Xingzhi Pan wrote:
On Tue, Jan 26, 2010 at 11:24 PM, Eduard Sergeev
<eduard.serg...@gmail.com> wrote:
Xingzhi Pan wrote:
The first argument to foldr is of type (a -> b -> a), which takes 2
arguments. But 'step' here is defined as a function taking 3
arguments. What am I missing here?
You can think of step as a function of two arguments which returns a
function with one argument (although in reality, as any curried function,
'step' is _one_ argument function anyway):
step :: b -> (a -> c) -> (b -> c)
e.g. 'step' could have been defined as such:
step x g = \a -> g (f a x)
to save on lambda 'a' was moved to argument list.
Right. But then step is of the type "b -> (a -> c) -> (b -> c)". But
as the first argument to foldr, does it agree with (a -> b -> a),
which was what I saw when I type ":t foldr" in ghci?
step is of type b -> (a -> a) -> (a -> a), which does agree with (a -> b
-> b), the first argument to foldr (what you posted, both times, a -> b
-> a, is the type of the first argument of *foldl* not foldr). The code
is building up a function (type: a -> a) from the list items, which it
then applies to the initial value given to foldl.
Thanks,
Neil.
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
