[Haskell-beginners] which Applicative instance?
Graham Gill
math.simplex at gmail.com
Sun Nov 15 02:29:21 UTC 2015
I'm following the CIS194 lectures and exercises. In particular, in one
of the Applicative lectures [0] we're asked to implement mapA and
sequenceA, from the name and given type signature alone. I renamed them
mapA' and sequenceA' to avoid name clashes, and defined:
import Control.Applicative
mapA' :: Applicative f => (a -> f b) -> ([a] -> f [b])
mapA' f = sequenceA' . map f
sequenceA' :: Applicative f => [f a] -> f [a]
sequenceA' = foldr (\fa fla -> (:) <$> fa <*> fla) (pure [])
These implementations seem to be correct, though undoubtedly not the nicest.
Now, I ask for a type in ghci:
> :t mapA' pure [1..5]
mapA' pure [1..5] :: (Enum b, Num b, Applicative f) => f [b]
And so I try the following:
> mapA' pure [1..5] :: [[Int]]
[[1,2,3,4,5]]
> take 5 . getZipList $ (mapA' pure [1..5] :: ZipList [Int])
[[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]]
> mapA' pure [1..5] :: Maybe [Int]
Just [1,2,3,4,5]
That makes sense. But I don't understand the following:
> mapA' pure [1..5]
[1,2,3,4,5]
Which Applicative instance is being used here? The type test above says
my expression should have type f [b], where f is an Applicative and b is
a Num, but all I see is a result of type [b] when I compute mapA' pure
[1..5]. What's going on?
More simply
> :t pure
pure :: Applicative f => a -> f a
> pure [1..5]
[1,2,3,4,5]
a must match a list of Nums, but then where did f go? Is ghci using some
default instance?
Actually, after I add the following lines to my source file:
ggg = pure [1..5]
hhh = mapA' pure [1..5]
I get the errors:
No instance for (Applicative f0) arising from a use of ‘pure’
No instance for (Applicative f1) arising from a use of ‘mapA'’
respectively. That makes sense to me.
Graham
[0]
http://www.seas.upenn.edu/~cis194/fall14/spring13/lectures/11-applicative2.html
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