[Haskell-beginners] foldl by foldr

[John Bender]

Matt,

I was having issues with the very same problem not too long ago and I wrote
a blog post about how it illustrates the power of partial application.

http://nickelcode.com/2009/04/12/haskell-learnings/

Sorry for the blog plug, I just didn't see any reason to copy and paste the
content. I make no claims about the quality of my writing but it does
include some basic expansions of the execution/thunks for understanding.

Best

On Fri, May 14, 2010 at 6:58 AM, Brent Yorgey <byorgey at seas.upenn.edu>wrote:

> On Fri, May 14, 2010 at 12:29:01PM +1000, Matt Andrew wrote:
> >
> > The thing I am having trouble understanding is what the 'id'
> > function is doing in a function that expects 3 arguments and is
> > given 4 (foldr).
>
> "Number of arguments" in Haskell is a red herring.  In fact, every
> Haskell function takes exactly *one* argument.  Functions which appear
> to "take more than one argument" are really functions which take one
> argument and return another function (which takes the next argument,
> and so on).  That's why the type of a "multi-argument" function is written
> like
>
>  X -> Y -> Z -> ...
>
> which can also be written more explicitly as
>
>  X -> (Y -> (Z -> ...))
>
> Polymorphic functions (like foldr) can also be deceiving as far as
> "number of arguments" goes.  For example, consider id:
>
>  id :: a -> a
>
> Looks like this takes only one argument, right?  Well, what if  a = (Int ->
> Int):
>
>  id :: (Int -> Int) -> (Int -> Int)
>
> which can also be written
>
>  id :: (Int -> Int) -> Int -> Int
>
> so now it looks like id "takes two arguments" -- an (Int -> Int)
> function, and an Int.  Of course, the real answer is that id always
> takes exactly one argument, just like any other function; but
> sometimes that argument may itself be a function, in which case
> the result can be applied to additional argument(s).
>
> -Brent
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