Yeah, the lifetimes on a struct are the lifetimes of things that the struct points to, not an instance of the struct itself. For an instance to be valid, everything it points to must be valid, so if `foo: Foo<'c>`, `foo` cannot be stored for longer than `'c`. E.g. this code is invalid, because the `Foo` instance would live longer than the thing it points to:
let foo = {
let age = 0;
Foo { age: &age }
};
The `age` variable goes out of scope at the } (i.e. the maximum possible lifetime for &age ends there), and hence letting the `Foo` escape from that scope would be a bad idea.
(NB. this applies to the "core" lifetime'd types like `&'c T` too: that is really just a nicer way to write a parameterised type like `Ref<'c, T>`. So the rule "cannot return a reference to a local variable" is the same rule that stops a `Foo<'c>` living too long.)
(NB. this applies to the "core" lifetime'd types like `&'c T` too: that is really just a nicer way to write a parameterised type like `Ref<'c, T>`. So the rule "cannot return a reference to a local variable" is the same rule that stops a `Foo<'c>` living too long.)