4 ms·

Minor nitpick. Technically, this: #(->> % (drop 3) (filter p) (take 5)) Is equivalent to: take 5 . filter p . drop 3

fnordsensei 11y ago
Is take 5 . filter p . drop 3 An anonymous function call in Haskell? I don't know the language (although it's on my to-learn sequence).
- catnaroek 11y ago
  It's an anonymous function, which it not being called.
  - deleted 11y ago
    [deleted]
    - catnaroek 11y ago
      You probably meant to reply to mietek.
- mietek 11y ago
  Yes, written in pointless style. You could make it less pointless by replacing the (.)s with a point. \xs -> take 5 (filter p (drop 3 xs)) https://wiki.haskell.org/Pointfree https://wiki.haskell.org/Pointfree
- marcosdumay 11y ago
  Yes. It's three anonymous ("take 5", "filter p" and "drop 3") functions composed by the "." operator, creating another anonymous function. One of the types, for example: take :: Int -> [a] -> [a] What means a function that takes an Int argument and returns another function, that takes a list and returns a list of the same type.
- pja 11y ago
  ghci will (usually) tell you: phil@arod:~$ ghci GHCi, version 7.6.3: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Prelude> :t take 5 . filter (\ _ -> True) . drop 3 take 5 . filter (\ _ -> True) . drop 3 :: [a] -> [a] So it’s a function from List of a to List of a
deleted 11y ago
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takle 11y ago
One could also change the Haskell. Data.Function (&); Control.Arrow (>>>) -- the pipe operator seems to be all the rage these days (|>) = (&) -- forward application + forward composition xs |> (drop 3 >>> filter p >>> take 5) Only function application xs |> drop 3 |> filter p |> take 5
- catnaroek 11y ago
  I never understood the appeal of the pipe operator. It's a kludge to work around the need to eta-expand chained function compositions, which in turn is a kludge to work around the value restriction, which in turn is a kludge to prevent the combination of polymorphism and storage effects from breaking type soundness. Haskell doesn't need any of this.
  - takle 11y ago
    I agree with you. I'm going to write a post about function application and composition (right-to-left / left-to-right) so other communities can stop discussing non-essential matters. I do like the fact that the `pipe operator` is often mentioned along the bash '|' which is used in a pointfree style. Well it might make sense sometimes, so its good to know one's options :) p x y z = f (g x y) z p = ((.) f) . g
    - catnaroek 11y ago
      Of course, something is very wrong if you ever use a section of the function composition operator. Use point-free within reason... :-p
      - klibertp 11y ago
        Or go completely, absolutely nuts with point-free and use J (http://www.jsoftware.com/ http://www.jsoftware.com/).

Is take 5 . filter p . drop 3 An anonymous function call in Haskell? I don't know the language (although it's on my to-learn sequence).

catnaroek 11y ago
It's an anonymous function, which it not being called.
- deleted 11y ago
  [deleted]
  - catnaroek 11y ago
    You probably meant to reply to mietek.
mietek 11y ago
Yes, written in pointless style. You could make it less pointless by replacing the (.)s with a point. \xs -> take 5 (filter p (drop 3 xs)) https://wiki.haskell.org/Pointfree https://wiki.haskell.org/Pointfree
marcosdumay 11y ago
Yes. It's three anonymous ("take 5", "filter p" and "drop 3") functions composed by the "." operator, creating another anonymous function. One of the types, for example: take :: Int -> [a] -> [a] What means a function that takes an Int argument and returns another function, that takes a list and returns a list of the same type.
pja 11y ago
ghci will (usually) tell you: phil@arod:~$ ghci GHCi, version 7.6.3: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Prelude> :t take 5 . filter (\ _ -> True) . drop 3 take 5 . filter (\ _ -> True) . drop 3 :: [a] -> [a] So it’s a function from List of a to List of a

deleted 11y ago

[deleted]

takle 11y ago

One could also change the Haskell. Data.Function (&); Control.Arrow (>>>) -- the pipe operator seems to be all the rage these days (|>) = (&) -- forward application + forward composition xs |> (drop 3 >>> filter p >>> take 5) Only function application xs |> drop 3 |> filter p |> take 5

catnaroek 11y ago
I never understood the appeal of the pipe operator. It's a kludge to work around the need to eta-expand chained function compositions, which in turn is a kludge to work around the value restriction, which in turn is a kludge to prevent the combination of polymorphism and storage effects from breaking type soundness. Haskell doesn't need any of this.
- takle 11y ago
  I agree with you. I'm going to write a post about function application and composition (right-to-left / left-to-right) so other communities can stop discussing non-essential matters. I do like the fact that the `pipe operator` is often mentioned along the bash '|' which is used in a pointfree style. Well it might make sense sometimes, so its good to know one's options :) p x y z = f (g x y) z p = ((.) f) . g
  - catnaroek 11y ago
    Of course, something is very wrong if you ever use a section of the function composition operator. Use point-free within reason... :-p
    - klibertp 11y ago
      Or go completely, absolutely nuts with point-free and use J (http://www.jsoftware.com/ http://www.jsoftware.com/).