将泛型结构视为特征 object

Question

I have a struct with generic T for which I want to create a vector of instances of this struct, where for each instance T can be different. I realise I probably need to use a box and possibly need to treat my struct as a trait object, but I am not sure how, since it is not a collection of different structs which implement the same trait, but instead the same struct with a different generic. Below is what I have so far and hopefully illustrates what I am trying to achive, it doesn't work because the dyn keyword seems to expect a trait rather than a generic struct.

struct Point<T> {
    x: T,
    y: T,
}

struct Points {
    points: Vec<Box<dyn Point>>,
}
fn main() {
    let points = Points {
        points: vec![Point { x: 1, y: 2 }, Point { x: 1.1, y: 2.2 }],
    };
}

Answer 1

"The same struct with a different generic" behaves very similarly to "a different struct". Here, when dealing with trait objects, the concrete type doesn't matter, which is the key idea behind a trait object.

For example:

trait Trait {
  fn foo(&self) -> &'static str;
}

struct Point<T> {
  x: T,
  y: T,
}

// you may have different impls for different generics
impl<T> Trait for Point<T> {
  fn foo(&self) -> &'static str {
    "a point"
  }
}

fn main() {
  let points: Vec<Box<dyn Trait>> = vec![
    Box::new(Point {x: 1, y: 2}),
    Box::new(Point {x: 1.1, y: 2.2}),
    Box::new(Point {x: (), y: ()}),
  ];

  for point in points {
    println!("{}", point.foo());
  }
}

If you're already on nightly, and have the box_syntax feature enabled, you can also write box Point { x: 1, y: 2 } to save some boilerplate, although likely not worth switching to nightly for.

Answer 2

You can wrap the points in an enum :

struct Point<T> {
    x: T,
    y: T,
}

struct Points {
    points: Vec<PointType>,
}

enum PointType {
    U64(Point<u64>),
    F32(Point<f32>),
}

fn main() {
    let points = Points {
        points: vec![
            PointType::U64(Point { x: 1, y: 2 }),
            PointType::F32(Point { x: 1.1, y: 2.2 }),
        ],
    };
}

Playground