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.