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

David Banas capn.freako at gmail.com
Fri Oct 30 20:03:49 UTC 2015

```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 =>
-- 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

```