What does it mean when we let a trait inherits 'static? - rust

Rust supports trait inheritance, as follows:
pub trait A {}
pub trait B: A {}
B: A means that if some type T implements B, it also needs to implement all the methods in A.
But today I see the following code:
trait Display: 'static {
fn print(&self);
}
What does it mean? It doesn't seem to be trait inheritance.

Rust supports trait inheritance, as follows [...] B: A means that if some type T implements B, it also needs to implement all the methods in A.
Technically, that is not inheritance but requirement. It is a trait bound not entirely dissimilar to one you'd have in a function: it constraints the type on which B is implementable to only types on which A is already implemented.
With that change in wording, the second version is much easier to understand: it's a lifetime bound, meaning it constraints the type on which B is implementable to only types with 'static lifetime, meaning if you're trying to implement B on a type, that must either have no lifetime at all, or have a 'static lifetime (or the implementation must have a lifetime bound aka only work for some uses of the type).
You can see that if you try to implement the trait on a lifetime-generic structure:
struct A<'a>(&'a str);
trait Display: 'static {
fn print(&self);
}
impl <'a>Display for A<'a> {
fn print(&self) { todo!() }
}
will yield
error[E0478]: lifetime bound not satisfied
That is because 'a can be anything, so implementing Display for A<'a> means it is also implemented for non-'static instances, which is not valid.
By adding the relevant lifetime bound on the impl, and thus limiting the implementation to A<'static> instances:
struct A<'a>(&'a str);
trait Display: 'static {
fn print(&self);
}
impl <'a: 'static>Display for A<'a> {
fn print(&self) { todo!() }
}
the requirements of the trait are satisfied, and the impl is valid (nb: the 'a is not necessary here you can just impl ... for A<'static>, I'm showing it for regularity).
And if your struct has no lifetime it works by default, because no lifetime ~ 'static:
struct A(String);
trait Display: 'static {
fn print(&self);
}
impl Display for A {
fn print(&self) { todo!() }
}

Rust doesn't have inheritance.
What it has is a way to define constraints. For example a trait may be constrained to only be implemented by types which implement another trait.
In your case the constraint is a lifetime bound.
To implement your Display trait, an object may contain references but in this case their lifetime must respect this constraint.
Let's suppose you have this type:
struct S<'a> {
s: &'a str,
}
Then you can't implement the trait for any lifetime, but only 'static.
impl Display for S<'static> {
fn print(&self){}
}
fn main() {
let s1 = "test";
let a = S { s: s1 };
a.print(); // compiles
let s2 = "test".to_string();
let a = S { s: &s2 };
a.print(); // doesn't compile because s doesn't live long enough
}

Related

Rust: additional trait bounds in conjunction with higher-ranked lifetime bounds

I want trait implementations in Rust to be able to return arbitrary iterators (of specific item type) that may reference the original object with a lifetime 'a without having to explicitly mention 'a in the trait generics and everywhere where the trait is used or otherwise introducing significant trait bound bloat to user code. The only simple way I've figured to do this is that the trait has to be implemented for &'a MyStruct instead of MyStruct (this approach is used in some places in the standard library), but the significant drawback is that in generic code wrappers cannot “own” implementations of the trait (MyStruct) without exposing the lifetime in trait bounds all over the code. So nothing gained when ownership is needed.
Another way I figured out that should work (just done the simple test below so far) is to use higher-ranked lifetime bounds on a “base trait” that would implement the iterator-generation functions. In the code below Foo is the main trait, FooInterfaceGen is the iterator-generator trait that has its lifetime “hidden” through for <'a> when assumed as a super-trait of Foo. The FooInterface generated by FooInterfaceGen would be the trait for an appropriate type of iterator when modified to that application of the idea. However, it seems impossible to make additional trait bounds on the specific implementation FooInterfaceGen::Interface. The code below works, but when you uncomment the Asdf trait bound in the function footest, the compiler complains that
the trait `for<'a> Asdf` is not implemented for `<_ as FooInterfaceGen<'a>>::Interface
But I have implemented Asdf! It's as if the compiler is ignoring the 'a in the expression <T as FooInterfaceGen<'a>> and just applying for<'a> to the right-hand-side. Any ideas if this is a compiler bug, a known restriction, or of any ways around it?
trait FooInterface<'a> {
fn foo(&self) -> u32;
}
trait FooInterfaceGen<'a> {
type Interface : FooInterface<'a>;
fn gen(&'a self) -> Self::Interface;
}
trait Foo : for<'a> FooInterfaceGen<'a> { }
struct S2;
struct S1(S2);
impl<'a> FooInterfaceGen<'a> for S1 {
type Interface = &'a S2;
fn gen(&'a self) -> Self::Interface { &self.0 }
}
impl Foo for S1 { }
impl<'a> FooInterface<'a> for &'a S2 {
fn foo(&self) -> u32 { 42 }
}
trait Asdf {}
impl<'a> Asdf for &'a S2 {}
fn footest<T : Foo>(a : &T) -> u32
/* where for<'a> <T as FooInterfaceGen<'a>>::Interface : Asdf */ {
a.gen().foo()
}
fn main() {
let q = S1(S2);
println!("{}", footest(&q));
}
(Regarding some alternative implementations, maybe there's a technical reason for it, but otherwise I really don't understand the reason behind the significant trait bound bloat that Rust code easily introduces. Assuming a trait should in any reasonable situation automatically assume all the trait bound as well, also in generic code, not just specific code, without having to copy-paste an increasing number of where-clauses all over the code.)
The error seems to be a known compiler bug: https://github.com/rust-lang/rust/issues/89196

Understanding "the trait X cannot be made into an object" for `&mut Box<Self>` parameter

I've got this code snippet (playground):
struct TeddyBear {
fluffiness: u8,
}
trait Scruffy {
fn scruff_up(self: &mut Box<Self>) -> Box<dyn Scruffy>;
}
impl Scruffy for TeddyBear {
fn scruff_up(self: &mut Box<Self>) -> Box<dyn Scruffy> {
// do something about the TeddyBear's fluffiness
}
}
It doesn't compile. The error is:
the trait Scruffy cannot be made into an object
, along with the hint:
because method scruff_up's self parameter cannot be dispatched on.
I checked the "E0038" error description, but I haven't been able to figure out which category my error falls into.
I also read the "object-safety" entry in "The Rust Reference", and I believe this matches the "All associated functions must either be dispatchable from a trait object", but I'm not sure, partly because I'm not sure what "receiver" means in that context.
Can you please clarify for me what's the problem with this code and why it doesn't work?
The problem is when you pass it in as a reference, because the inner type may not be well-sized (e.g. a trait object, like if you passed in a Box<Fluffy>) the compiler doesn't have enough information to figure out how to call methods on it. If you restrict it to sized objects (like your TeddyBear) it should compile
trait Scruffy {
fn scruff_up(self: &mut Box<Self>) -> Box<dyn Scruffy> where Self: Sized;
}
A receiver is the self (&self, &mut self, self: &mut Box<Self> and so on).
Note that the list you cited from the reference lists both Box<Self> and &mut Self, but does not list &mut Box<Self> nor it says that combinations of these types are allowed.
This is, indeed, forbidden. As for the why, it is a little more complex.
In order for a type to be a valid receiver, it needs to hold the following condition:
Given any type Self that implements Trait and the receiver type Receiver, the receiver type should implement DispatchFromDyn<dyn Trait> for itself with all Self occurrences of Self replaced with dyn Trait.
For instance:
&self (has the type &Self) has to implement DispatchFromDyn<&dyn Trait>, which it does.
Box<Self> has to implement DispatchFromDyn<Box<dyn Trait>>, which it does.
But in order for &mut Box<Self> to be an object-safe receiver, it would need to impl DispatchFromDyn<&mut Box<dyn Trait>>. What you want is kind of blanket implementation DispatchFromDyn<&mut T> for &mut U where U: DispatchFromDyn<T>.
This impl will never exist. Because it is unsound (even ignoring coherence problems).
As explained in the code in rustc that calculates this:
The only case where the receiver is not dispatchable, but is still a valid receiver type (just not object-safe), is when there is more than one level of pointer indirection. E.g., self: &&Self, self: &Rc<Self>, self: Box<Box<Self>>. In these cases, there is no way, or at least no inexpensive way, to coerce the receiver from the version where Self = dyn Trait to the version where Self = T, where T is the unknown erased type contained by the trait object, because the object that needs to be coerced is behind a pointer.
The problem is inherent to how Rust handles dyn Trait.
dyn Trait is a fat pointer: it is actually two words sized. One is a pointer to the data, and the other is a pointer to the vtable.
When you call a method on dyn Trait, the compiler looks up in the vtable, find the method for the concrete type (which is unknown at compilation time, but known at runtime), and calls it.
This all may be very abstract without an example:
trait Trait {
fn foo(&self);
}
impl Trait for () {
fn foo(&self) {}
}
fn call_foo(v: &dyn Trait) {
v.foo();
}
fn create_dyn_trait(v: &impl Trait) {
let v: &dyn Trait = v;
call_foo(v);
}
The compiler generates code like:
trait Trait {
fn foo(&self);
}
impl Trait for () {
fn foo(&self) {}
}
struct TraitVTable {
foo: fn(*const ()),
}
static TRAIT_FOR_UNIT_VTABLE: TraitVTable = TraitVTable {
foo: unsafe { std::mem::transmute(<() as Trait>::foo) },
};
type DynTraitRef = (*const (), &'static TraitVTable);
impl Trait for dyn Trait {
fn foo(self: DynTraitRef) {
(self.1.foo)(self.0)
}
}
fn call_foo(v: DynTraitRef) {
v.foo();
}
fn create_dyn_trait(v: &impl Trait) {
let v: DynTraitRef = (v as *const (), &TRAIT_FOR_UNIT_VTABLE);
call_foo(v);
}
Now suppose that the pointer to the value is behind an indirection. I'll use Box<&self> because it's simple but demonstrates the concept best, but the concept applies to &mut Box<Self> too: they have the same layout. How will we write foo() for impl Trait for dyn Trait?
trait Trait {
fn foo(self: Box<&Self>);
}
impl Trait for () {
fn foo(self: Box<&Self>) {}
}
struct TraitVTable {
foo: fn(Box<*const ()>),
}
static TRAIT_FOR_UNIT_VTABLE: TraitVTable = TraitVTable {
foo: unsafe { std::mem::transmute(<() as Trait>::foo) },
};
type DynTraitRef = (*const (), &'static TraitVTable);
impl Trait for dyn Trait {
fn foo(self: Box<DynTraitRef>) {
let concrete_foo: fn(Box<*const ()>) = self.1.foo;
let data: *const () = self.0;
concrete_foo(data) // We need to wrap `data` in `Box`! Error.
}
}
You may think "then the compiler should just insert a call to Box::new()!" But besides Box not being the only one here (what with Rc, for example?) and we will need some trait to abstract over this behavior, Rust never performs any hard work implicitly. This is a design choice, and an important one (as opposed to e.g. C++, where an innocent-looking statement like auto v1 = v; can allocate and copy 10GB by a copy constructor). Converting a type to dyn Trait and back is done implicitly: the first one by a coercion, the second one when you call a method of the trait. Thus, the only thing that Rust does for that is attaching a VTable pointer in the first case, or discarding it in the second case. Even allowing only references (&&&Self, no need to call a method, just take the address of a temporary) exceeds that. And it can have severe implications in unexpected places, e.g. register allocation.
So, what to do? You can take &mut self or self: Box<Self>. Which one to choose depends on whether you need ownership (use Box) or not (use a reference). And anyway, &mut Box<Self> is not so useful (its only advantage over &mut T is that you can replace the box and not just its contents, but when you do that that's usually a mistake).

How do I specify lifetime parameters in an associated type?

I have this trait and simple structure:
use std::path::{Path, PathBuf};
trait Foo {
type Item: AsRef<Path>;
type Iter: Iterator<Item = Self::Item>;
fn get(&self) -> Self::Iter;
}
struct Bar {
v: Vec<PathBuf>,
}
I would like to implement the Foo trait for Bar:
impl Foo for Bar {
type Item = PathBuf;
type Iter = std::slice::Iter<PathBuf>;
fn get(&self) -> Self::Iter {
self.v.iter()
}
}
However I'm getting this error:
error[E0106]: missing lifetime specifier
--> src/main.rs:16:17
|
16 | type Iter = std::slice::Iter<PathBuf>;
| ^^^^^^^^^^^^^^^^^^^^^^^^^ expected lifetime parameter
I found no way to specify lifetimes inside that associated type. In particular I want to express that the iterator cannot outlive the self lifetime.
How do I have to modify the Foo trait, or the Bar trait implementation, to make this work?
Rust playground
There are a two solutions to your problem. Let's start with the simplest one:
Add a lifetime to your trait
trait Foo<'a> {
type Item: AsRef<Path>;
type Iter: Iterator<Item = Self::Item>;
fn get(&'a self) -> Self::Iter;
}
This requires you to annotate the lifetime everywhere you use the trait. When you implement the trait, you need to do a generic implementation:
impl<'a> Foo<'a> for Bar {
type Item = &'a PathBuf;
type Iter = std::slice::Iter<'a, PathBuf>;
fn get(&'a self) -> Self::Iter {
self.v.iter()
}
}
When you require the trait for a generic argument, you also need to make sure that any references to your trait object have the same lifetime:
fn fooget<'a, T: Foo<'a>>(foo: &'a T) {}
Implement the trait for a reference to your type
Instead of implementing the trait for your type, implement it for a reference to your type. The trait never needs to know anything about lifetimes this way.
The trait function then must take its argument by value. In your case you will implement the trait for a reference:
trait Foo {
type Item: AsRef<Path>;
type Iter: Iterator<Item = Self::Item>;
fn get(self) -> Self::Iter;
}
impl<'a> Foo for &'a Bar {
type Item = &'a PathBuf;
type Iter = std::slice::Iter<'a, PathBuf>;
fn get(self) -> Self::Iter {
self.v.iter()
}
}
Your fooget function now simply becomes
fn fooget<T: Foo>(foo: T) {}
The problem with this is that the fooget function doesn't know T is in reality a &Bar. When you call the get function, you are actually moving out of the foo variable. You don't move out of the object, you just move the reference. If your fooget function tries to call get twice, the function won't compile.
If you want your fooget function to only accept arguments where the Foo trait is implemented for references, you need to explicitly state this bound:
fn fooget_twice<'a, T>(foo: &'a T)
where
&'a T: Foo,
{}
The where clause makes sure that you only call this function for references where Foo was implemented for the reference instead of the type. It may also be implemented for both.
Technically, the compiler could automatically infer the lifetime in fooget_twice so you could write it as
fn fooget_twice<T>(foo: &T)
where
&T: Foo,
{}
but it's not smart enough yet.
For more complicated cases, you can use a Rust feature which is not yet implemented: Generic Associated Types (GATs). Work for that is being tracked in issue 44265.
Use a wrapper type
If the trait and all its implementations are defined in one crate, a helper type can be useful:
trait Foo {
fn get<'a>(&'a self) -> IterableFoo<'a, Self> {
IterableFoo(self)
}
}
struct IterableFoo<'a, T: ?Sized + Foo>(pub &'a T);
For a concrete type that implements Foo, implement the iterator conversion on the IterableFoo wrapping it:
impl Foo for Bar {}
impl<'a> IntoIterator for IterableFoo<'a, Bar> {
type Item = &'a PathBuf;
type IntoIter = std::slice::Iter<'a, PathBuf>;
fn into_iter(self) -> Self::IntoIter {
self.0.v.iter()
}
}
This solution does not allow implementations in a different crate. Another disadvantage is that an IntoIterator bound cannot be encoded into the definition of the trait, so it will need to be specified as an additional (and higher-rank) bound for generic code that wants to iterate over the result of Foo::get:
fn use_foo_get<T>(foo: &T)
where
T: Foo,
for<'a> IterableFoo<'a, T>: IntoIterator,
for<'a> <IterableFoo<'a, T> as IntoIterator>::Item: AsRef<Path>
{
for p in foo.get() {
println!("{}", p.as_ref().to_string_lossy());
}
}
Associated type for an internal object providing desired functionality
The trait can define an associated type that gives access to a part of the object that, bound in a reference, provides the necessary access traits.
trait Foo {
type Iterable: ?Sized;
fn get(&self) -> &Self::Iterable;
}
This requires that any implementation type contains a part that can be so exposed:
impl Foo for Bar {
type Iterable = [PathBuf];
fn get(&self) -> &Self::Iterable {
&self.v
}
}
Put bounds on the reference to the associated type in generic code that uses the the result of get:
fn use_foo_get<'a, T>(foo: &'a T)
where
T: Foo,
&'a T::Iterable: IntoIterator,
<&'a T::Iterable as IntoIterator>::Item: AsRef<Path>
{
for p in foo.get() {
println!("{}", p.as_ref().to_string_lossy());
}
}
This solution permits implementations outside of the trait definition crate.
The bound work at generic use sites is as annoying as with the previous solution.
An implementing type may need an internal shell struct with the only purpose of providing the associated type, in case when the use-site bounds are not as readily satisfied as with Vec and IntoIterator in the example discussed.
In future, you'll want an associated type constructor for your lifetime 'a but Rust does not support that yet. See RFC 1598

What is 'core::kinds::Sized` is not implemented for the type `Self' in rust?

This used to work:
struct Foo<'a, T> {
parent:&'a (Array<T> + 'a)
}
impl<'a, T> Foo<'a, T> { //'
pub fn new<T>(parent:&Array<T>) -> Foo<T> {
return Foo {
parent: parent
};
}
}
trait Array<T> {
fn as_foo(&self) -> Foo<T> {
return Foo::new(self);
}
}
fn main() {
}
Now it errors:
:15:21: 15:25 error: the trait core::kinds::Sized is not implemented for the type Self
:15 return Foo::new(self);
I can kind of guess what's wrong; it's saying that my impl of Foo<'a, T> is for T, not Sized? T, but I'm not trying to store a Sized? element in it; I'm storing a reference to a Sized element in it. That should be a pointer, fixed size.
I don't see what's wrong with what I'm doing, or why it's wrong?
For example, I should (I think...) be able to store a &Array in my Foo, no problem. I can't see any reason this would force my Foo instance to be unsized.
playpen link: http://is.gd/eZSZYv
There's two things going on here: trait objects coercions (the error), and object safety (fixing it).
The error
As suggested by the error message, the difficult part of the code is the Foo::new(self), and this is because pub fn new<T>(parent: &Array<T>) -> ..., that is, self is being coerced to an &Array<T> trait object. I'll simplify the code to:
trait Array {
fn as_foo(&self) {
let _ = self as &Array; // coerce to a trait object
}
}
fn main() {}
which gives the same thing:
<anon>:3:13: 3:27 error: the trait `core::kinds::Sized` is not implemented for the type `Self`
<anon>:3 let _ = self as &Array; // coerce to a trait object
^~~~~~~~~~~~~~
Self is the stand-in name for the type that implements the trait. Unlike most generic parameters, Self is possibly-unsized (?Sized) by default, since RFC 546 and #20341 for the purposes of allowing e.g. impl Array<T> for Array<T> to work by default more often (we'll come to this later).
The variable self has type &Self. If Self is a sized type, then this is a normal reference: a single pointer. If Self is an unsized type (like [T] or a trait), then &Self (&[T] or &Trait) is a slice/trait object: a fat pointer.
The error appears because the only references &T that can be cast to a trait object are when T is sized: Rust doesn't support making fat pointers fatter, only thin pointer → fat pointer is valid. Hence, since the compiler doesn't know that Self will always be Sized (remember, it's special and ?Sized by default) it has to assume the worst: that the coercion is not legal, and so it's disallowed.
Fixing it
It seems logical that the fix we're looking for is to ensure that Self: Sized when we want to do a coercion. The obvious way to do this would be to make Self always Sized, that is, override the default ?Sized bound as follows:
trait Array: Sized {
fn as_foo(&self) {
let _ = self as &Array; // coerce to a trait object
}
}
fn main() {}
Looks good!
Except there's the small point that it doesn't work; but at least it's for a difference reason, we're making progress! Trait objects can only be made out of traits that are "object safe" (i.e. safe to be made into a trait object), and having Sized Self is one of the things that breaks object safety:
<anon>:3:13: 3:17 error: cannot convert to a trait object because trait `Array` is not object-safe [E0038]
<anon>:3 let _ = self as &Array; // coerce to a trait object
^~~~
<anon>:3:13: 3:17 note: the trait cannot require that `Self : Sized`
<anon>:3 let _ = self as &Array; // coerce to a trait object
^~~~
<anon>:3:13: 3:17 note: the trait cannot require that `Self : Sized`
<anon>:3 let _ = self as &Array; // coerce to a trait object
^~~~
(I filed the double printing of the note as #20692.)
Back to the drawing board. There's a few other "easy" possibilities for a solution:
define an extension trait trait ArrayExt: Sized + Array { fn as_foo(&self) { ... } } and implement it for all Sized + Array types
just use a free function fn array_as_foo<A: Array>(x: &A) { ... }
However, these don't necessarily work for every use case, e.g. specific types can't customise the behaviour by overloading the default method. However, fortunately there is a fix!
The Turon Trick
(Named for Aaron Turon, who discovered it.)
Using generalised where clauses we can be highly specific about when Self should implement Sized, restricting it to just the method(s) where it is required, without infecting the rest of the trait:
trait Array {
fn as_foo(&self) where Self: Sized {
let _ = self as &Array; // coerce to a trait object
}
}
fn main() {}
This compiles just fine! By using the where clause like this, the compiler understands that (a) the coercion is legal because Self is Sized so self is a thin pointer, and (b) that the method is illegal to call on a trait object anyway, and so doesn't break object safety. To see it being disallowed, changing the body of as_foo to
let x = self as &Array; // coerce to a trait object
x.as_foo();
gives
<anon>:4:7: 4:15 error: the trait `core::kinds::Sized` is not implemented for the type `Array`
<anon>:4 x.as_foo();
^~~~~~~~
as expected.
Wrapping it all up
Making this change to the original unsimplified code is as simple adding that where clause to the as_foo method:
struct Foo<'a, T> { //'
parent:&'a (Array<T> + 'a)
}
impl<'a, T> Foo<'a, T> {
pub fn new(parent:&Array<T>) -> Foo<T> {
return Foo {
parent: parent
};
}
}
trait Array<T> {
fn as_foo(&self) -> Foo<T> where Self: Sized {
return Foo::new(self);
}
}
fn main() {
}
which compiles without error. (NB. I had to remove the unnecessary <T> in pub fn new<T> because that was causing inference failures.)
(I have some in-progress blog posts that go into trait objects, object safety and the Turon trick, they will appear on /r/rust in the near future: first one.)

Why "explicit lifetime bound required" for Box<T> in struct?

Editor's note: This code no longer produces the same error after RFC 599 was implemented, but the concepts discussed in the answers are still valid.
I'm trying to compile this code:
trait A {
fn f(&self);
}
struct S {
a: Box<A>,
}
and I'm getting this error:
a.rs:6:13: 6:14 error: explicit lifetime bound required
a.rs:6 a: Box<A>,
I want S.a to own an instance of A, and don't see how that lifetime is appropriate here. What do I need to do to make the compiler happy?
My Rust version:
rustc --version
rustc 0.12.0-pre-nightly (79a5448f4 2014-09-13 20:36:02 +0000)
(Slightly pedantic point: that A is a trait, so S is not owning an instance of A, it is owning an boxed instance of some type that implements A.)
A trait object represents data with some unknown type, that is, the only thing known about the data is that it implements the trait A. Because the type is not known, the compiler cannot directly reason about the lifetime of the contained data, and so requires that this information is explicitly stated in the trait object type.
This is done via Trait+'lifetime. The easiest route is to just use 'static, that is, completely disallow storing data that can become invalid due to scopes:
a: Box<A + 'static>
Previously, (before the possibility of lifetime-bounded trait objects and this explicit lifetime bound required error message was introduced) all boxed trait objects were implicitly 'static, that is, this restricted form was the only choice.
The most flexible form is exposing the lifetime externally:
struct S<'x> {
a: Box<A + 'x>
}
This allows S to store a trait object of any type that implements A, possibly with some restrictions on the scopes in which the S is valid (i.e. for types for which 'x is less than 'static the S object will be trapped within some stack frame).
The problem here is that a trait can be implemented for references too, so if you don't specify the required lifetime for Box anything could be stored in there.
You can see about lifetime requirements in this rfc.
So one possible solution is to bind the lifetime so Send (we put I in S):
trait A {
fn f(&self);
}
struct I;
impl A for I {
fn f(&self) {
println!("A for I")
}
}
struct S {
a: Box<A + Send>
}
fn main() {
let s = S {
a: box I
};
s.a.f();
}
The other is setting the lifetime to 'a (we can put a reference &I or I to S):
trait A {
fn f(&self);
}
struct I;
impl A for I {
fn f(&self) {
println!("A for I")
}
}
impl <'a> A for &'a I {
fn f(&self) {
println!("A for &I")
}
}
struct S<'a> {
a: Box<A + 'a>
}
fn main() {
let s = S {
a: box &I
};
s.a.f();
}
Note that this is more general and we can store both references and owned data (Send kind that has a lifetime of 'static) but you require a lifetime parameter everywhere the type is used.

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