在借来的 RefCell<&mut T>
上调用方法(即 Ref<&mut T>
)按预期工作,但我似乎无法将其传递给函数。考虑以下代码:
use std::cell::RefCell;
fn main() {
let mut nums = vec![1, 2, 3];
foo(&mut nums);
println!("{:?}", nums);
}
fn foo(nums: &mut Vec<usize>) {
let num_cell = RefCell::new(nums);
num_cell.borrow_mut().push(4);
push_5(*num_cell.borrow_mut());
}
fn push_5(nums: &mut Vec<usize>) {
nums.push(4);
}
num_cell.borrow_mut().push(4)
有效,但是 push_5(*num_cell.borrow_mut())
错误:
error[E0389]: cannot borrow data mutably in a `&` reference
--> src/main.rs:14:12
|
14 | push_5(*num_cell.borrow_mut());
| ^^^^^^^^^^^^^^^^^^^^^^ assignment into an immutable reference
取消引用 Ref
后,我希望在内部获得可变引用,所以这个错误对我来说真的没有意义。给了什么?
最佳答案
push_5(*num_cell.borrow_mut());
删除*
,编译器建议
error[E0308]: mismatched types
--> src/main.rs:14:12
|
14 | push_5(num_cell.borrow_mut());
| ^^^^^^^^^^^^^^^^^^^^^
| |
| expected mutable reference, found struct `std::cell::RefMut`
| help: consider mutably borrowing here: `&mut num_cell.borrow_mut()`
|
= note: expected type `&mut std::vec::Vec<usize>`
found type `std::cell::RefMut<'_, &mut std::vec::Vec<usize>>`
push_5(&mut num_cell.borrow_mut());
编译。
push_5(num_cell.borrow_mut().as_mut());
也是
关于rust - 将 RefCell<&mut T> 的内容传递给函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49841847/