如何克隆存储盒装特征对象的结构？

Question

我编写了一个程序，该程序具有 Animal 特征和实现该特征的结构 Dog。它还有一个结构体AnimalHouse，将动物存储为特征对象Box<Animal>。

trait Animal {
    fn speak(&self);
}

struct Dog {
    name: String,
}

impl Dog {
    fn new(name: &str) -> Dog {
        return Dog {
            name: name.to_string(),
        };
    }
}

impl Animal for Dog {
    fn speak(&self) {
        println!{"{}: ruff, ruff!", self.name};
    }
}

struct AnimalHouse {
    animal: Box<Animal>,
}

fn main() {
    let house = AnimalHouse {
        animal: Box::new(Dog::new("Bobby")),
    };
    house.animal.speak();
}

它返回“鲍比：拉夫，拉夫！”正如预期的那样，但如果我尝试克隆house，编译器会返回错误：

fn main() {
    let house = AnimalHouse {
        animal: Box::new(Dog::new("Bobby")),
    };
    let house2 = house.clone();
    house2.animal.speak();
}

error[E0599]: no method named `clone` found for type `AnimalHouse` in the current scope
  --> src/main.rs:31:24
   |
23 | struct AnimalHouse {
   | ------------------ method `clone` not found for this
...
31 |     let house2 = house.clone();
   |                        ^^^^^
   |
   = help: items from traits can only be used if the trait is implemented and in scope
   = note: the following trait defines an item `clone`, perhaps you need to implement it:
           candidate #1: `std::clone::Clone`

我尝试在struct AnimalHouse 之前添加#[derive(Clone)] 并得到另一个错误：

error[E0277]: the trait bound `Animal: std::clone::Clone` is not satisfied
  --> src/main.rs:25:5
   |
25 |     animal: Box<Animal>,
   |     ^^^^^^^^^^^^^^^^^^^ the trait `std::clone::Clone` is not implemented for `Animal`
   |
   = note: required because of the requirements on the impl of `std::clone::Clone` for `std::boxed::Box<Animal>`
   = note: required by `std::clone::Clone::clone`

如何使结构AnimalHouse 可克隆？一般来说，积极使用 trait 对象是惯用的 Rust 吗？

Answer 1

有几个问题。首先是没有什么要求Animal 也实现Clone。您可以通过更改特征定义来解决此问题：

trait Animal: Clone {
    /* ... */
}

这将导致 Animal 不再是对象安全的，这意味着 Box<dyn Animal> 将变得无效，所以这不是很好。

你可以做的是插入一个额外的步骤。 To whit（来自@ChrisMorgan's comment 的补充）。

trait Animal: AnimalClone {
    fn speak(&self);
}

// Splitting AnimalClone into its own trait allows us to provide a blanket
// implementation for all compatible types, without having to implement the
// rest of Animal.  In this case, we implement it for all types that have
// 'static lifetime (*i.e.* they don't contain non-'static pointers), and
// implement both Animal and Clone.  Don't ask me how the compiler resolves
// implementing AnimalClone for Animal when Animal requires AnimalClone; I
// have *no* idea why this works.
trait AnimalClone {
    fn clone_box(&self) -> Box<dyn Animal>;
}

impl<T> AnimalClone for T
where
    T: 'static + Animal + Clone,
{
    fn clone_box(&self) -> Box<dyn Animal> {
        Box::new(self.clone())
    }
}

// We can now implement Clone manually by forwarding to clone_box.
impl Clone for Box<dyn Animal> {
    fn clone(&self) -> Box<dyn Animal> {
        self.clone_box()
    }
}

#[derive(Clone)]
struct Dog {
    name: String,
}

impl Dog {
    fn new(name: &str) -> Dog {
        Dog {
            name: name.to_string(),
        }
    }
}

impl Animal for Dog {
    fn speak(&self) {
        println!("{}: ruff, ruff!", self.name);
    }
}

#[derive(Clone)]
struct AnimalHouse {
    animal: Box<dyn Animal>,
}

fn main() {
    let house = AnimalHouse {
        animal: Box::new(Dog::new("Bobby")),
    };
    let house2 = house.clone();
    house2.animal.speak();
}

通过引入clone_box，我们可以解决试图克隆特征对象的问题。

Answer 2

我的dyn-clone crate 实现了DK.'s answer 的可重用版本。有了它，您只需进行最少的更改即可使您的原始代码正常工作。

• 一行添加DynClone 作为Animal 的超特征，要求每个动物实现都是可克隆的。
• 一行代码为Box<dyn Animal> 生成标准库Clone 的实现。

---

// [dependencies]
// dyn-clone = "1.0"

use dyn_clone::{clone_trait_object, DynClone};

trait Animal: DynClone {
    fn speak(&self);
}

clone_trait_object!(Animal);

#[derive(Clone)]
struct Dog {
    name: String,
}

impl Dog {
    fn new(name: &str) -> Dog {
        Dog { name: name.to_owned() }
    }
}

impl Animal for Dog {
    fn speak(&self) {
        println!{"{}: ruff, ruff!", self.name};
    }
}

#[derive(Clone)]
struct AnimalHouse {
    animal: Box<dyn Animal>,
}

fn main() {
    let house = AnimalHouse {
        animal: Box::new(Dog::new("Bobby")),
    };
    let house2 = house.clone();
    house2.animal.speak();
}

Answer 3

previous answer 正确回答了有关存储盒装特征对象的问题。

关于标题的话题，但不是关于使用 trait 对象的惯用方式，另一种解决方案是使用 Rc 智能指针而不是 Box：这避免了绕过对象的解决方法安全：

#[derive(Clone)]
struct AnimalHouse {
    animal: Rc<Animal>,
}

fn main() {
    let house = AnimalHouse { animal: Rc::new(Dog::new("Bobby")) };
    let house2 = house.clone();
    house2.animal.speak();
}

注意：Rc<T>仅用于单线程场景；还有Arc<T>。

Answer 4

我尝试使用 Dk 和 dtolnay 中的解决方案，在这样的情况下，我需要一个结构，其中包含一个带有框的成员在一个衍生任务中（通过 tokio）。在那里我收到结构未发送和同步的错误。为了避免这种情况，可以在 Dk 克隆特征中添加发送和同步。也许这也可以添加到 dyn_clone 中。