[Haskell-cafe] foldl in terms of foldr
vengeance.storm at gmail.com
Tue Jan 26 10:44:11 EST 2010
On Tue, Jan 26, 2010 at 11:24 PM, Eduard Sergeev
<Eduard.Sergeev at 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?
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