[jhanschoo]

Parent wasn't slightly wrong, you just tunneled in on the signature of the operator function, when the reason why parent comment brought it up was to illustrate that the operator function is not necessarily associative, and that the function signature was meant to suggest the difference in associativity.

You're wrong about the operations possibly not being commutative; as you pointed out, whether the parameter list is (A, B) or (B, A) doesn't really matter, so it doesn't pose a problem.

An example is the subtraction operator: with just two elements, I can change `(a, b) => b - a` to `(b, a) => b - a` with with no difference in result. But there is no way I can write a subtraction function for foldLeft that gives the same results as a foldRight on subtraction in general. That is, I can write op such that a op b = b - a, but not that (a op b) op c = (c - b) - a. Nevertheless, I can still write op such that a op (b op c) = (c - b) - a due to some dual notion of commutativity.

[leafboi]

Function composition is associative but not commutative.

So in essence given three functions z,g,f and composition operator <.>

   f . g . z != z . g . f (commutativity)

which is sort of what fold left or right is doing (but with z g and f being the same function).

but:

   f . g . z == z . (g . f) (associativity)

My edit is right about the operations not being commutative. Parent is wrong about associativity as it has nothing to do with this, but he is right that the codomains of left and right are not equal.

Function composition isn't completely accurate to what's going on, it's a more higher order form of composition going on with fold but the rules remain the same.

Whatever, either way, Overall I'm wrong

[roflc0ptic]

But sincerely, thanks for commenting. If I'm walking around self righteously asserting incorrect stuff I much appreciate people pointing it out

[jhanschoo]

You're right.

> which is sort of what fold left or right is doing (but with z g and f being the same function).

> it's a more higher order form of composition

More precisely, a fold performs function composition on the provided operator curried with the respective elements, so that z g and f above are different functions (hence not commutative in general, but associative in general, wrt folding).