Why is foldl defined in a strange way in Racket?

Question

In Haskell, like in many other functional languages, the function foldl is defined such that, for example, foldl (-) 0 [1,2,3,4] = -10.

Thi

Answer 1

Racket's foldl and foldr (and also SRFI-1's fold and fold-right) have the property that

(foldr cons null lst) = lst
(foldl cons null lst) = (reverse lst)

I speculate the argument order was chosen for that reason.