Question

我想创建一个闭包，该闭包以对某项的可变引用为基础，然后返回从原始引用创建的可变引用。

我已将代码最小化为此功能（playground）：

fn produce_closure<'a>(string: &'a mut String) -> impl FnMut() -> &'a mut String + 'a {
    move || &mut *string
}

当我尝试编译以上代码时，出现此错误：

error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
 --> src/lib.rs:2:13
  |
2 |     move || &mut *string
  |             ^^^^^^^^^^^^
  |
note: first, the lifetime cannot outlive the lifetime '_ as defined on the body at 2:5...
 --> src/lib.rs:2:5
  |
2 |     move || &mut *string
  |     ^^^^^^^^^^^^^^^^^^^^
note: ...so that reference does not outlive borrowed content
 --> src/lib.rs:2:13
  |
2 |     move || &mut *string
  |             ^^^^^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the function body at 1:20...
 --> src/lib.rs:1:20
  |
1 | fn produce_closure<'a>(string: &'a mut String) -> impl FnMut() -> &'a mut String + 'a {
  |                    ^^
note: ...so that reference does not outlive borrowed content
 --> src/lib.rs:2:13
  |
2 |     move || &mut *string
  |             ^^^^^^^^^^^^

为了弄清楚这种闭合是否可行，我尝试手工编写（playground）：

struct HandClosure<'a>(&'a mut String);

impl<'a> HandClosure<'a> {
    fn produce(capture: &'a mut String) -> Self {
        Self(capture)
    }

    fn call_mut(&'a mut self) -> &'a mut String {
        self.0
    }

    fn call_once(self) -> &'a mut String {
        self.0
    }
}

该代码可以正常编译。这是编译器错误吗？我想念什么？

为什么我不能创建一个封闭的东西，它产生对封闭东西的可变引用？

0 个答案: