The `&mut list` been released or not

help
1 
struct Manager<'a> {
    text: &'a str,
}

struct List<'a> {
    manager: Manager<'a>,
}

impl<'a> List<'a> {
    pub fn xxx(self: &'a mut List<'a>) {}

    pub fn yyy<'b>(self: &'b mut List<'a>) {}
}

fn main() {
    let mut list = List {
        manager: Manager { text: "hello" }
    };
    list.xxx();			// Produced one &'a mut list<'a >
    list.text = 20;		// error
}

This code produce a permanent borrow &'a mut List<'a>, and the list can no longer be accessed.
My question is: does this mutable borrow &'a mut List<'a> actually last as long as the lifetime of list, or has this mutable borrow &'a mut List<'a> already been released, but the borrow checker notices that the lifetime annotation of the mutable borrow is the same as 'a, so the borrow checker rejects the code?

2

Yes

The borrow checker does not reason this way. It uses lifetime annotations to create constraints for the actual lifetimes, and then solve them. Types like &'a mut List<'a> generate almost conflicting requirements, with the only solution being borrowing as long as it exists.

4 Likes
3

Thank you once again for your answer

4

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