Re: [Haskell-beginners] Making sense of currying in the context of the Hask Category

On Fri, Sep 28, 2012 at 08:53:11PM -0400, Lino Rosa wrote:

Thanks for the responses :) I've read on cartesian closed categories, but I guess I need more time for it to sink in - my knowlege on category theory is *very* limited.

Still, continuing on the Int + Int example. Would this be the sequence of transformations, then?

You seem quite set on this idea of there being "a sequence of transformations", I am not sure why. When thinking about (+) there are many different related arrows; some of them form sequences (can be composed) and some can't. So this is not necessarily a helpful way of thinking about things.

Int -> (Int -> Int) -> Int (+) (z) (?)

What do I label the arrow (?) ?

That said, you could label the arrow (?) with something like ($2) which means "apply a function to the argument 2".

I understand the first transformation as "function (+) when applied to an Int will result in (Int -> Int)". How would I describe function (?) similarly ? It's almost as it the arrow is z and (?) is the Int and I'm applying z to (?), but that makes no sense!

If I understood correctly, 'z' would be a special type, one which when applied (?) would return the final Int. This comes Hask being cartesian closed. Still I'm clueless on the arrow (?)

I think you actually had it right here. Indeed it does not make sense for (?) to be an Int, but it can indeed be a function like "apply the input to the argument 2". Having an "apply" operation is one of the criteria for being cartesian closed. -Brent

On Fri, Sep 28, 2012 at 8:07 AM, Kim-Ee Yeoh wrote:

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Hi Lino,

Brent gave an excellent answer. Looking up "cartesian closed category" should yield even more insights.

On Fri, Sep 28, 2012 at 9:07 AM, Lino Rosa wrote:

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That's the function resulting from the previous partial application of (+), but that fuction only exists at run time, after you apply the first one.

When you speak of a function that "only exists at run time", I think you're alluding to partial evaluation. Haskell doesn't do that, although many have wished for it.

-- Kim-Ee

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