Hacker News new | ask | show | jobs
by bodhi 4051 days 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]
3 comments
kimundi 4051 days 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.

link
bodhi 4051 days ago
Aha, thanks (to twic also). Now I get it!
link
dbaupp 4051 days 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.)
link
evanpw 4051 days 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. One other difference is that in the Haskell showStatic, static dispatch is not a guarantee, just an optimization.
link
bodhi 4050 days 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%20...

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

link
bodhi 4050 days 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]
link
twic 4051 days 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.
link