[Haskell-cafe] On to applicative
Alexander Solla
ajs at 2piix.com
Thu Aug 26 14:15:42 EDT 2010
On Aug 26, 2010, at 9:27 AM, michael rice wrote:
> Some functions just happen to map to other functions.
>
> <$> is flip fmap. f <$> functor = fmap f functor #### Brent
> Yorgey's post noted.
>
> #### "map to"? Take as arguments?
"maps to" as in "outputs".
> pure f <*> functor = f <$> functor
>
> #### Prelude Control.Applicative> pure double <*> (Just 5)
> #### Just 10
>
> #### Not so, the f got applied or where did we get the 10? Not sure,
> is this the #### "mistake" you point out in your second post?
double is getting applied in that expression, but it isn't because of
pure. The <*> operator is pulling double out of pure double (which
equals Just double in this case), and applying it to Just 5.
My correction was to point out that pure's type is more general than I
had said. Instead of
pure :: (a -> b) -> f (a -> b), it is pure :: a -> f a -- which
includes (a -> b) -> f (a -> b) as a special case. In fact, pure and
return are "essentially equivalent" in form. So you could write out
your verification case as
(pure double <*> pure 5) :: Just Int
to further decouple your code from the particular functor you're
working with. (You need the type annotation to run it in GHCi)
>
> #### Two ways of doing the same thing?
>
> #### 1)
> #### Prelude Control.Applicative> (double <$> (Just 7))
> #### Just 14
>
> #### 2)
> #### Prelude Control.Applicative> pure double <*> (Just 7)
> #### Just 14
Indeed.
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