CODEWITHSUNDEEP
haskellfoldmonoids
fragant
fragant

Reputation: 362

Folds above functions - monoids

Why is the fold above the length of a list no monoid?

length = foldr (\_ n -> 1+n) 0

Isn't it associative and has the neutral element 0 so that it should be a monoid?

Upvotes: 0

Views: 126

Answers (1)

Ørjan Johansen
Ørjan Johansen

Reputation: 18189

It is hard for me to understand what you're really asking, and I have a suspicion that the exact phrasing of the claim from your lecture might matter.

But by the most intuitive interpretation I can think of you are right, as mathematically, the length function is a "monoid homomorphism", mapping the monoid of lists with the concatenation operation to the monoid of integers with the addition operation.

It's technically not so in Haskell, but mostly for the reason that number types haven't been given a Monoid instance because there are two obvious operations to choose between, addition and multiplication. Another reason is that lists in Haskell can be infinite, so length doesn't always give a result.

Upvotes: 3

Related Questions