在 Rust 中将盒装特征转换为可变特征引用 [英] Convert boxed trait to mutable trait reference in Rust

问题描述

我在 Rust 中的动态调度指针类型方面遇到了一些问题.我想将 Box 类型的值转换为 &mut MyTrait 以传递给函数.例如，我试过:

I am having some trouble with the dynamic dispatch pointer types in Rust. I want to convert a value of type Box<MyTrait> to &mut MyTrait to pass to a function. For example, I tried:

use std::borrow::BorrowMut;

trait MyTrait {
    fn say_hi(&mut self);
}

struct MyStruct { }

impl MyTrait for MyStruct {
    fn say_hi(&mut self) {
        println!("hi");
    }
}

fn invoke_from_ref(value: &mut MyTrait) {
    value.say_hi();
}

fn main() {
    let mut boxed_trait: Box<MyTrait> = Box::new(MyStruct {});
    invoke_from_ref(boxed_trait.borrow_mut());
}

失败并出现以下错误:

error: `boxed_trait` does not live long enough
  --> <anon>:22:5
   |
21 |         invoke_from_ref(boxed_trait.borrow_mut());
   |                         ----------- borrow occurs here
22 |     }
   |     ^ `boxed_trait` dropped here while still borrowed
   |
   = note: values in a scope are dropped in the opposite order they are created

奇怪的是，这适用于 &MyTrait 但不适用于 &mut MyTrait.有什么办法可以让这种转换在可变情况下工作吗?

Strangely enough, this works for &MyTrait but not for &mut MyTrait. Is there any way I can get this conversion to work in the mutable case?

推荐答案

我认为您遇到了当前编译器生命周期处理的限制.borrow_mut 作为一个函数，对生命周期的要求比必要的要严格.

I think you're running into a limitation of the current compiler's lifetime handling. borrow_mut, being a function, imposes stricter lifetime requirements than necessary.

相反，您可以通过首先取消引用框来对框的内部进行可变借用，如下所示:

Instead, you can take a mutable borrow to the box's interior by first dereferencing the box, like this:

fn main() {
    let mut boxed_trait: Box<MyTrait> = Box::new(MyStruct {});
    invoke_from_ref(&mut *boxed_trait);
}

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