Is it possible to cast a trait object to another trait object? [duplicate]

Answer 1

Another option is to create a trait that uses both TraitA and TraitB as supertraits and provides a cast to each type:

trait TraitC: TraitA + TraitB {
    fn as_trait_a(&self) -> &dyn TraitA;
    fn as_trait_b(&self) -> &dyn TraitB;
}

Then have MyType implement it:

impl TraitC for MyType {
    fn as_trait_a(&self) -> &dyn TraitA {
        self
    }
    fn as_trait_b(&self) -> &dyn TraitB {
        self
    }
}

Once you do that, you can use TraitC for your Box and your program logic that uses both TraitA and TraitB together.

Sample main to show various ways to use:

fn test_a(a: &TraitA) {
    a.say_hello();
}
fn test_b(b: &TraitB) {
    b.say_hello();
}

fn main() {
    let c: Box<dyn TraitC> = Box::new(MyType {});

    TraitA::say_hello(&*c);
    TraitB::say_hello(&*c);

    c.as_trait_a().say_hello();
    c.as_trait_b().say_hello();

    test_a(c.as_trait_a());
    test_b(c.as_trait_b());

    let a: &dyn TraitA = c.as_trait_a();
    a.say_hello();
    let b: &dyn TraitB = c.as_trait_b();
    b.say_hello();
}

Rust Playground

If A and B do truly belong together, this better represents that and still gives you the freedom to use them separately if you desire.

Answer 2

Using Box<MyType> instead of Box<dyn Trait> solves this problem.

fn main() {
    let a = Box::new(MyType {});

    TraitA::say_hello(&*a);
    TraitB::say_hello(&*a);
}

In this case there's no need to use trait objects. Rust has a different paradigm from C++. In most cases you can usually use generic types to solve problems. If your problem is really suitable to solve with trait objects, you can refer to the OOP chapter in the book.