为什么通过DerefMut可变地借用闭包是行不通的？ [英] Why does a mutable borrow of a closure through DerefMut not work?

问题描述

我正在尝试可变地借用可变变量。 Foo 实现了 Deref 和 DerefMut ，但是编译失败：

I am trying to mutably borrow a mutable variable. Deref and DerefMut are implemented for Foo, but compilation fails:

use std::ops::{Deref, DerefMut};

struct Foo;

impl Deref for Foo {
    type Target = FnMut() + 'static;
    fn deref(&self) -> &Self::Target {
        unimplemented!()
    }
}

impl DerefMut for Foo {
    fn deref_mut(&mut self) -> &mut Self::Target {
        unimplemented!()
    }
}

fn main() {
    let mut t = Foo;
    t();
}

error[E0596]: cannot borrow immutable borrowed content as mutable
  --> src/main.rs:20:5
   |
20 |     t();
   |     ^ cannot borrow as mutable

较新版本的编译器具有更新的错误消息：

Newer versions of the compiler have an updated error message:

error[E0596]: cannot borrow data in a dereference of `Foo` as mutable
  --> src/main.rs:20:5
   |
20 |     t();
   |     ^ cannot borrow as mutable
   |
   = help: trait `DerefMut` is required to modify through a dereference, but it is not implemented for `Foo`

而言则未实现

推荐答案

这是 a有关如何通过 Deref 推断功能特征的已知问题。解决方法是，您需要通过进行可变的重新借用来明确获得可变的引用：

This is a known issue regarding how the function traits are inferred through Deref. As a workaround, you need to explicitly get a mutable reference by doing a mutable reborrow:

let mut t = Foo;
(&mut *t)();

或调用 DerefMut :: deref_mut ：

let mut t = Foo;
t.deref_mut()();

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