Lifetime(a) = lifetime(b) iff (a=b)?

help
5

Hi! Thanks for your demonstration of two situation of same lifetime.

Recently I found a situation in which two references of different variables share same lifetime and that makes me confused.

        let mut x2: u8 = 1;
        {
            let mut x = 42;
            let mut y = &x;
            let mut yt: InvariantLifetime;
            yt = get_lifetime(&mut y);

            let mut y2 = &x2;
            let mut y2t: InvariantLifetime;
            y2t = get_lifetime(&mut y2);
            y2t = yt;
        }

These code could be compiled, which imply that yt and y2t carry same lifetime.
In above code, some type and function is implemented as(ref):

pub struct Invariant<T>(PhantomData<*mut T>);
pub struct InvariantLifetime<'id>(Invariant<&'id ()>);
fn get_lifetime<'a, T>(_: &mut &'a T) -> InvariantLifetime<'a> { InvariantLifetime::new() }

Does these code shows that two references(yt,y2t) have save timelife?