当两个泛型类型相同时告诉 Rust 编译器

Question

I'm wondering if it is possible to tell to the Rust compiler that two generic types are the same.

In the specific case, I have a trait with a generic method and I want to implement a struct that has the same generic type as field, like in the following example:

trait Trait {
    fn foo<T>(&self, t: T) -> T;
}

struct Struct<T> {
    t: T
}

impl<T> Trait for Struct<T> {
    fn foo<U>(&self, t: U) {
        self.t
    }
}

Ofc here I have a compiler error because it expects U but have a T. How can I handle this case?

Moving the generic outside foo creating Trait<T> is not an option.

Answer 1

You could probably use associated types:

trait Trait {
    type U;
    fn foo(&self, t: Self::U) -> Self::U;
}

struct Struct<T> {
    t: T
}

impl<T: Copy> Trait for Struct<T> {
    type U = T;
    fn foo(&self, t: T) -> T {
        self.t
    }
}

Playground

PS: I added an extra T: Copy because T needs to be Copy to return self.t as it is a shared ref.

Answer 2

fn foo<T>(&self, t: T) -> T;

in a trait means the method must work for all T , and your struct's field can't possibly have all of those types at the same time!

That is, this code will compile given the trait and struct definition:

fn main() {
    let x: Struct<i32> = Struct { t: 0 };
    let y: &str = x.foo("abc");
    let z: [i32; 2] = x.foo([0,0]);
}

and obviously y and z can't be xt .

(after writing the answer, I see the point has already been made in mcarton's comment; I will delete it if he asks, but hopefully it's a bit clearer this way without requiring dyn ).