CODEWITHSUNDEEP
rust
SBSTP
SBSTP

Reputation: 3639

Alias generic trait with default types

I have a trait that is generic: trait Trait<T> and I want to create another trait that specifies the generics: type Alias = Trait<String>. This would allow impl Alias for T and not have to specify the type parameters. I tried a couple ways of doing this and haven't found any that works.

This is not a duplicate of Type alias for multiple traits or Aliasing trait with associated types because doing trait Alias: Trait<T> requires people to implement Trait<T> anyway. I want to offer a trait that hides the generics.

A clearer code sample:

trait DefaultEvents = Events<UserStruct, ChannelStruct, IrcStruct>;

struct MyHandler;

impl DefaultEvents for MyHandler {
    ...
}

Upvotes: 4

Views: 1042

Answers (1)

ampron
ampron

Reputation: 3636

Here's my best suggestion, it's going to mean a bit more work on your part (with lots of manual trait inheritance), but it should achieve the user convenience that you want.

pub mod user_friendly {
    pub trait GivesNum<T> {
        fn get_num(&self) -> T;
    }

    pub trait GivesDouble {
        fn get_double(&self) -> f64;
    }

    impl<S> GivesNum<f64> for S where S: GivesDouble {
        fn get_num(&self) -> f64 { self.get_double() }
    }
}

// now your library's user needs to do less
use user_friendly::*;

struct MyStruct { num: f64 }

impl GivesDouble for MyStruct {
    fn get_double(&self) -> f64 { 2.0 * self.num }
}

fn main() {
    let s = MyStruct{ num: 5.0 };
    println!("MyStruct.get_num() = {}", s.get_num());
}

Try it on Rust Playground

Upvotes: 1

Related Questions