仅当自我：大小时才具有超级特质

Question

Lets suppose I have a trait

trait A {
    fn new() -> Self where Self : Sized;
    fn foo(&self) -> i32;
}
struct B {
   data : i32
}
impl A for B {
    fn new() -> Self {
        B {data : 42}
    }
    fn foo(&self) -> i32 {
        self.data
    }
}

Now I can use Box<dyn A> , I just do not have the new() method available. And when I have a generic T: A , I can do T::new() . So I can use A as a trait object (without the functions which would prevent this) and I can use it in a templated code and use all functions on it.

My question is, is it possible to get this behavior when having for example Clone as supertrait? In the dyn A case A does not implement Clone. In the generic case is does.

Answer 1

You can do like that:

trait A {
    fn new() -> Self where Self : Sized;
    fn foo(&self) -> i32;
}

#[derive(Clone)]
struct B {
   data : i32
}

impl A for B {
    fn new() -> Self {
        B {data : 42}
    }
    fn foo(&self) -> i32 {
        self.data
    }
}


fn make_clone<T: Clone + A>(toc: &T) -> T {
    dbg!(toc.foo());
    toc.clone()
}

fn main() {
    let b = B{data: 0};
    make_clone(&b);
}

so you can access A method and the Clone supertrait from make_clone , you can obviously do the same thing with a supertrait instead of a generic method