Rust 为什么我在这里得到一个闭包实现`FnMut`，所以对捕获变量的引用不能逃脱闭包

我有一个普通的Interface号特征.Interface有一个method，它将使用动态调度来调用.具体地说，我将在Rc<RefCell<dyn Interface>>岁左右通过.

我有一个用例，其中我需要将对method的调用包装在闭包中.所需的行为是使用input调用闭包将与使用input调用obj.method相同，其中obj实现Interface.

以下是我到目前为止的try .

use std::cell::RefCell;
use std::rc::Rc;

pub trait Interface<'p, T> {
    fn method<'s: 'p>(&'p self, input: &'s str) -> T;
}

fn wrap_interface_with_closure<'p, 's: 'p, T: 'p>(
    instance: Rc<RefCell<dyn Interface<'p, T>>>,
) -> impl FnMut(&'s str) -> T + 'p {
    move |input| (*instance).borrow().method(input)
}

这会给我带来以下错误:

error[E0716]: temporary value dropped while borrowed
   |
13 |     instance: Rc<RefCell<dyn Interface<'p, T>>>,
   |     -------- lifetime `'2` appears in the type of `instance`
14 | ) -> impl FnMut(&'s str) -> T + 'p {
15 |     move |input| (*instance).borrow().method(input)
   |                  ^^^^^^^^^^^^^^^^^^^^--------------
   |                  |                                |
   |                  |                                temporary value is freed at the end of this statement
   |                  creates a temporary which is freed while still in use
   |                  argument requires that borrow lasts for `'2`

error: lifetime may not live long enough
   |
13 |     instance: Rc<RefCell<dyn Interface<'p, T>>>,
   |     -------- lifetime `'2` appears in the type of `instance`
14 | ) -> impl FnMut(&'s str) -> T + 'p {
15 |     move |input| (*instance).borrow().method(input)
   |     ------------   ^^^^^^^^ closure capture requires that `'1` must outlive `'2`
   |     |
   |     lifetime `'1` represents this closure's body
   |
   = note: closure implements `FnMut`, so references to captured variables can't escape the closure

error: aborting due to 2 previous errors; 2 warnings emitted

当它说closure implements FnMut, so references to captured variables can't escape the closure时，它指的是哪个捕获的变量？是不是说instance号逃走了？这是否意味着借款正在逃逸？