RE: [Haskell-cafe] Haskell's currying and partial application

On Tue, 3 Jul 2007, Peter Verswyvelen wrote:

Let's see

id :: a -> a

curry :: ((a,b) -> c) -> a -> b -> c

=> curry id :: ((a,b) -> (a,b)) -> a -> b -> (a,b)

So basically if

f = curry id

then

f x y = (x,y)

which means

curry id = (,)

something like this?

You got it!

Do I win a price now? ;)

I hoped that nobody would care about the puzzle, thus I have not thought about a price so far. :-]