问题
Considering the following code:
fn foo<'a, T: 'a>(t: T) -> Box<Fn() -> &'a T + 'a> {
Box::new(move || &t)
}
What I expect:
- The type T has lifetime
'a. - The value
tlive as long asT. tmoves to the closure, so the closure live as long ast- The closure returns a reference to
twhich was moved to the closure. So the reference is valid as long as the closure exists. - There is no lifetime problem, the code compiles.
What actually happens:
- The code does not compile:
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
--> src/lib.rs:2:22
|
2 | Box::new(move || &t)
| ^^
|
note: first, the lifetime cannot outlive the lifetime as defined on the body at 2:14...
--> src/lib.rs:2:14
|
2 | Box::new(move || &t)
| ^^^^^^^^^^
note: ...so that closure can access `t`
--> src/lib.rs:2:22
|
2 | Box::new(move || &t)
| ^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the function body at 1:8...
--> src/lib.rs:1:8
|
1 | fn foo<'a, T: 'a>(t: T) -> Box<Fn() -> &'a T + 'a> {
| ^^
= note: ...so that the expression is assignable:
expected std::boxed::Box<(dyn std::ops::Fn() -> &'a T + 'a)>
found std::boxed::Box<dyn std::ops::Fn() -> &T>
I do not understand the conflict. How can I fix it?
回答1:
Very interesting question! I think I understood the problem(s) at play here. Let me try to explain.
tl;dr: move closures cannot return references to their environment as that would be a reference to self. Such a reference cannot be returned because the Fn* traits don't allow us to express that. This is basically the same as the streaming iterator problem and could be fixed via GATs (generic associated types).
Implementing it manually
As you probably know, when you write a closure, the compiler will generate a struct and impl blocks for the appropriate Fn traits, so closures are basically syntax sugar. Let's try to avoid all that sugar and build your type manually.
What you want is a type which owns another type and can return references to that owned type. And you want to have a function which returns a boxed instance of said type.
struct Baz<T>(T);
impl<T> Baz<T> {
fn call(&self) -> &T {
&self.0
}
}
fn make_baz<T>(t: T) -> Box<Baz<T>> {
Box::new(Baz(t))
}
This is pretty equivalent to your boxed closure. Let's try to use it:
let outside = {
let s = "hi".to_string();
let baz = make_baz(s);
println!("{}", baz.call()); // works
baz
};
println!("{}", outside.call()); // works too
This works just fine. The string s is moved into the Baz type and that Baz instance is moved into the Box. s is now owned by baz and then by outside.
It gets more interesting when we add a single character:
let outside = {
let s = "hi".to_string();
let baz = make_baz(&s); // <-- NOW BORROWED!
println!("{}", baz.call()); // works
baz
};
println!("{}", outside.call()); // doesn't work!
Now we cannot make the lifetime of baz bigger than the lifetime of s, since baz contains a reference to s which would be an dangling reference of s would go out of scope earlier than baz.
The point I wanted to make with this snippet: we didn't need to annotate any lifetimes on the type Baz to make this safe; Rust figured it out on its own and enforces that baz lives no longer than s. This will be important below.
Writing a trait for it
So far we only covered the basics. Let's try to write a trait like Fn to get closer to your original problem:
trait MyFn {
type Output;
fn call(&self) -> Self::Output;
}
In our trait, there are no function parameters, but otherwise it's fairly identical to the real Fn trait.
Let's implement it!
impl<T> MyFn for Baz<T> {
type Output = ???;
fn call(&self) -> Self::Output {
&self.0
}
}
Now we have a problem: what do we write instead of ???? Naively one would write &T... but we need a lifetime parameter for that reference. Where do we get one? What lifetime does the return value even have?
Let's check the function we implemented before:
impl<T> Baz<T> {
fn call(&self) -> &T {
&self.0
}
}
So here we use &T without lifetime parameter too. But this only works because of lifetime elision. Basically, the compiler fills in the blanks so that fn call(&self) -> &T is equivalent to:
fn call<'s>(&'s self) -> &'s T
Aha, so the lifetime of the returned reference is bound to the self lifetime! (more experienced Rust users might already have a feeling where this is going...).
(As a side note: why is the returned reference not dependent on the lifetime of T itself? If T references something non-'static then this has to be accounted for, right? Yes, but it is already accounted for! Remember that no instance of Baz<T> can ever live longer than the thing T might reference. So the self lifetime is already shorter than whatever lifetime T might have. Thus we only need to concentrate on the self lifetime)
But how do we express that in the trait impl? Turns out: we can't (yet). This problem is regularly mentioned in the context of streaming iterators -- that is, iterators that return an item with a lifetime bound to the self lifetime. In today's Rust, it is sadly impossible to implement this; the type system is not strong enough.
What about the future?
Luckily, there is an RFC "Generic Associated Types" which was merged some time ago. This RFC extends the Rust type system to allow associated types of traits to be generic (over other types and lifetimes).
Let's see how we can make your example (kinda) work with GATs (according to the RFC; this stuff doesn't work yet ☹). First we have to change the trait definition:
trait MyFn {
type Output<'a>; // <-- we added <'a> to make it generic
fn call(&self) -> Self::Output;
}
The function signature hasn't changed in the code, but notice that lifetime elision kicks in! The above fn call(&self) -> Self::Output is equivalent to:
fn call<'s>(&'s self) -> Self::Output<'s>
So the lifetime of the associated type is bound to the self lifetime. Just as we wanted! The impl looks like this:
impl<T> MyFn for Baz<T> {
type Output<'a> = &'a T;
fn call(&self) -> Self::Output {
&self.0
}
}
To return a boxed MyFn we would need to write this (according to this section of the RFC:
fn make_baz<T>(t: T) -> Box<for<'a> MyFn<Output<'a> = &'a T>> {
Box::new(Baz(t))
}
And what if we want to use the real Fn trait? As far as I understand, we can't, even with GATs. I think it's impossible to change the existing Fn trait to use GATs in a backwards compatible manner. So it's likely that the standard library will keep the less powerful trait as is. (side note: how to evolve the standard library in backwards incompatible ways to use new language features is something I wondered about a few times already; so far I haven't heard of any real plan in this regards; I hope the Rust team comes up with something...)
Summary
What you want is not technically impossible or unsafe (we implemented it as a simple struct and it works). However, unfortunately it is impossible to express what you want in the form of closures/Fn traits in Rust's type system right now. This is the same problem streaming iterators are dealing with.
With the planned GAT feature, it is possible to express all of this in the type system. However, the standard library would need to catch up somehow to make your exact code possible.
回答2:
What I expect:
- The type
Thas lifetime'a.- The value
tlive as long asT.
This makes no sense. A value cannot "live as long" as a type, because a type doesn't live. "T has lifetime 'a" is a very imprecise statement, easy to misunderstand. What T: 'a really means is "instances of T must stay valid at least as long as lifetime 'a. For example, T must not be a reference with a lifetime shorter than 'a, or a struct containing such a reference. Note that this has nothing to do with forming references to T, i.e. &T.
The value t, then, lives as long as its lexical scope (it's a function parameter) says it does, which has nothing to do with 'a at all.
tmoves to the closure, so the closure live as long ast
This is also incorrect. The closure lives as long as the closure does lexically. It is a temporary in the result expression, and therefore lives until the end of the result expression. t's lifetime concerns the closure not at all, since it has its own T variable inside, the capture of t. Since the capture is a copy/move of t, it is not in any way affected by t's lifetime.
The temporary closure is then moved into the box's storage, but that's a new object with its own lifetime. The lifetime of that closure is bound to the lifetime of the box, i.e. it is the return value of the function, and later (if you store the box outside the function) the lifetime of whatever variable you store the box in.
All of that means that a closure that returns a reference to its own capture state must bind the lifetime of that reference to its own reference. Unfortunately, this is not possible.
Here's why:
The Fn trait implies the FnMut trait, which in turn implies the FnOnce trait. That is, every function object in Rust can be called with a by-value self argument. This means that every function object must be still valid being called with a by-value self argument and returning the same thing as always.
In other words, trying to write a closure that returns a reference to its own captures expands to roughly this code:
struct Closure<T> {
captured: T,
}
impl<T> FnOnce<()> for Closure<T> {
type Output = &'??? T; // what do I put as lifetime here?
fn call_once(self, _: ()) -> Self::Output {
&self.captured // returning reference to local variable
// no matter what, the reference would be invalid once we return
}
}
And this is why what you're trying to do is fundamentally impossible. Take a step back, think of what you're actually trying to accomplish with this closure, and find some other way to accomplish it.
回答3:
You expect the type T to have lifetime 'a, but t is not a reference to a value of type T. The function takes ownership of the variable t by argument passing:
// t is moved here, t lifetime is the scope of the function
fn foo<'a, T: 'a>(t: T)
You should do:
fn foo<'a, T: 'a>(t: &'a T) -> Box<Fn() -> &'a T + 'a> {
Box::new(move || t)
}
回答4:
The other answers are top-notch, but I wanted to chime in with another reason your original code couldn't work. A big problem lies in the signature:
fn foo<'a, T: 'a>(t: T) -> Box<Fn() -> &'a T + 'a>
This says that the caller may specify any lifetime when calling foo and the code will be valid and memory-safe. That cannot possibly be true for this code. It wouldn't make sense to call this with 'a set to 'static, but nothing about this signature would prevent that.
来源：https://stackoverflow.com/questions/49806352/cannot-infer-an-appropriate-lifetime-for-a-closure-that-returns-a-reference