[Haskell-beginners] Dependent and independent variables in foldl and foldr

[Lawrence Bottorff]

This is the definition of list foldr

foldr            :: (a -> b -> b) -> b -> [a] -> b
foldr _ z []     =  z
foldr f z (x:xs) =  f x (foldr f z xs)

In both foldl and foldr in the OP the n variable in lambda functions would
seem to be for the accumulator, hence, I assume the n is considered the
free variable? And then the wildcard in each lambda function refers to the
bound variable, i.e., the list argument's elements to be folded. So I can
recreate

foldr (+) 5 [1,2,3,4]

with

foldr (\x n -> x + n) 5 [1,2,3,4]

They both return 15. The first one results in

(+) 1 ((+) 2 ((+) 3 ((+) 4 5)))

but the second example I'm not sure how the (\x n -> x + n) is being
applied in the form . . . f x (foldr f z xs) It obviously must be doing the
same (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) but how the function is being applied
I don't understand.  Beta reduction doesn't get me very far

\x n -> x + n (5)([1,2,3,4])
\x 5 -> x + 5 ([1,2,3,4])

but obviously the enclosing lambda calc for foldr is doing something to
create the (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) form.

BTW, is the t a format in

:type foldr
foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b

something from category theory, i.e., for the list instance, t a is [a] What
is the algebraic syntax where t a becomes [a] in the case of lists? It
would be nice to understand some day exactly what :i Foldable is saying





On Sat, Jan 16, 2021 at 4:36 PM Francesco Ariis <fa-ml at ariis.it> wrote:

> Il 16 gennaio 2021 alle 16:10 Lawrence Bottorff ha scritto:
> > I have this
> >
> > myLength1 = foldl (\n _ -> n + 1) 0
> >
> > and this
> >
> > myLength2 = foldr (\_ n -> n + 1) 0
> >
> > I am guessing that foldl knows to assign the accumulator-seed argument to
> > the dependent variable and the list argument's elements recursively to
> the
> > independent variable; and with foldr to do the opposite. Is this a fair
> > assumption? BTW, where can I get a look at the code for fold functions;
> or
> > does the type definition answer my original question? Not really able to
> > decipher it so well
> >
> >  :t foldl
> > foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
>
> foldl and foldr have slightly different signatures,
>
>     λ> :t +d foldl
>     foldl :: (b -> a -> b) -> b -> [a] -> b
>     λ> :t +d foldr
>     foldr :: (a -> b -> b) -> b -> [a] -> b
>
> (Notice  `b -> a -> b`  vs.  `a -> b -> b`), hence the lambdas have a
> different non-matched parameter.
> Does this answer your question? —F
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