In foldl definition possible wrong in SML/NJ 110.75, I found that the relation foldl (op -) 2 [1] = foldr (op -) 2 [1] holds. But when I tried the above in Haskell I found that the above relation rewritten in Haskell as foldl (-) 2 [1] == foldr (-) 2 [1] doesn't hold. Why is this? Does Haskell have different definition for fold than SML/NJ?
Thanks
Long story short, they are essentially the same, with one minor difference: the order of the arguments passed to the operator (the combining function you pass to fold) are flipped. And since subtraction is not commutative, it will produce different results.
In Haskell (as well as OCaml, C++, Clojure, Common Lisp, Erlang, F#, JavaScript, PHP, Python, Ruby, Scala, and many others), for foldl, the supplied function's first argument is the initial value, or the "folded value so far", while the second argument is an element from the list.
However, in Standard ML, the supplied function's first argument is the element from the list, and the second argument is the initial value, or the "folded value so far".
Neither is "correct" or "incorrect". The order of arguments is purely a design decision. The way Haskell does it is more commonly used today across languages. And in a certain "graphical" way of looking at folding, it makes more sense. Why did SML define theirs the way they did? I am not sure. Perhaps so that the signatures of foldl and foldr will be the same.