存储具有相同特征和不同输出类型的结构[重复]

Question

我试图在Vec中存储具有共同特征的不同结构

游乐场

fn main()
{
    let p = TheirPoint ...
    let l = TheirLine ...
    let vec_shapes: Vec<Box<dyn TheirShadow>> = vec![Box::new(l), Box::new(p)];
}

impl TheirShadow for TheirPoint {
    type Output = (TheirPoint, TheirPoint);
...
}

impl TheirShadow for TheirLine {
    type Output = (TheirPoint, TheirPoint, TheirPoint, TheirPoint);
...
}

pub trait TheirShadow {
    type Output;
...
}

error[E0191]: the value of the associated type `Output` (from trait `TheirShadow`) must be specified
  --> src/main.rs:32:33
   |
32 |     let vec_shapes: Vec<Box<dyn TheirShadow>> = vec![Box::new(l), Box::new(p)];
   |                                 ^^^^^^^^^^^ help: specify the associated type: `TheirShadow<Output = Type>`
...
73 |     type Output;
   |     ------------ `Output` defined here

这可以实现吗，或者是否像在另一个问题答案中一样，集合中所有项目的所有方法的签名必须相同，以便您可以互换使用它们？




Answer 1

匿名用户

由于您不知道< code>Output，您实际上不能使用返回< code>Output的函数，因为它们的大小都不同，Rust无法知道返回的是哪一个。解决这个问题最简单的方法是使用枚举:

enum Shadow {
    Tuple2(TheirPoint, TheirPoint),
    Tuple4(TheirPoint, TheirPoint, TheirPoint, TheirPoint)
}

impl TheirShadow for TheirPoint {
    type Output = Shadow;
    fn cast_shadow(self) -> Self::Output {
        Shadow::Tuple2(some_point, some_other_point)
    }
}

// and so on

您甚至可以消除输出类型，只需在trait中强制返回Shadow。