On 2008 Oct 25, at 18:44, Glurk wrote:
Or, does it take the only valid parameter I give it, the a, and then form a function out of this (a==), and then compose this new function with filter, which does take a function of this new type. Then, there is one parameter left, [a], which is what this new function needs...and so it all works out !
This is it exactly.
Why doesn't it try to take the 2 parameters for (==)
In reality there are no two-parameter functions. They are actually single-parameter functions which return functions. Thus only one argument will ever be consumed by a particular application. Symbolically: (\a b -> f a b) = \a -> (\b -> f a b) taking argument a and returning the function (\b -> f a b) with a having whatever value it was given in the first application (in other languages this is often referred to as a closure). -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery@kf8nh.com system administrator [openafs,heimdal,too many hats] allbery@ece.cmu.edu electrical and computer engineering, carnegie mellon university KF8NH