Edit: Sorry, just stupidity on my part. I 'm using the 7.6.3 Platform, but reading online docs for 'latest', and I just did this:
>:i Seq
newtype Seq a
= Data.Sequence.Seq (Data.Sequence.FingerTree
(Data.Sequence.Elem a))
-- Defined in `Data.Sequence'
instance Eq a => Eq (Seq a) -- Defined in `Data.Sequence'
instance Monad Seq -- Defined in `Data.Sequence'
instance Functor Seq -- Defined in `Data.Sequence'
instance Ord a => Ord (Seq a) -- Defined in `Data.Sequence'
instance Read a => Read (Seq a) -- Defined in `Data.Sequence'
instance Show a => Show (Seq a) -- Defined in `Data.Sequence'
and look, no applicative instance. Sorry for the bother, and thanks to bheklilr for asking me about versioning, which I just spaced on.
The following confuses me. There certainly seems to be an instance defining (<*>) sensibly for Seq:
instance Applicative Seq where
pure = singleton
fs <*> xs = foldl' add empty fs
where add ys f = ys >< fmap f xs
Yet, this happens:
>:m + Data.Sequence Control.Applicative
>(*) <$> [1..4] <*> [3..4]
[3,4,6,8,9,12,12,16]
>(*) <$> fromList [1..4] <*> fromList [3..4]
<interactive>:58:25:
No instance for (Applicative Seq) arising from a use of `<*>'
Possible fix: add an instance declaration for (Applicative Seq)
In the expression: (*) <$> fromList [1 .. 4] <*> fromList [3 .. 4]
In an equation for `it':
it = (*) <$> fromList [1 .. 4] <*> fromList [3 .. 4]
>:t (*) <$> fromList [1..4]
(*) <$> fromList [1..4] :: (Enum a, Num a) => Seq (a -> a)
Using a singleton produces slightly different typing (no Enum qualification) and hence a slightly different error.
>:t (*) <$> singleton 3
(*) <$> singleton 3 :: Num a => Seq (a -> a)
>:t (*) <$> singleton 3 <*> fromList [3 .. 4]
<interactive>:1:21:
Could not deduce (Applicative Seq) arising from a use of `<*>'
from the context (Enum b, Num b)
bound by the inferred type of it :: (Enum b, Num b) => Seq b
at Top level
Possible fix: add an instance declaration for (Applicative Seq)
In the expression: (*) <$> singleton 3 <*> fromList [3 .. 4]
The type Seq (a -> a) would seem to me to conform to the f (a -> b) required by (<*>) :: Applicative f => f (a -> b) -> f a -> f b.
I'm sure this is simple and I'll be embarrassed for asking, but thanks in advance.
