Return reference with lifetime of self

问题

I'd like to write some code like the following:

struct Foo {
    foo: usize
}

impl Foo {
    pub fn get_foo<'a>(&'a self) -> &'self usize {
        &self.foo
    }
}

But this doesn't work, failing with invalid lifetime name: 'self is no longer a special lifetime.

How can I return a reference that lives as long as the object itself?

回答1:

You don't want the reference to live exactly as long as the object. You just want a borrow on the object (quite possibly shorter than the entire lifetime of the object), and you want the resulting reference to have the lifetime of that borrow. That's written like this:

pub fn get_foo<'a>(&'a self) -> &'a usize {
    &self.foo
}

Additionally, lifetime elision makes the signature prettier:

pub fn get_foo(&self) -> &usize {
    &self.foo
}

回答2:

In your example the lifetime of self is 'a so the lifetime of the returned reference should be 'a:

pub fn get_foo<'a>(&'a self) -> &'a usize {
    &self.foo
}

However the compiler is able to deduce (lifetime elision) the correct lifetime in simple cases like that, so you can avoid to specify lifetime at all, this way:

pub fn get_foo(&self) -> &usize {
    &self.foo
}

Look here for lifetime elision rules

来源：https://stackoverflow.com/questions/31641610/return-reference-with-lifetime-of-self