用了谁的 borrow_mut

use std::cell::RefCell;
use std::rc::Rc;
// use std::borrow::BorrowMut;
fn main(){
    let value = Rc::new(RefCell::new(5));
    *value.borrow_mut() += 10;
    println!("{:?}", value);
}

当打开注释的 use 时，返回错误如下：

Compiling playground v0.0.1 (/playground)
error[E0368]: binary assignment operation `+=` cannot be applied to type `Rc<RefCell<{integer}>>`
 --> src/main.rs:7:5
  |
7 |     *value.borrow_mut() += 10;
  |     -------------------^^^^^^
  |     |
  |     cannot use `+=` on type `Rc<RefCell<{integer}>>`
error: aborting due to previous error
For more information about this error, try `rustc --explain E0368`.

BorrowMut 为所有类型都实现了：

#[stable(feature = "rust1", since = "1.0.0")]
impl<T: ?Sized> BorrowMut<T> for T {
    fn borrow_mut(&mut self) -> &mut T {
        self
    }
}

当引入 BorrowMut 之后，impl生效也是理所当然的事，T会特化为Rc<_>。
*(res.borrow_mut())（此处加不加括号效果是一样的）的解引用类型是Rc<_>，一般来说原类型就是&Rc<_>或&mut Rc<_>。

RefCell borrow_mut 定义如下：

    #[stable(feature = "rust1", since = "1.0.0")]
    #[inline]
    #[track_caller]
    pub fn borrow_mut(&self) -> RefMut<'_, T> {
        self.try_borrow_mut().expect("already borrowed")
    }

可以通过显示解引用来避免歧义：

use std::ops::Deref;
*value.deref().borrow_mut() += 10;

https://rustcc.cn/article?id=843f8aa9-d549-4ac7-9646-05dd6202fe85