在 Rust 的 trait 对象上克隆 Rc 指针？

Question

我正在学习 Rust，但不明白为什么以下内容不起作用。我想我们无法在 trait 对象上克隆 Rc 指针？如some_function 中所尝试的，我如何将此类引用传递给仅由特征定义的函数？

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，您会看到：

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>`

这意味着编译器错误地将 s 的类型推断为 Rc<SomeThing> 而不是您正在寻找的 Rc<dyn SomeTrait>，这可以通过向 let s 提供明显不正确的类型的常用技巧来证实：

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 invariant、Rc<SomeThing> 和 Rc<dyn SomeTrait> 是完全不兼容的值，因此无法将一个用于另一个。

解决方案是简单地正确输入s：

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，中间的两个调用不起作用，因为第一个将移动本地 Rc，而第二个（也是最后一个调用）仍然需要它。