[英]How can I make tree-like structure where the nodes are trait objects in Rust?
I have a rust-playground link here , and the full code for my basic example is:我在这里有一个 rust-playground 链接,我的基本示例的完整代码是:
trait Obj {
fn hit(&self, t0: f64, t1: f64) -> bool;
}
struct Leaf {
data: f64,
}
impl Obj for Leaf {
fn hit(&self, t0: f64, t1: f64) -> bool { return t0 < self.data && self.data < t1 }
}
struct Node<'a> {
data: f64,
left: &'a Box<dyn Obj>,
right: &'a Box<dyn Obj>,
}
impl<'a> Obj for Node<'a> {
fn hit(&self, t0: f64, t1: f64) -> bool {
self.left.hit(t0, t1) || self.right.hit(t0, t1)
}
}
impl<'a> Node<'a> {
fn new(tmin: f64, tmax: f64) -> Self {
let mid = (tmin + tmax) / 2.;
if tmax - tmin < 1. {
return Self { data: tmin, left: &Box::new(Leaf {data: tmin }), right: &Box::new(Leaf {data :tmax} ) };
}
let left = Node::new(tmin, mid);
let right = Node::new(mid, tmax);
Self {data: tmin, left: &Box::new(left), right: &Box::new(right) }
}
}
fn main() {
let node = Node::new(0., 100.);
()
}
It's a contrived example, but it gives the same compiler errors I get in my real version.这是一个人为的例子,但它给出了我在真实版本中遇到的相同编译器错误。 In the constructor for Node, I get the following errors:
在 Node 的构造函数中,出现以下错误:
error[E0308]: mismatched types
--> src/main.rs:29:83
|
29 | return Self { data: tmin, left: &Box::new(Leaf {data: tmin }), right: &Box::new(Leaf {data :tmax} ) };
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected trait object `dyn Obj`, found struct `Leaf`
|
= note: expected reference `&'a std::boxed::Box<(dyn Obj + 'static)>`
found reference `&std::boxed::Box<Leaf>`
error[E0308]: mismatched types
--> src/main.rs:33:33
|
33 | Self {data: tmin, left: &Box::new(left), right: &Box::new(right) }
| ^^^^^^^^^^^^^^^ expected trait object `dyn Obj`, found struct `Node`
|
= note: expected reference `&'a std::boxed::Box<(dyn Obj + 'static)>`
found reference `&std::boxed::Box<Node<'_>>`
Why do I get these errors?为什么我会收到这些错误? Both
Leaf
and Node
implement Obj
, and my understanding is that they could both be of type dyn Obj
. Leaf
和Node
都实现了Obj
,我的理解是它们都可以是dyn Obj
类型。 I think it has something to do with lifetimes, but I can't think of how to fix this.我认为它与生命有关,但我想不出如何解决这个问题。 What's the proper way of making this binary tree structure in Rust?
在 Rust 中制作这种二叉树结构的正确方法是什么?
Box<..>
is an owned type, so it can live on its own, been moved without a lifetime track. Box<..>
是一个拥有的类型,因此它可以独立存在,在没有生命周期的情况下被移动。 In your example you simply can remove all of the references and a lifetime, so the struct definition becomes:在您的示例中,您只需删除所有引用和生命周期,因此结构定义变为:
struct Node {
data: f64,
left: Box<dyn Obj>,
right: Box<dyn Obj>,
}
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