Michael Mossey mpm at alumni.caltech.edu
Sun Mar 29 02:33:13 EDT 2009

```I have a question about a problem in Yet Another Haskell Tutorial
(problem 7.1). My answers seems to disagree with Hal's, and it fact
something looks wrong in Hal's answer (maybe it's an error in his
paper).  The problem is to write the following function in "point-free"
style:

func5 f list = foldr (\x y -> f(y,x)) 0 list

Here's how I approached it. It's easy to drop 'list' by the
eta-reduction (using partial application):

func5 f = foldr (\x y -> f(y,x)) 0

But how to get rid of f? First I reasoned that f is a function of a
tuple. That is, it is not curried. So to curry it:

func5 f = foldr (\x y -> (curry f) y x) 0

The second argument to foldr is obviously flipping the arugments, so

func5 f = foldr (flip \$ curry f) 0

Now I want to use the eta-reduction again, but I have to transform this
expression into something that takes f as its last argument instead of
the 0. I can use flip again, this time on foldr:

func5 f = flip foldr 0 (flip \$ curry f)

Now f can be dropped:

THE FINAL ANSWER: func5 = flip foldr 0 . flip. curry

This works, in a couple of my test cases.  Now Hal gives this as the answer:

func5 = foldr (uncurry \$ flip f) 0

The first problem is that there's no argument f, though he refers to it.
So maybe he meant

func5 f = foldr (uncurry \$ flip f) 0

But more problems. He's applying flip to f, but f takes only one
argument (a 2-tuple). Then he's uncurry-ing it, but I thought it needed
currying, not uncurry-ing.

Can anyone figure out what Hal is up to, or does it look like a simple
mistake?

Thanks,
Mike

```