Re: [Haskell-cafe] foldl in terms of foldr

On Tue, Jan 26, 2010 at 11:51 PM, Neil Brown wrote:

Xingzhi Pan wrote:

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On Tue, Jan 26, 2010 at 11:24 PM, Eduard Sergeev wrote:

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Xingzhi Pan wrote:

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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