Reverse a list in Scheme with foldl and foldr

Question

How can you define a function to reverse a list in Scheme by using foldr and foldl?

What we want is a succinct solution to reverse a list i

Answer 1

Your rev1 'solution' doesn't compile... it would if you replaced list with l

> (define (rev1 l) 
    (foldl cons l '()))
> (rev1 '(1 2 3))
(3 2 1)

For your rev2 this works as the body:

> (foldr (lambda (a b) (append b (list a))) '(1 2 3) '())
(3 2 1)