关于rust：为什么通过DerefMut可变地借用一个关闭操作无效？

Why does a mutable borrow of a closure through DerefMut not work?

我试图可变地借用一个可变变量。 Deref和DerefMut为Foo实现，但编译失败：

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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();
}

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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：

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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`

相关讨论

这是有关如何通过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显式地获取可变的引用：

1 2	let mut t = Foo; (&mut *t)();

或致电DerefMut::deref_mut：

1 2	let mut t = Foo; t.deref_mut()();