Free monads from scratch
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Free monads from scratch
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Re: Free monads from scratch
#2Re: Free monads from scratch
#3Yes so why doesn’t some GHC extension do this?
The issue I have with this and Functors and Monads and Monad transformers is that there is a lot of boilerplate. I wish i could apply f x y instead of lift $ f x pure y. It would be awesome if there was a ghc extension which, whenever it encounters something like ‘(a -> b -> c) applied to (f a and f b, expected to return g (f c))’ it automatically adds the appropriate ‘fmap’, ‘bind’, ‘pure’, ‘lift’, ‘inj’, ‘Impure’ etc. Then combined with do-notation, you can write truly imperative-looking code, without having to manually convert ‘a’ and ‘State a’ and ‘Reader a’ and ‘Exception a’ into ‘(State :+: Reader :+: Exception) a’.
IMO free monads are an implementation detail for how to enable typed effects and subtypes in a pure language. They’re still kind of important for users to understand, but they would be easier to explain if you let users write imperative-looking code first and then convert it into free monads. That would fix the question most people have when learning about category theory, “why?” - most people don’t know what to make with category theory abstractions because they don’t know what they’re for.
Re: Free monads from scratch
#4> While it looks like boilerplate, we can more or less mechanically write out the instances… Yes so why doesn’t some GHC extension do this? The issue I have with this and Functors and Monads and Monad transformers is that there is a lot of boilerplate. I wish i could apply f x y instead of lift $ f x pure y. It would be awesome if there was a ghc extension which, whenever it encounters something like ‘(a -> b -> c) a…
In "Applicative Programming with Effects"[0], the authors use such a notation, and even leave it as an exercise to implement it in Haskell using MultiParamTypeClasses:
> Given Haskell extended with multi-parameter type classes, enthusiasts for overloading may replace ‘⟦’ and ‘⟧’ by identifiers i[ and ]i with the right behaviour.
So they have code that looks like
eval :: Exp v -> Env v -> Int
eval (Var x) = fetch x
eval (Val i) = ⟦ i ⟧
eval (Add p q) = ⟦ (+) (eval p) (eval q) ⟧
> Yes so why doesn’t some GHC extension do this?It's possible to derive the functor instance automatically if the data type is built from products, sums and arrows. However even for Applicatives you can't do this in general because there can be many lawful Applicative instances for the same type, for instance[1] for lists.
[0] http://strictlypositive.org/IdiomLite.pdf
[1] https://hackage.haskell.org/package/base-4.16.1.0/docs/Contr...
Re: Free monads from scratch
#5Monads are hard. I read many articles explaining what a monad is and what is good for but I never really grasped the concept. I probably have to learn Haskell.
Re: Free monads from scratch
#6Monads are hard. I read many articles explaining what a monad is and what is good for but I never really grasped the concept. I probably have to learn Haskell.
Re: Free monads from scratch
#7Monads are hard. I read many articles explaining what a monad is and what is good for but I never really grasped the concept. I probably have to learn Haskell.
https://fsharpforfunandprofit.com/rop/ is by far my most recommended starting point, using a railway analogy. To grossly oversimplify, as far as I can tell, a monad is a type well-defined enough that you can hook up functions that work with it in a “parallel tracks” manner. So it’s great for chained functions, and reusing and weaving those functions, where you accumulate information other than the “obvious” thing to…
Re: Free monads from scratch
#8Monads are hard. I read many articles explaining what a monad is and what is good for but I never really grasped the concept. I probably have to learn Haskell.
They are a more general construction in category theory but knowing that doesn't tell you how to use them (because the full generality is not often needed).
Re: Free monads from scratch
#9Monads are hard. I read many articles explaining what a monad is and what is good for but I never really grasped the concept. I probably have to learn Haskell.
Re: Free monads from scratch
#10Monads are hard. I read many articles explaining what a monad is and what is good for but I never really grasped the concept. I probably have to learn Haskell.
If you've internalized reasoning about a program's execution symbolically and in a time-independent way, then monads solve the problem of how you enforce a correct sequencing of otherwise time-independent operations at a library level without any fancy modifications to the type system. And oh by the way, this structure shows up everywhere and isn't that pretty cool. People who hang around in the formal PL world tend to assume this already because it emerges naturally from how we talk about the semantics of languages via symbolic manipulation.
But if you haven't grasped that yet, then monads solve a problem that you probably don't even realize exists and no amount of rephrasing the monad laws will help you.
I'd almost always recommend anyone new to Haskell ignore monads as much as possible. Fiddle around with evaluating toy functions with pen and paper symbolically to reason about the semantics, and then try to imagine how you could represent the state of a program changing across the page by creating a new object representing the state of the program from the old state. Then, dig in to the RealWorld type and how the IO Monad actually works (not at a type level).