[Haskell-cafe] foldl in terms of foldr

[Xingzhi Pan]

On Tue, Jan 26, 2010 at 11:51 PM, Neil Brown <nccb2 at kent.ac.uk> wrote:
> Xingzhi Pan wrote:
>>
>> 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?
>>
>
> 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.
>

My mistake with the foldr signature.

I'm a little confused with the type of step here.  Can it be
considered as taking 2 or 3 arguments and then the compiler has to
infer to decide?  Say if I, as a code reader, meet such a function
defined with three formal parameters, how can I draw the conclusion of
its type (and it takes 2 arguments actually)?

Thanks.
-- 
Pan, Xingzhi
http://www.panxingzhi.net