Home Reading Searching Subscribe Sponsors Statistics Posting Contact Spam Lists Links About Hosting Filtering Features Download Marketing Archives FAQ Blog From: Daniel Fischer web.de> Subject: Re: question from Yet Another Haskell Tutorial Newsgroups: gmane.comp.lang.haskell.beginners Date: Sunday 29th March 2009 12:45:22 UTC (over 9 years ago) ```Am Sonntag 29 März 2009 08:33:13 schrieb Michael Mossey: > 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. As it should, since it is correct. A simple way to check for sanity of the results is: Prelude> :t flip foldr 0 . flip. curry flip foldr 0 . flip. curry :: (Num c) => ((c, b) -> c) -> [b] -> c Prelude> :t \f list -> foldr (\x y -> f (y,x)) 0 list \f list -> foldr (\x y -> f (y,x)) 0 list :: (Num b) => ((b, a) -> b) -> [a] -> b Prelude> :t \f -> foldr (uncurry \$ flip f) 0 \f -> foldr (uncurry \$ flip f) 0 :: (Num b1) => (b -> a -> b1 -> b1) -> [(a, b)] -> b1 So you see that your result has the correct type, while Hal's hasn't. > 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? Simple mistake. YAHT was never systematically proofread to catch all such mistakes, so there are several still in. Nevertheless, it is an excellent tutorial. > > Thanks, > Mike >```
CD: 3ms