Re: [Haskell-cafe] Determining the length of a Foldable Applicative.

30 Oct 2015


      Maybe what you are missing is an argument to the app_len function?

Am Freitag, 30. Oktober 2015 schrieb David Banas :
...
Hi all,
I thought I had a simple way to determine the “length" (i.e. - number of
elements in) of a Foldable Applicative container:
import Prelude hiding (sum)
    import Data.Foldable (Foldable(..), sum)
    import Control.Applicative
-- Calculate the "length" (i.e. - number of elements in) an
Applicative container.
    app_len :: (Applicative f, Foldable f) => f a -> Int
    app_len = sum $ pure 1
but I didn’t:
app_len_test.hs:9:11:
    Could not deduce (Foldable t0) arising from a use of ‘sum’
    from the context (Applicative f, Foldable f)
      bound by the type signature for
                 app_len :: (Applicative f, Foldable f) => f a -> Int
      at app_len_test.hs:8:12-52
    The type variable ‘t0’ is ambiguous
    Note: there are several potential instances:
      instance Foldable ((,) a) -- Defined in ‘Data.Foldable’
      instance GHC.Arr.Ix i => Foldable (GHC.Arr.Array i)
        -- Defined in ‘Data.Foldable’
      instance Foldable (Const m) -- Defined in ‘Data.Foldable’
      ...plus four others
    In the expression: sum
    In the expression: sum $ pure 1
    In an equation for ‘app_len’: app_len = sum $ pure 1
app_len_test.hs:9:17:
    Could not deduce (Applicative t0) arising from a use of ‘pure’
    from the context (Applicative f, Foldable f)
      bound by the type signature for
                 app_len :: (Applicative f, Foldable f) => f a -> Int
      at app_len_test.hs:8:12-52
    The type variable ‘t0’ is ambiguous
    Note: there are several potential instances:
      instance Data.Monoid.Monoid a => Applicative ((,) a)
        -- Defined in ‘Control.Applicative’
      instance Applicative ((->) a) -- Defined in ‘Control.Applicative’
      instance Control.Arrow.Arrow a =>
               Applicative (Control.Arrow.ArrowMonad a)
        -- Defined in ‘Control.Applicative’
      ...plus 14 others
    In the second argument of ‘($)’, namely ‘pure 1’
    In the expression: sum $ pure 1
    In an equation for ‘app_len’: app_len = sum $ pure 1
Can anyone help me understand what I’m missing?
Thanks, and have a great weekend,
-db
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Re: [Haskell-cafe] Determining the length of a Foldable Applicative.

Janis Voigtländer