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Well (.) g f x = g (f x) in Haskell is just sugar for (.) g f = \x -> g (f x), which ultimately is turned into (.) = \g -> \f -> \x -> g (f x)
by orbifold 12y ago
Well (.) g f x = g (f x) in Haskell is just sugar for
(.) g f = \x -> g (f x), which ultimately is turned into (.) = \g -> \f -> \x -> g (f x)