'static lifetime bounds are kind of confusing in Rust, but here goes: A 'static lifetime constraint on T says: types that satisfy T must not contain any borrowed value/content whose lifetime will be shorter than the life of the program. So suppose you have a string s, and Option<&String> q. If q has the constraint 'static, and you do q = Some(&s) this should be rejected by the compiler since a reference to s gets a shorter lifetime than 'static by default (you could do q = Some(&'static s) if s is say a string literal or a leaked box but not for a normal string). In the examples it looks like you have to use a tuple to trick the borrow checker but the concept is the same: Rust allowed violation of the 'static lifetime bounds. The drop implemenation demonstrates it: drop happens in reverse declaration order, so the string s gets dropped first. Now q has a dangling reference, a pointer to dropped memory essentially, and we get a pretty classic use after free bug.

Rust is supposed to guarantee statically that a reference cannot outlive its referent. The `T: 'static` bound means that the value that Q holds must have a 'static lifetime (e.g. either be an owned value or a reference to a static value). So this program should not compile since a reference to l is not 'static. This code is unsound because rust will drop the string l before q so when the Q destructor runs the underlying string has already been freed yet we have a valid reference to it.