在 Rust 中克隆一个 Rc 指针到一个特征对象上？

Question

我正在学习 Rust，不明白为什么以下内容不起作用。我想我们无法在特征对象上克隆 Rc 指针？我如何将这样的引用传递给仅由特征定义的函数，如尝试的那样some_function？

use std::rc::Rc;

trait SomeTrait {}
struct SomeThing {}
impl SomeTrait for SomeThing {}

fn some_function(s: Rc<dyn SomeTrait>) {}
fn another_function(s: &Rc<dyn SomeTrait>) {}

fn main() {
    let s = Rc::new(SomeThing{});

    // This doesnt work
    some_function(Rc::clone(&s));

    // I could do this
    some_function(s);

    // But then I could not do this
    some_function(s);
    
    // For that matter, neither can I do this
    another_function(&s);
}

Answer 1

如果您查看编译器的错误消息，您将看到以下内容：

error[E0308]: mismatched types
   |
14 |     some_function(Rc::clone(&s));
   |                             ^^ expected trait object `dyn SomeTrait`, found struct `SomeThing`
   |
   = note: expected reference `&Rc<dyn SomeTrait>`
              found reference `&Rc<SomeThing>`

s这意味着编译器错误地推断了to的类型，Rc<SomeThing>而不是Rc<dyn SomeTrait>您正在寻找的类型，这可以通过提供明显错误的类型 to 的常见技巧来确认let s：

error[E0308]: mismatched types
  --> src/main.rs:11:17
   |
11 |     let s: () = Rc::new(SomeThing{});
   |            --   ^^^^^^^^^^^^^^^^^^^^ expected `()`, found struct `Rc`
   |            |
   |            expected due to this
   |
   = note: expected unit type `()`
                 found struct `Rc<SomeThing>`

由于 Rust 包装器是不变的，anRc<SomeThing>和 anRc<dyn SomeTrait>是完全不兼容的值，因此无法仅将一个值用于另一个值。

解决方案是简单地正确明确s地键入：

use std::rc::Rc;

trait SomeTrait {}
struct SomeThing {}
impl SomeTrait for SomeThing {}

fn some_function(s: Rc<dyn SomeTrait>) {}
fn another_function(s: &Rc<dyn SomeTrait>) {}

fn main() {
    let s: Rc<dyn SomeTrait> = Rc::new(SomeThing{});

    // This doesnt work
    some_function(Rc::clone(&s));
    
    // For that matter, neither can I do this
    another_function(&s);
}

显然，中间的两个调用无法工作，因为第一个调用将移动local Rc，而第二个调用（和最后一个调用）仍然想要移动 local 。