CODEWITHSUNDEEP
haskell
Niccolo M.
Niccolo M.

Reputation: 3413

Why does Haskell's "flip id" has this type?

I'm curious about the expression flip id (It's not homework: I found it in the getOpt documentation).

I wonder why it has this type:

Prelude> :t (flip id)
(flip id) :: b -> (b -> c) -> c

For example, (flip id) 5 (+6) gives 11.

I know why id (+6) 5 gives 11, but I don't "get" the flip id thing.

I tried to figure this out myself using pen and paper but couldn't. Could anybody please explain this to me? I mean, how does flip id come to have the type b -> (b -> c) -> c ?

Upvotes: 50

Views: 2855

Answers (1)

ertes
ertes

Reputation: 826

The id function has this type:

id :: a -> a

You get an instance of this type, when you replace a by a -> b:

id :: (a -> b) -> (a -> b)

which, because of currying, is the same as:

id :: (a -> b) -> a -> b

Now apply flip to this and you get:

flip id :: a -> (a -> b) -> b

In the case of id (+) the instance is:

id :: (Num a) => (a -> a) -> (a -> a)

Now flip id gives you:

flip id :: (Num a) => a -> (a -> a) -> a

Side note: This also shows you how ($) is the same as id, just with a more restricted type:

($) :: (a -> b) -> a -> b
($) f x = f x
-- unpoint:
($) f   = f
-- hence:
($)     = id

Upvotes: 81

Related Questions