Comment by thinkharderdev

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.