> No, that's not sufficient if done naively as I think you're describing. You seem to be missing the context of this discussion as described by the article. Other code already depends on the current compiled abstraction, and you want to extend it in various ways, safely, so that existing code continues to work without recompilation, but you can extend the abstraction that code is already safely handling with new types and new operations without modifying the original source code.

That is exactly what a new trait would accomplish. I remain confused as to the distinction you are making.

▲ naasking 4 days ago | parent [-]

Read the Haskell-specific follow-up which I mentioned, type classes have a correspondance with traits so the problems with this approach are the same:

https://eli.thegreenplace.net/2018/more-thoughts-on-the-expr...

▲ adastra22 4 days ago | parent [-]

Unfortunately I don’t speak Haskell. As none of the terminology used by Haskell folks align with languages I do know, I really can’t make head or tails of that article.

▲ gf000 4 days ago | parent | next [-]

I tried to translate the article to Rust, but seems that the article's gotcha in Haskell is not necessarily an issue in Rust if we return `dyn OurTrait`, so now I'm even more confused. If anyone could take a look what I'm missing that would be welcome:

------------

Say, you want to write a simple language interpreter.

You have an `Expr` enum (sum type), with say, a `Constant(f64)` and a `BinaryPlus(Expression, Expression)`.

You can easily add a new function expecting an `Expr` indeed, but if you were to add a new variant to the `Expr` enum, you would have to go through the code and change every use site.

You can solve the issue by simply making a `struct Constant` and a `struct BinaryPlus`. Now you can just define a new trait for both of them, and you can use that `dyn trait` in your code -- you can add new functions and also new types without code change at use site!

So what's the issue?

In Haskell, a logic like

```

func example(runtime_val: f64) -> Expr {

  if (runtime_val is someRuntimeCheck())
    return Constant(..)

  else
    return BinaryPlus(.., ..)

}

```

can't compile in itself as `Expr` is a type class (=trait). Basically, in this mode Haskell awaits a concrete implementation (we actually get the exact same behavior with `impl Expr` in Rust), but here is my confusion: this can be circumvented in Rust with dyn traits..

Here is my (very ugly due to just hacking something together while pleasing the borrow checker) code showing it in Rust: https://play.rust-lang.org/?version=stable&mode=debug&editio...

	▲	adastra22 4 days ago \| parent [-]
		Your confusion is my confusion: Rust supports this.

▲ internet_points 4 days ago | parent | prev [-]

There's data definitions:

    data Constant           = Constant Double deriving (Show)

is a bit like `#[derive(Debug)] struct Constant(f64);`, ie. it's just a wrapper around a double and you can print it.

And there's typeclasses. Show is a typeclass. You can think of them as interfaces, so `instance Sqlite DB where open dsn = ...` is a bit like saying Sqlite is an implementation (instance) of the DB interface (typeclass). The typeclass itself could be defined like `class DB where open :: String -> IO ()` meaning the interface requires a function taking a string and having IO access (and of course you can require more than one function in your interface/typeclass).

The article also uses typeclasses with parameters. Parameters (and functions and variables) are written lowercase, while classes and constructors and such are capitalized, so

    class (Expr e) => Stringify e where
      stringify :: e -> String

    instance Stringify Constant where
      stringify (Constant x) = show x

means there's an interface Stringify that's parametrized over some type e (and that e in turn has to be in the Expr typeclass).

▲

adastra22 4 days ago | parent [-]

Thank you. Now I understand the article. And the problem identified for type classes isn’t an issue in Rust. You’d return a ‘dyn Trait’ and be done. It means the object would have to be heap allocated, but that is already a given because you are wanting to define a function that can work with any type including future defined types. I guess Haskell doesn’t support the equivalent of trait objects for type classes?

I’m not trying to prove that rust is better or something. People who do that are annoying. It’s just weird to me that this is being presented as a fundamental and largely unsolved challenge when there is a simple solution at the heart of a widely deployed and well known language, which in turn stole it from elsewhere.

	▲	naasking 3 days ago \| parent [-]
		dyn traits have quite a few restrictions if I'm reading the docs right. Like, dispatchable functions can't even have type parameters [1]. I wouldn't exactly call that "solved", although maybe a language with traits and dyn traits that used GC wouldn't have such onerous restrictions. [1] https://doc.rust-lang.org/reference/items/traits.html#dyn-co...