Nested async function call causes lifetime issue - rust

Question

I feel like I am totally lost; what does "'1" mean in the error message?
error[E0597]: `o` does not live long enough
--> src/main.rs:32:19
|
31 | async fn foo6(&mut self, mut o: &'a mut Outer) {
| --------- lifetime `'1` appears in the type of `self`
32 | self.foo5(&mut o).await
| ----------^^^^^^-
| | |
| | borrowed value does not live long enough
| argument requires that `o` is borrowed for `'1`
33 | }
| - `o` dropped here while still borrowed
What can I do to make o live long enough? I think I'm using futures 03.
use futures::future::BoxFuture;
use futures::FutureExt;
use futures::stream::FuturesOrdered;
struct Inner {}
impl Inner {
async fn foo3(&mut self) -> Result<u32, ()> {
Ok(8)
}
}
// --- 1
struct Outer {
i: Inner,
}
impl Outer {
fn foo4(&mut self) -> BoxFuture<'_, Result<u32, ()>> {
self.i.foo3().boxed()
}
}
/// --- 2
struct Outer2<'a> {
futures_list: FuturesOrdered<BoxFuture<'a, Result<u32, ()>>>,
}
impl <'a> Outer2<'a> {
async fn foo6(&mut self, mut o: &'a mut Outer) {
self.foo5(&mut o).await
}
async fn foo5(&mut self, o: &'a mut Outer) {
self.futures_list.push(o.foo4());
}
}
#[tokio::main]
async fn main() {
let mut o = Outer { i: Inner {} };
let mut o2 = Outer2 { futures_list: FuturesOrdered::new() };
o2.foo5(&mut o).await;
}
playground

Answer 1

The '1 represents the anonymous lifetime of the self parameter for function foo5.
Note that in foo6, o is already a mutable reference to Outer, so writing &mut o actually gives you a mutable reference to a mutable reference to Outer with one too many indirections. You can fix your code by removing the extra reference:
async fn foo6(&mut self, mut o: &'a mut Outer) {
self.foo5(o).await
}
Playground