Haskell function to reverse function call

Question

I have a lambda \x f -> f x that is being used in a foldM operation, where x is a value and f :: a -> b.

Is there a built-in function that does this?

Can I replace

foldM (\x f -> f x) ...

with some f'

foldM f' ...

I thought that flip would do this, but it takes three arguments (flip :: (a -> b -> c) -> b -> a -> c)

It is probably similar to |> in F#.

Niklas B. · Accepted Answer · 2012-05-30 13:45:42Z

22

You can use flip id or flip ($) (as ($) is just a special id for functions):

Prelude> flip id 3 (+2)
5
Prelude> flip ($) 7 (>10)
False

This is an interesting use of partial application: id f x with f being a function is just f x. Obviously, this is also the same as (flip id) x f, so flip id is the function you are looking for.

If you feel adventurous, try inferring the type of flip id or flip ($) manually. It's fun :)

edited May 30, 2012 at 13:45

answered May 30, 2012 at 12:07

Niklas B.

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9 Comments

Asherah Over a year ago

Wow, that took me about 5 minutes with paper and pen to work out how id's a -> a signature could somehow map onto flip's expecting something with a -> b -> c. It might be worth going into that a little bit!

Niklas B. Over a year ago

@Len: It becomes clearer once you see that when applied to functions, id is the same as ($)

Asherah Over a year ago

That's very true! The "a-ha" moment for me was realising that I should see a -> b -> c as the a -> (b -> c) that it is, and then matched that to id's a' -> a' and found that a = a' = (b -> c)—I was looking too much at the differences in arity and couldn't get past it.

Ralph Over a year ago

Thanks. Haskell is a cool language, but I feel like my brain is in a maze of twisty little passages, all alike ...

Vitus Over a year ago

@Ralph: There's also pointfree package on Hackage. cabal install pointfree and pointfree "\a b -> b a".

|

Riccardo T. · Accepted Answer · 2012-05-30 12:19:36Z

Yes, it's called flip :: (a -> b -> c) -> b -> a -> c, e.g. flip (>) 3 5 == True. More infos and source on hackage: flip.

What you want is simply to reverse the arguments of function application, right? Well, since ($) is function application, by using flip you may write flip ($) :: b -> (b -> c) -> c. Let see what happens. Here is the source for the two prelude functions:

-- from Hackage:
($)                     :: (a -> b) -> a -> b
f $ x                   =  f x

-- from Hackage:
flip                    :: (a -> b -> c) -> b -> a -> c
flip f x y              =  f y x

So, basically if you put together the types, flip ($) becomes

flip ($) :: 
  b        ->    -- type of x, argument of y and second argument of ($)
  (b -> c) ->    -- type of y, function applied by ($) as its first argument
  c        ->    -- result of the application "y x"

If you follow the actual definitions of the functions:

flip ($) = (\f x y -> f y x) ($)    -- from flip's def.
         = \x y -> ($) y x          -- partial application
         = y x                      -- from ($)'s def.

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