Rust LinkedList 第二个可变借用，同时将第一项移动到末尾

Question

Good day. I was trying to move the first item to the end in a linked list: I have a linked list like this:

let mut list = LinkedList::new();
list.push_back('a');
list.push_back('b');
list.push_back('c');

If I move item like this:

list.push_back(list.pop_front().unwrap());

Rust would throw an "second mutable borrow occurs" error.

But If a get the first item at another line of code:

let str = list.pop_front().unwrap();
list.push_back(str);

Then it's fine.

Why these two pieces of code behave differently? I totally don't understand..

Answer 1

Both push_back and pop_front are methods that keep a mutable reference to list, so you can't do this:

list.push_back(list.pop_front().unwrap());

Doing it the other way that you pointed out works because you drop the mutable reference at the end of the first line:

let str = list.pop_front().unwrap(); // mutable reference dropped here, so it's fine to have another mutable reference afterwards
list.push_back(str);