a' = [1] Patrick
Thank you
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-- ===================== Patrick LeBoutillier Rosemère, Québec, Canada