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Comment by dbaupp | Hacker News Reader

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.)

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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. One other difference is that in the Haskell showStatic, static dispatch is not a guarantee, just an optimization.

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%20...

Which seems to me to be effectively the same as `Barrable` in this case, although it's more generic.

> 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]

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