CODEWITHSUNDEEP
rust
Oleg Antonyan
Oleg Antonyan

Reputation: 3113

Wrong number of lifetime parameters when returning Vec of structs with reference

I have a struct with a reference in it:

pub struct ScheduledItem<'a> {
    pub item: &'a item::Item,
    pub timeshift: i32
}

Now I want to write a function which return a Vec of references to this struct:

pub fn items_with_times<'a>(items: &Vec<ScheduledItem>) -> Vec<(u32, &'a ScheduledItem)> {

But what I get is an error:

src/scheduled_item.rs:25:74: 25:87 error: wrong number of lifetime parameters: expected 1, found 0 [E0107]
src/scheduled_item.rs:25 pub fn items_with_times<'a>(items: &Vec<ScheduledItem>) -> Vec<(u32, &'a ScheduledItem)> {
                                                                                                  ^~~~~~~~~~~~~

Isn't &'a enough? What's wrong here?

Upvotes: 1

Views: 259

Answers (1)

oli_obk
oli_obk

Reputation: 31263

Your struct has a generic lifetime parameter. In Rust you need to specify all generic parameters (e.g. you can't return a Vec, just a Vec<T>). So your return type should be Vec<(u32, ScheduledItem<'a>)> and your argument type should be &[ScheduledItem<'a>], since there's no benefit of &Vec<T> over &[T]

Isn't &'a enough?

&'a T specifies that it's a reference with the lifetime 'a to a T, meaning that the object it is pointing to doesn't outlive 'a.

T<'a> on the other hand specifies that your T<'a> type doesn't live longer than 'a. Which in turn means that any object of that type won't outlive 'a and that the object cannot contain references to objects that live shorter than 'a.

Upvotes: 2

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