Chapter 4. Why You Can’t Store a Value and a Reference to that Value in the Same Struct
Let’s say we have a value and we want to store that value and a reference to something inside that value in our own type:
struct Thing {
count: u32,
}
struct Combined<‘a>(Thing, &’a u32);
fn make_combined<‘a>() -> Combined<‘a> {
let thing = Thing { count: 42 };
Combined(thing, &thing.count)
}
Or say we have a value and we want to store that value and a reference to that value in the same structure:
struct Combined<‘a>(Thing, &’a Thing);
fn make_combined<‘a>() -> Combined<‘a> {
let thing = Thing::new();
Combined(thing, &thing)
}
Or a case when we’re not even taking a reference of the value:
struct Combined<‘a>(Parent, Child<‘a>);
fn make_combined<‘a>() -> Combined<‘a> {
let parent = Parent::new();
let child = parent.child();
Combined(parent, child)
}
In each of these cases, we get an error that one of the values “does not live long enough”. What does this error mean?
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 { parent, child }
}
}