There is actually more than one reason why the code above fails. Let's break it down a little and explore a few options on how to fix it.
First let's remove the new
and try building an instance of A
directly in main
, so that you see that the first part of the problem has nothing to do with lifetimes:
struct C;
struct B<'b> {
c: &'b C,
}
struct A<'a> {
b: B<'a>,
c: C,
}
fn main() {
// I copied your new directly here
// and renamed c1 so we know what "c"
// the errors refer to
let c1 = C;
let _ = A {
c: c1,
b: B { c: &c1 },
};
}
this fails with:
error[E0382]: use of moved value: `c1`
--> src/main.rs:20:20
|
19 | c: c1,
| -- value moved here
20 | b: B { c: &c1 },
| ^^ value used here after move
|
= note: move occurs because `c1` has type `C`, which does not implement the `Copy` trait
what it says is that if you assign c1
to c
, you move its ownership to c
(i.e. you can't access it any longer through c1
, only through c
). This means that all the references to c1
would be no longer valid. But you have a &c1
still in scope (in B), so the compiler can't let you compile this code.
The compiler hints at a possible solution in the error message when it says that type C
is non-copyable. If you could make a copy of a C
, your code would then be valid, because assigning c1
to c
would create a new copy of the value instead of moving ownership of the original copy.
We can make C
copyable by changing its definition like this:
#[derive(Copy, Clone)]
struct C;
Now the code above works. Note that what @matthieu-m comments is still true: we can't store both the reference to a value and the value itself in B (we're storing a reference to a value and a COPY of the value here). That's not just for structs, though, it's how ownership works.
Now, if you don't want to (or can't) make C
copyable, you can store references in both A
and B
instead.
struct C;
struct B<'b> {
c: &'b C,
}
struct A<'a> {
b: B<'a>,
c: &'a C, // now this is a reference too
}
fn main() {
let c1 = C;
let _ = A {
c: &c1,
b: B { c: &c1 },
};
}
All good then? Not really... we still want to move the creation of A
back into a new
method. And THAT's where we will run in trouble with lifetimes. Let's move the creation of A
back into a method:
impl<'a> A<'a> {
fn new() -> A<'a> {
let c1 = C;
A {
c: &c1,
b: B { c: &c1 },
}
}
}
as anticipated, here's our lifetime error:
error[E0597]: `c1` does not live long enough
--> src/main.rs:17:17
|
17 | c: &c1,
| ^^ borrowed value does not live long enough
...
20 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the lifetime 'a as defined on the impl at 13:1...
--> src/main.rs:13:1
|
13 | impl<'a> A<'a> {
| ^^^^^^^^^^^^^^
error[E0597]: `c1` does not live long enough
--> src/main.rs:18:24
|
18 | b: B { c: &c1 },
| ^^ borrowed value does not live long enough
19 | }
20 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the lifetime 'a as defined on the impl at 13:1...
--> src/main.rs:13:1
|
13 | impl<'a> A<'a> {
| ^^^^^^^^^^^^^^
this is because c1
is destroyed at the end of the new
method, so we can't return a reference to it.
fn new() -> A<'a> {
let c1 = C; // we create c1 here
A {
c: &c1, // ...take a reference to it
b: B { c: &c1 }, // ...and another
}
} // and destroy c1 here (so we can't return A with a reference to c1)
One possible solution is to create C
outside of new
and pass it in as a parameter:
struct C;
struct B<'b> {
c: &'b C,
}
struct A<'a> {
b: B<'a>,
c: &'a C
}
fn main() {
let c1 = C;
let _ = A::new(&c1);
}
impl<'a> A<'a> {
fn new(c: &'a C) -> A<'a> {
A {c: c, b: B{c: c}}
}
}
playground