Mutable references in field

[tprodanov]

Sorry, maybe it is hundred times solved problem, but I cannot use structure filed, which is mutable reference with longer lifetime, than the structure itself. In this code

struct A<'r> {
    v: &'r mut Vec<int>
}

impl<'r> A<'r> {
    fn create(v: &'r mut Vec<int>) -> A<'r> {
        A::<'r> {
            v: v
        }
    }

    fn get_mut(&mut self, i: uint) -> &'r mut int {
        self.v.get_mut(i)
    }
}

in function get_mut I got an error, lifetime of 'self' is too short. But without all mut keywords code runs without errors.

[alexcrichton]

cc @nikomatsakis

[prajwalkman]

Not sure if this solves it, but

    fn get_mut(&'r mut self, i: uint) -> &'r mut int {
        self.v.get_mut(i)
    }

compiles fine. Is the explicit lifetime on self supposed to be elided?

[tprodanov]

@prajwalkman, Yes, but for example if I want to implement mutable iterator Iterator<&'r mut T> I have to write function fn next(&mut self) -> Option<&'a mut T>, so I have &mut self in argument. Maybe there is another way to do it but it is confusing that for immutable iterator Iterator<&'r T> everything works

[alexcrichton]

Hm upon re-reflection I believe this is actually intended behavior. You cannot take out a mutable borrow longer than the lifetime of the mutable borrow which granted mutable access. In this case the function is trying to borrow &'a mut &'r mut int to &'r mut int (essentially), but this is can lead to safety violations later on down the road, so closing this an expected error.