- Point-free define: foo :: (Ord a, Ord b) => [([a], [b])] -> (Set a, Set b)
- Андреев Кирилл @nonaem00 foo = (Set.fromList . concat *** Set.fromList . concat) . unzip
- point-free-itize computeTotalWithTax :: Num b => ((a, b), b) -> b computeTotalWithTax ((a, b), c) = b + c
- point-free-itize foo (k,v) m = Map.insert k b m with obvs types for k, v, and m.
- @SlussNoah foo = uncurry Map.insert
- point-free-itize: shower :: forall a. forall b. Show a => [b -> a] -> b -> [a] shower fns thing = map (app . flip (,) thing) fns
- @gautier_difolco shower = flip $ map . flip ($)
- row :: String -> (Item, (USD, Measure)) given csv :: String -> [String] and line is = "apple,$1.99 Lb" hint: words "a b" = ["a","b"] ... all types mentioned above are in today's @1HaskellADay problem at http://lpaste.net/4698665561507233792
- For Read a, point-free-itize: f a list = read a:list (f is used in a foldr-expression)
- Or you could just do: map read
- point-free-itize f such that: f a b c = a + b + c
- @phollow f = ((+) .) . (+)

## Thursday, September 3, 2015

### 1Liners Pre-July 2015

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