On Sep 30, 2007, at 11:57 , PR Stanley wrote:
Well, note that foldr takes a function of x, which produces a
function of xs. This function of xs either conses x onto it, or
leaves it unchanged. We can write this down explicitly by
removing the xs parameter and just writing what function should be
produced:
filter f = foldr (\x -> if (f x) then (x:) else id) []
That's one neat solution but it also raises a few questions for me:
foldr :: (a -> b -> b) -> b -> [a] -> b
yet you've managed to squeeze in a function that takes only one
argument. How is this possible without GHCI blowing its top?
2. Could you please explain the role of the identity function (id)
in your code? Where's its argument, or is it sort of implicitly
noted? For example, is it the case that the second argument is held
in reserve and passed to the first function which would take it,
hence (x:) and id?
Many thanks, Paul
Those are the same question, actually. He is returning a function
which takes one argument (either the partial application / section
(x:) which expects a list argument, or the function "id" which
expects any argument and returns it unmodified); since one argument
is left unconsumed, that typechecks. (This is, after all, functional
programming; returning functions is not only possible, but encouraged.)
This follows from the fact that all (normal) Haskell functions are
curried, so (\x y -> something) is the same as (\x -> (\y ->
something)), i.e. a 2-argument function is really a 1-argument
function which returns a 1-argument function. Contrariwise, any time
a 2-argument function is expected you can instead pass a 1-argument
function which returns a 1-argument function.
--
brandon s. allbery [solaris,freebsd,perl,pugs,haskell] [EMAIL PROTECTED]
system administrator [openafs,heimdal,too many hats] [EMAIL PROTECTED]
electrical and computer engineering, carnegie mellon university KF8NH
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