为什么通过DerefMut对一个闭包的可变借用不起作用？

Question

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

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

Answer 1

这是关于如何通过Deref推断出功能特征的a known issue。作为一种解决方法，您需要显式获取可变引用：

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

或

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