Check out the entire error: error[E0310]: the parameter type `U` may not live long enough –> src/main.rs:9:24 | 8 | fn add<U: Bar<T>>(&mut self, x: U) { | — help: consider adding an explicit lifetime bound `U: ‘static`… 9 | self.data.push(Box::new(x)); | ^^^^^^^^^^^ | note: …so that the type `U` will meet its required lifetime … Read more
The lifetime of function static variables begins the first time[0] the program flow encounters the declaration and it ends at program termination. This means that the run-time must perform some book keeping in order to destruct it only if it was actually constructed. Additionally, since the standard says that the destructors of static objects must … Read more
The thread support in the standard library allows the created threads to outlive the thread that created them; that’s a good thing! However, if you were to pass a reference to a stack-allocated variable to one of these threads, there’s no guarantee that the variable will still be valid by the time the thread executes. … Read more
It’s easiest to understand what non-lexical lifetimes are by understanding what lexical lifetimes are. In versions of Rust before non-lexical lifetimes are present, this code will fail: fn main() { let mut scores = vec![1, 2, 3]; let score = &scores[0]; scores.push(4); } The Rust compiler sees that scores is borrowed by the score variable, … Read more
The question you asked TL;DR: No, you cannot return a reference to a variable that is owned by a function. This applies if you created the variable or if you took ownership of the variable as a function argument. Solutions Instead of trying to return a reference, return an owned object. String instead of &str, … Read more
Let’s look at a simple implementation of this: struct Parent { count: u32, } struct Child<‘a> { parent: &’a Parent, } struct Combined<‘a> { parent: Parent, child: Child<‘a>, } impl<‘a> Combined<‘a> { fn new() -> Self { let parent = Parent { count: 42 }; let child = Child { parent: &parent }; Combined { … Read more