View source for Talk:Foldr Foldl Foldl'

In Section ''Rules of Thumb for Folds'' the following is written:<br>
''... you do not care about the implicit reversal (for example, because your combining function is commutative like ...''<br>
Does this suggest that when your binary operator is commutative, foldl and fold r give the same result?<br>
'''But that is not true!''' If the operation is associative then you get always the same result (if both deliver a result).<br>
I can provide a commutative example where foldl and foldr return different result.

Moreover ''foldl' conceptually reverses the order of the list'' is a misleading formulation:<br>
<code>divisel :: [Double] -> Double
divisel = foldl (/) 1

diviser :: [Double] -> Double
diviser = foldr (/) 1

divisel [1,2] = diviser [1,2] = 0.5
</code><br>
Both divide 1 by 2.

I think it would be better to say ''foldl/foldr associates to the left/right.

--[[User:Klt|Klt]] ([[User talk:Klt|talk]]) 14:28, 19 July 2024 (UTC)

As the original author of the erroneous language, I think you are right.

[[User:CarlEdman|CarlEdman]] ([[User talk:CarlEdman|talk]]) 15:07, 19 July 2024 (UTC)