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#.

Answer 1

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 🙂