Hacker News Reader

Comment by jerf | Hacker News Reader

In my opinion, in a programming context, that's Haskell specific. Most programmers in other languages won't worry about the laws. The laws are less useful and less relevant in strict languages, as is the case with many laws, which is why Haskell programmers are just about the only ones discussing them.

I label this my opinion quite on purpose, because I see the alternative viewpoint. I think you'll be more successful taking about lawfulness once someone gets the mechanics, given how much that has been a stopper over the years!

View on HN

Don't get me wrong - I'm not saying the article is bad or not useful.

However, Monad and Functor are well defined terms. If you write an article where you complain about confusing explanations and try to do better, then you must start with a disclaimer: "I'm not explaining Monads as they are defined in math. I'm explaining something that is quite similar to it to make it more practical and later explain the actual term" or so.

Otherwise you are just adding to the confusion that people experience.

> The laws are less useful and less relevant in strict languages, as is the case with many laws, which is why Haskell programmers are just about the only ones discussing them.

I'm not sure if I would agree with that. But even if I would, the laws are still really important, even in strict languages. I've implemented quite a few monad instances in Scala (which is a strict language) and I'm speaking from experience.

I don't know why just saying the following block of text combined with walking through a bunch of examples and how each of the examples fulfills those properties, wouldn't be sufficient : ' If F is a functor, then for any A,B,C,f,g, such that f : A -> B , and g : B -> C, (where we write composition with a semicolon so that the composition of f and g is written f;g : A -> C), F(f) : F(A) -> F(B) and F(g) : F(B) -> F(C) and F(f;g) : F(A) -> F(C) and F(f;g) = F(f) ; F(g) (also, if id_A : A -> A is the identity on A, the F(id_A) : F(A) -> F(A) is the identity on F(A) ) '

(in case they are unfamiliar with the notation f : A -> B , then this can be explained in each of the first few examples.)

I guess it might also be good to give some counterexamples, and point out how the counterexamples fail to be functors, and why they are also not nice.

Reply on news.ycombinator.com