I was bored one day and wanted to exercise my brain, so I decided to do the 99 Haskell Problems but restricted myself to doing them in point-free style. A problem that seems to crop up a lot when I'm doing things in point-free style is this: How do you apply multiple functions to the same value while keeping each result as an independent entity? Using pointed notation:
foobar x = [id x, reverse x]
And what I've come up with so far in point-free notation:
foobar' = `map` [id, reverse] ($ x)
I can't seem to get that x off the end of there.
Others have already posted how you can do this using the Reader monad, but that's not the only way. It turns out that your second function is pretty close. I think you meant to post
foobar' x = (`map` [id, reverse]) ($ x)
Since the x is already near a rightmost position, you're almost there. First, transform the section ($ x) into a function, because it's a bit easier to work with:
-- by the definition of a right operator section
foobar'2 x = (`map` [id, reverse]) (\y -> ($) y x)
Next remove the x from the lambda body by bringing a new variable into scope, and applying the function to x
-- lambda abstraction I think...
foobar'2 x = (`map` [id, reverse]) $ (\z y -> ($) y z) x
Rewrite this application as a function composition, and then you can eta reduce:
-- by definition of '.'
foobar'3 x = (`map` [id, reverse]) . (\z y -> ($) y z) $ x
-- eta reduction
foobar'4 = (`map` [id, reverse]) . (\z y -> ($) y z)
Finally, notice that we can replace the lambda with a function
-- by definition of `flip`
foobar'5 = (`map` [id,reverse]) . flip ($)
and you have a point-free form.