6 ms·

This is also how interfaces are implemented in Go.

by rwj 11y ago

nostrademons 11y ago
And typeclasses in Haskell as well.
- evanpw 11y ago
  I don't think that's true. Dictionary-passing (or something equivalent) attaches the table of methods to the functions which take typeclass instances as arguments (with extra, hidden arguments), while Rust's Box<Trait> attaches the table of methods to the object pointer itself. So you can have a list of Box<Show> in Rust, but you can't have a list of Show in Haskell. You have to set up the extra indirection yourself using something like this: https://wiki.haskell.org/Existential_type#Dynamic_dispatch_mechanism_of_OOP https://wiki.haskell.org/Existential_type#Dynamic_dispatch_m.... I'm not an expert in Rust or Haskell, though, so I welcome corrections.
  - bodhi 11y ago
    > but you can't have a list of Show in Haskell I'm not sure exactly what you mean, but my naive interpretation is that you can: Prelude> :t map show map show :: Show a => [a] -> [String]
    - dbaupp 11y ago
      I'm not exactly sure what they mean, since the following works, and is pretty much a drop-in replacement for a Box<...> trait object in Rust. data Showable = forall a. Show a => Showable a This allows for [Showable 1, Showable "foo", Showable 'x'] (It requires the ExistentialQuantification language feature.)
      - evanpw 11y ago
        Right, this is exactly what I meant by setting up the indirection yourself. The main difference seems to be that in Haskell, you have to write this code separately for each typeclass, and wrap/unwrap it manually. In Rust, you just change Show to &Show and you get dynamic dispatch with no extra code. Here's a gist of the two approaches: https://gist.github.com/evanpw/89d89aae1159c608c476 https://gist.github.com/evanpw/89d89aae1159c608c476. One other difference is that in the Haskell showStatic, static dispatch is not a guarantee, just an optimization.
        bodhi 11y ago
        This works on single instances, but I'm trying to see how it plays out with lists of `Bar`. As far as I can see, you need to use `Box`: https://play.rust-lang.org/?code=struct%20Foo%3B%0Astruct%20Baz%3B%0A%20%0Atrait%20Bar%20%7B%0A%20%20%20%20fn%20baz(%26self)%20-%3E%20i64%3B%0A%7D%0A%20%0Aimpl%20Bar%20for%20Foo%20%7B%0A%20%20%20%20fn%20baz(%26self)%20-%3E%20i64%20%7B%2010%20%7D%0A%7D%0A%0Aimpl%20Bar%20for%20Baz%20%7B%0A%20%20%20%20fn%20baz(%26self)%20-%3E%20i64%20%7B%2020%20%7D%0A%7D%0A%20%0Afn%20show_static%3CT%3A%20Bar%3E(t%3A%20%26T)%20-%3E%20String%20%7B%0A%20%20%20%20t.baz().to_string()%0A%7D%0A%20%0Afn%20show_dynamic(t%3A%20%26Bar)%20-%3E%20String%20%7B%0A%20%20%20%20t.baz().to_string()%0A%7D%0A%20%0Afn%20main()%20%7B%0A%20%20%20%20let%20foo%20%3D%20Foo%3B%0A%20%20%20%20let%20baz%20%3D%20Baz%3B%0A%20%20%20%20let%20k%3A%5BBox%3C%26Bar%3E%3B%202%5D%20%3D%20%5BBox%3A%3Anew(%26foo)%2C%20Box%3A%3Anew(%26baz)%5D%3B%0A%20%20%20%20println!(%22%7B%7D%20%7B%7D%22%2C%20show_static(%26foo)%2C%20show_dynamic(%26foo))%3B%0A%7D&version=beta https://play.rust-lang.org/?code=struct%20Foo%3B%0Astruct%20... Which seems to me to be effectively the same as `Barrable` in this case, although it's more generic.
        bodhi 11y ago
        > As far as I can see, you need to use `Box` Which is exactly what you said in the first place! > So you can have a list of Box<Show> in Rust, but you can't have a list of Show in Haskell. But isn't this comparing two different things? You can't have a list of trait/typeclass in either language (excuse the pseudo-syntax): Rust: [Bar] Haskell: [Bar a => a] But you can (with some extensions in Haskell) have: Rust: [Box<Bar>] Haskell: [Box Bar]
    - twic 11y ago
      That takes a list of a, where a is some type which is Show. You can't have a grab-bag of different Show things in there.
    - kimundi 11y ago
      That's not a list of Show, that's a list of a single type that instantiates Show. The difference being that in your code you can only put in a single type at a time, eg [Int] or [String], but not both Int and String under the common Show interface type.
      - bodhi 11y ago
        Aha, thanks (to twic also). Now I get it!

This is also how interfaces are implemented in Go.

by rwj 11y ago

This is also how interfaces are implemented in Go.

nostrademons 11y ago

And typeclasses in Haskell as well.

evanpw 11y ago
I don't think that's true. Dictionary-passing (or something equivalent) attaches the table of methods to the functions which take typeclass instances as arguments (with extra, hidden arguments), while Rust's Box<Trait> attaches the table of methods to the object pointer itself. So you can have a list of Box<Show> in Rust, but you can't have a list of Show in Haskell. You have to set up the extra indirection yourself using something like this: https://wiki.haskell.org/Existential_type#Dynamic_dispatch_mechanism_of_OOP https://wiki.haskell.org/Existential_type#Dynamic_dispatch_m.... I'm not an expert in Rust or Haskell, though, so I welcome corrections.
- bodhi 11y ago
  > but you can't have a list of Show in Haskell I'm not sure exactly what you mean, but my naive interpretation is that you can: Prelude> :t map show map show :: Show a => [a] -> [String]
  - dbaupp 11y ago
    I'm not exactly sure what they mean, since the following works, and is pretty much a drop-in replacement for a Box<...> trait object in Rust. data Showable = forall a. Show a => Showable a This allows for [Showable 1, Showable "foo", Showable 'x'] (It requires the ExistentialQuantification language feature.)
    - evanpw 11y ago
      Right, this is exactly what I meant by setting up the indirection yourself. The main difference seems to be that in Haskell, you have to write this code separately for each typeclass, and wrap/unwrap it manually. In Rust, you just change Show to &Show and you get dynamic dispatch with no extra code. Here's a gist of the two approaches: https://gist.github.com/evanpw/89d89aae1159c608c476 https://gist.github.com/evanpw/89d89aae1159c608c476. One other difference is that in the Haskell showStatic, static dispatch is not a guarantee, just an optimization.
      - bodhi 11y ago
        This works on single instances, but I'm trying to see how it plays out with lists of `Bar`. As far as I can see, you need to use `Box`: https://play.rust-lang.org/?code=struct%20Foo%3B%0Astruct%20Baz%3B%0A%20%0Atrait%20Bar%20%7B%0A%20%20%20%20fn%20baz(%26self)%20-%3E%20i64%3B%0A%7D%0A%20%0Aimpl%20Bar%20for%20Foo%20%7B%0A%20%20%20%20fn%20baz(%26self)%20-%3E%20i64%20%7B%2010%20%7D%0A%7D%0A%0Aimpl%20Bar%20for%20Baz%20%7B%0A%20%20%20%20fn%20baz(%26self)%20-%3E%20i64%20%7B%2020%20%7D%0A%7D%0A%20%0Afn%20show_static%3CT%3A%20Bar%3E(t%3A%20%26T)%20-%3E%20String%20%7B%0A%20%20%20%20t.baz().to_string()%0A%7D%0A%20%0Afn%20show_dynamic(t%3A%20%26Bar)%20-%3E%20String%20%7B%0A%20%20%20%20t.baz().to_string()%0A%7D%0A%20%0Afn%20main()%20%7B%0A%20%20%20%20let%20foo%20%3D%20Foo%3B%0A%20%20%20%20let%20baz%20%3D%20Baz%3B%0A%20%20%20%20let%20k%3A%5BBox%3C%26Bar%3E%3B%202%5D%20%3D%20%5BBox%3A%3Anew(%26foo)%2C%20Box%3A%3Anew(%26baz)%5D%3B%0A%20%20%20%20println!(%22%7B%7D%20%7B%7D%22%2C%20show_static(%26foo)%2C%20show_dynamic(%26foo))%3B%0A%7D&version=beta https://play.rust-lang.org/?code=struct%20Foo%3B%0Astruct%20... Which seems to me to be effectively the same as `Barrable` in this case, although it's more generic.
        bodhi 11y ago
        > As far as I can see, you need to use `Box` Which is exactly what you said in the first place! > So you can have a list of Box<Show> in Rust, but you can't have a list of Show in Haskell. But isn't this comparing two different things? You can't have a list of trait/typeclass in either language (excuse the pseudo-syntax): Rust: [Bar] Haskell: [Bar a => a] But you can (with some extensions in Haskell) have: Rust: [Box<Bar>] Haskell: [Box Bar]
  - twic 11y ago
    That takes a list of a, where a is some type which is Show. You can't have a grab-bag of different Show things in there.
  - kimundi 11y ago
    That's not a list of Show, that's a list of a single type that instantiates Show. The difference being that in your code you can only put in a single type at a time, eg [Int] or [String], but not both Int and String under the common Show interface type.
    - bodhi 11y ago
      Aha, thanks (to twic also). Now I get it!