I'd like to make the following code compile:
struct Provider {}
impl Provider {
fn get_string<'a>(&'a self) -> &'a str { "this is a string" }
}
fn main() {
let provider = Provider{};
let mut vec: Vec<&str> = Vec::new();
// PROBLEM: how do I say that this reference s here
// needs to live as long as vec?
let fun = |s: &str| {
vec.push(s);
};
fun(provider.get_string());
}
Playground link
This is the compile error that I get:
error[E0495]: cannot infer an appropriate lifetime due to conflicting requirements
--> src/main.rs:9:22
|
9 | let mut vec: Vec<&str> = Vec::new();
| ^^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the block at 11:24...
--> src/main.rs:11:25
|
11| let fun = |s: &str| {
| ^
note: ...so that reference does not outlive borrowed content
--> src/main.rs:12:18
|
12| vec.push(s);
| ^
note: but, the lifetime must be valid for the block suffix following statement 2 at 13:6...
--> src/main.rs:13:7
|
13| };
| ^
note: ...so that variable is valid at time of its declaration
--> src/main.rs:11:9
|
11| let fun = |s: &str| {
| ^^^
Your code works just fine if remove all the lifetime annotations and let the compiler inference do its job:
struct Provider;
impl Provider {
fn get_string(&self) -> &str { "this is a string" }
}
fn main() {
let provider = Provider;
let mut vec = Vec::new();
let mut fun = |s| {
vec.push(s);
};
fun(provider.get_string());
}
In short, there's no way to explicitly refer to the lifetime of a local variable, only function arguments. The compiler knows how to do it, though.
If you truly needed it, you could create a function to allow annotating the lifetimes:
fn thing<'a>(provider: &'a Provider) -> Vec<&'a str> {
let mut vec: Vec<&'a str> = Vec::new();
{
let mut fun = |s: &'a str| vec.push(s);
fun(provider.get_string());
} // End mutable borrow of `vec`
vec
}
fn main() {
let provider = Provider;
thing(&provider);
}
why did the original annotations stop things from working?
Specifically, it's this bit:
let fun = |s: &str| {
vec.push(s);
};
This declares a new lifetime on the closure. Using a made-up syntax (you can't declare lifetimes on closure arguments), it would be equivalent to:
let fun = <'a> |s: &'a str| {
vec.push(s);
};
Which is why the compiler has the error:
the lifetime cannot outlive the anonymous lifetime #1 defined on [the closure's block]
There's no connection between that generated lifetime and that of the Provider. Leaving it out allows the compiler to insert the desired but unnamable lifetime.
Here's a version which compiles:
use std::marker::PhantomData;
struct Provider<'a> {
_dummy: PhantomData<&'a ()>,
}
impl<'a> Provider<'a> {
fn get_string(&self) -> &'a str {
"this is a string"
}
}
fn f<'b>() {
let provider = Provider { _dummy: PhantomData };
let mut vec: Vec<&str> = Vec::new();
// PROBLEM: how do I say that this reference s here
// needs to live as long as vec?
let mut fun = |s: &'b str| { vec.push(s); };
fun(provider.get_string());
}
fn main() {
f()
}
Playground link
I made the following changes:
Add a lifetime to Provider (I added a PhantomData, but I guess your provider already owns some data it'll provide).
Update the get_string method to show that it returns something with the provider's lifetime, not the input lifetime (ie based on the Provider's lifetime parameter).
Add a new lifetime parameter 'b to the function (which I renamed to f(), since main() can't have one), which I use to name the lifetime of the closure parameter.
The last one is slightly confusing, as apparently merely adding a name to a lifetime (without apparently adding any constraints) has made it work.
I think (but I'd love some documentation for this) that this is because of lifetime elision. A closure is really a hidden struct with a fn call(&self, s: &str) (in this case) method. According to the lifetime elision rules, the s parameter gains the same lifetime as &self, which is the closure itself. In this case, the closure is declared after vec, so the lifetime is too short. The explicit lifetime means that it is decoupled from the closure's own lifetime.
Related
I ran into a lifetime problem with a little game. The below code represents a very boiled down version of the update loop.
I need the container mutable reference to get references to other game objects or to create new ones or trigger a functionality.
For that reason, I need the Any trait to be able to cast the trait to a struct, so in my GameObj trait I added an as_any method, but this resulted in a lifetime issue.
use std::any::Any;
trait GameObj<'a> {
fn as_any<'b>(&'b self) -> &'b (dyn Any + 'a);
fn update(&mut self, cont: &mut container);
}
struct object<'a> {
content: &'a String,
}
impl<'a> GameObj<'a> for object<'a> {
fn as_any<'b>(&'b self) -> &'b (dyn Any + 'a) {
return self;
}
fn update(&mut self, cont: &mut container) {
let val = cont.get_obj().unwrap();
let any = val.as_any();
}
}
struct container<'a> {
data: Vec<Box<dyn GameObj<'a> + 'a>>,
}
impl<'a> container<'a> {
fn get_obj<'b>(&'b self) -> Option<&'b Box<dyn GameObj<'a> + 'a>> {
return Some(&self.data[0]);
}
}
pub fn main() {
let a = String::from("hallo");
let b = String::from("asdf");
{
let abc = object { content: &a };
let def = object { content: &b };
let mut cont = container { data: Vec::new() };
cont.data.push(Box::new(abc));
cont.data.push(Box::new(def));
loop {
for i in 0..cont.data.len() {
let mut obj = cont.data.remove(0);
obj.update(&mut cont);
cont.data.insert(i, obj);
}
}
}
}
playground
When I try to build the code, it results in the following error message.
If I comment out/delete let any = val.as_any(); in the update function it compiles fine.
error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'a` due to conflicting requirements
--> src/main.rs:18:24
|
18 | let val = cont.get_obj().unwrap();
| ^^^^^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #3 defined on the method body at 17:5...
--> src/main.rs:17:5
|
17 | / fn update(&mut self, cont: &mut container) {
18 | | let val = cont.get_obj().unwrap();
19 | | let any = val.as_any();
20 | | }
| |_____^
= note: ...so that the types are compatible:
expected &container<'_>
found &container<'_>
= note: but, the lifetime must be valid for the static lifetime...
note: ...so that the declared lifetime parameter bounds are satisfied
--> src/main.rs:19:23
|
19 | let any = val.as_any();
| ^^^^^^
How I can make this work without using 'static, or why is this impossible?
Any is declared trait Any: 'static and can only store 'static types. So in order to make dyn Any + 'a a well-formed type, your as_any method was given an implicit 'a: 'static bound, leading to the lifetime error you showed.
If not for this restriction, you would be able to break safety by putting in an 'a type into an Any and getting out a 'static type, because TypeId can’t tell the difference—lifetimes are erased during compilation. See the discussion on RFC 1849 for more information.
You should think more carefully about why you want to use Any. It’s almost never what you actually want. Perhaps something as simple as an enum type of all the different object types you might want to store would satisfy your use case better?
If you really want to use Any, then you’ll need to find a way to make your types 'static. Rc (or Arc, if threads are involved) is often helpful for this purpose; for example, you could have your object store Rc<String> (or better, Rc<str>) instead of &'a String.
I want to write a function A which takes as parameter a function B which takes as parameter a type which is parameterized by a reference type which lifetime is at least the lifetime of the local variables in A’s body.
Consider the following example:
struct Foo {}
fn consume(mut v: Vec<&Foo>) {
while let Some(..) = v.pop() {
// Do stuff
continue;
}
}
fn setup_and<F>(f: F)
where
F: FnOnce(&mut Vec<&Foo>) + Send,
{
let mut v: Vec<&Foo> = vec![];
let other_foo = Foo {};
f(&mut v);
v.push(&other_foo);
consume(v);
}
fn main() {
let foo = Foo {};
setup_and(|v| {
v.push(&foo);
});
}
rustc cannot infer lifetimes on its own. It complains:
error[E0597]: `foo` does not live long enough
--> src/main.rs:25:17
|
24 | setup_and(|v| {
| --- value captured here
25 | v.push(&foo);
| --------^^^-
| | |
| | borrowed value does not live long enough
| argument requires that `foo` is borrowed for `'static`
26 | });
27 | }
| - `foo` dropped here while still borrowed
I tried to specify a lifetime for the reference taken by setup_and like so:
fn setup_and<'a, F>(f: F)
where
F: FnOnce(&mut Vec<&'a Foo>) + Send,
{
let mut v: Vec<&'a Foo> = vec![];
Now rustc complains about the setup_and local reference other_foo not living long enough. I assume it is because it wants a larger lifetime than the scope of setup_and.
How would I bind lifetimes correctly in that case ? I would like to express that the references must be valid until the end of the consume call.
You have a serious, serious issue with conflicting lifetimes in your implementation, and there is no simple fix without at least a partial redesign of the outside signatures of your struct and methods. They all stem from the setup_and method, and are highlighted by the compiler when you explicitly described the lifetime bounds.
The body of your method is copied below, with annotations for you to understand the issue:
let mut v: Vec<&Foo> = vec![];
let other_foo = Foo {}; // other_foo is created here
f(&mut v);
v.push(&other_foo); // other_foo requires lifetime 'a to be added to this
consume(v); // consume does not restrict the lifetime requirement 'a
// other_foo is dropped here, at lifetime less than 'a
The easiest solution to this problem is to store an Arc<Foo>, like so (playground):
fn setup_and<F>(f: F)
where
F: FnOnce(&mut Vec<Arc<Foo>>) + Send,
{
let mut v: Vec<Arc<Foo>> = vec![];
let other_foo = Foo {};
f(&mut v);
v.push(Arc::new(other_foo));
consume(&mut v);
}
Arc is an Atomic Reference-Counting pointer. It is a clonable structure that works as a dynamic pointer to an object on the heap; for all intents and purposes, it works as a read-only reference, without the requirement for a lifetime. When all copies of an Arc are dropped, the item inside it is also dropped.
This solves two problems:
Your other_foo is now moved into the Arc, and no longer causes its lifetime issue
You can now access your objects just like you would a reference (Arc implements Deref)
The choice of Arc was made because your FnOnce requires Send, which Rc (the single-threaded variant of Arc) cannot provide.
This code compiles:
struct BufRef<'a> {
buf: &'a [u8],
}
struct Foo<'a> {
buf_ref: BufRef<'a>,
}
impl<'a> Iterator for Foo<'a> {
type Item = &'a [u8];
fn next(&mut self) -> Option<Self::Item> {
let result = &self.buf_ref.buf;
Some(result)
}
}
However, if I change BufRef to:
struct BufRef<'a> {
buf: &'a mut [u8],
}
The compiler says:
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
--> src\main.rs:13:16
|
13 | let result = &self.buf_ref.buf;
| ^^^^^^^^^^^^^^^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 12:5...
--> src\main.rs:12:5
|
12 | / fn next(&mut self) -> Option<Self::Item> {
13 | | let result = &self.buf_ref.buf;
14 | | Some(result)
15 | | }
| |_____^
note: ...so that reference does not outlive borrowed content
--> src\main.rs:13:16
|
13 | let result = &self.buf_ref.buf;
| ^^^^^^^^^^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 9:6...
--> src\main.rs:9:6
|
9 | impl<'a> Iterator for Foo<'a> {
| ^^
= note: ...so that the types are compatible:
expected std::iter::Iterator
found std::iter::Iterator
Why does changing the field to &'a mut [u8] cause the error?
Also, what does the compiler mean by this:
...so that the types are compatible:
expected std::iter::Iterator
found std::iter::Iterator
I think that what is misleading you is that your code has a collapsed reference.
Your next function is basically equivalent to this code:
fn next(&mut self) -> Option<&'a [u8]> {
let result: &&'a [u8] = &self.buf_ref.buf;
Some(result)
}
This works because the double reference collapses to a single reference. In this case the double reference only obfuscates the code. Just write:
fn next(&mut self) -> Option<Self::Item> {
Some(self.buf_ref.buf)
}
And this works because references are always Copy.
But now what happens when you change your definition to &'a mut? You are probably guessing right now... mutable references are not Copy, so the same simple code will give you an easy-to-read error message:
cannot move out of self.buf_ref.buf which is behind a mutable reference
Naturally, you can reborrow a mutable ref as a const one, and then try to return it, but unfortunately this will not work because the the re-borrow cannot use the same lifetime as the mutable variable, it must be strictly smaller (or you could alias the pointed values). The compiler assigns the lifetime of this re-borrow as that of the next function, but now you cannot return this borrow, because it is a local reference!
Unfortunately, I don't know of any safe way to make your code compile. In fact I'm quite sure that it would create an unsound API. That is, if you managed to compile your code, then this safe code would create undefined behavior:
fn main() {
let mut buf = vec![1,2,3];
let buf_ref = BufRef { buf: &mut buf };
let mut foo = Foo { buf_ref };
let x: &[u8] = foo.next().unwrap();
//note that x's lifetime is that of buf, foo is not borrowed
//x and foo.buf_ref.buf alias the same memory!
//but the latter is mutable
println!("{}", x[0]); //prints 1
foo.buf_ref.buf[0] = 4;
println!("{}", x[0]); //prints what?
}
In my tests I had a helper function that runs a given method on differently configured objects, with a simplified version looking like this:
fn run_method<F>(f: F)
where
F: Fn(&Foo),
{
let to_test = vec![0i32];
to_test
.iter()
.map(|param| {
let foo = Foo(*param);
f(&foo);
})
.for_each(drop);
}
// run_method(Foo::run);
This worked fine until I added references to the tested struct, making it "lifetime-annotated" (for lack of a proper term, I mean Foo<'a>).
Now I get an error indicating, I think, that Rust doesn't want to accept a Foo::method as a function that can be used with any given lifetime (i.e. F: for<'a> Fn(&Foo<'a>)), as required by the closure:
error[E0631]: type mismatch in function arguments
--> src/main.rs:54:5
|
3 | fn run(&self) {
| ------------- found signature of `for<'r> fn(&'r Foo<'_>) -> _`
...
54 | run_method(Foo::run);
| ^^^^^^^^^^ expected signature of `for<'r, 's> fn(&'r Foo<'s>) -> _`
|
note: required by `run_method`
--> src/main.rs:44:1
|
44 | fn run_method<F>(f: F) where F: Fn(&Foo) {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
error[E0271]: type mismatch resolving `for<'r, 's> <for<'t0> fn(&'t0 Foo<'_>) {Foo::<'_>::run} as std::ops::FnOnce<(&'r Foo<'s>,)>>::Output == ()`
--> src/main.rs:54:5
|
54 | run_method(Foo::run);
| ^^^^^^^^^^ expected bound lifetime parameter, found concrete lifetime
|
note: required by `run_method`
--> src/main.rs:44:1
|
44 | fn run_method<F>(f: F) where F: Fn(&Foo) {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
I can work around the problem by avoiding closures (though I don't really understand how 'a gets constrained to be local to run_method - isn't the lifetime parameter supposed to be chosen by the caller?):
fn run_method<'a, F>(f: F)
where
F: Fn(&Foo<'a>),
{
let to_test = vec![&0i32];
for param in to_test {
let foo = Foo(param);
f(&foo);
}
}
Can I fix this without rewriting? If not - is there a reason why this shouldn't work?
Complete code:
struct Foo<'a>(&'a i32);
impl<'a> Foo<'a> {
fn run(&self) {
println!("Hello {}", self.0);
}
}
fn run_method<F>(f: F)
where
F: Fn(&Foo),
{
//same as F: for<'a> Fn(&Foo<'a>) {
let to_test = vec![0i32];
to_test
.iter()
.map(|param| {
let foo = Foo(param);
f(&foo);
})
.for_each(drop);
}
fn main() {
run_method(Foo::run);
}
// This works:
// fn run_method<'a, F>(f: F)
// where
// F: Fn(&Foo<'a>),
// {
// let to_test = vec![&0i32];
// for param in to_test {
// let foo = Foo(param);
// f(&foo);
// }
// }
// The lifetime-less version:
// struct Foo(i32);
// impl Foo {
// fn run(&self) {
// println!("Hello {}", self.0);
// }
// }
//
// fn run_parser_method<F>(f: F)
// where
// F: Fn(&Foo),
// {
// let to_test = vec![0i32];
// to_test
// .iter()
// .map(|param| {
// let foo = Foo(*param);
// f(&foo);
// })
// .for_each(drop);
// }
//
// fn main() {
// run_parser_method(Foo::run);
// }
playground
An overview of other questions about the same error code:
Expected bound lifetime parameter, found concrete lifetime is about mismatch between trait definition and implementation (trait { fn handle<'a>(); } vs impl<'a> { fn handle() {} })
Function references: expected bound lifetime parameter , found concrete lifetime [E0271] as well as Expected bound lifetime parameter, found concrete lifetime [E0271] is about a closure |args| {...} without explicit type annotations (|args: &[&str]|) not being accepted as a Fn(&[&str]) -> (); the answers don't explain why (the latter hints that it was not implemented in 2015)
Type mismatch "bound lifetime parameter" vs "concrete lifetime" when filling a collection from a closure is again about a closure without explicit type annotations specifying that it accepts a reference (let mut insert = |k| seq.insert(k); (1..10).cycle().take_while(insert)), which masks a more useful "borrowed data cannot be stored outside of its closure" error.
I can work around the problem by avoiding closures (though I don't really understand how 'a gets constrained to be local to run_method - isn't the lifetime parameter supposed to be chosen by the caller?)
It is. But when you rewrite it without closures, you have also put the reference inside the vec! invocation, so it is no longer constructed at runtime. Instead, the compiler can infer that to_test has type Vec<&'static i32>, and as 'static outlives any caller-chosen lifetime, there's no violation.
This fails to compile as you expect:
let to_test = vec![0i32];
for param in to_test.iter() {
let foo = Foo(param);
f(&foo);
}
is there a reason why this shouldn't work?
Yes, because the type of run is constrained by the type of Foo. More specifically, you have a lifetime decided by the caller (implicitly, in the type for Foo), so you have to construct a Foo of that lifetime to invoke the given run reference on it.
Can I fix this without rewriting?
That depends.
If all your test values are literals, you can make 'static references.
If you're able and willing to rewrite run, it's possible to make it unconstrained by the type of Foo
impl<'a> Foo<'a> {
fn run<'s>(_self: &Foo<'s>) {
println!("Hello {}", _self.0);
}
}
But then it doesn't need to go in an impl block at all.
I think the solution provided in the comments is your best bet, because given that you don't care about your concrete Foo<'a> type, there's no need to give the method reference for that type.
I ran into a lifetime problem with a little game. The below code represents a very boiled down version of the update loop.
I need the container mutable reference to get references to other game objects or to create new ones or trigger a functionality.
For that reason, I need the Any trait to be able to cast the trait to a struct, so in my GameObj trait I added an as_any method, but this resulted in a lifetime issue.
use std::any::Any;
trait GameObj<'a> {
fn as_any<'b>(&'b self) -> &'b (dyn Any + 'a);
fn update(&mut self, cont: &mut container);
}
struct object<'a> {
content: &'a String,
}
impl<'a> GameObj<'a> for object<'a> {
fn as_any<'b>(&'b self) -> &'b (dyn Any + 'a) {
return self;
}
fn update(&mut self, cont: &mut container) {
let val = cont.get_obj().unwrap();
let any = val.as_any();
}
}
struct container<'a> {
data: Vec<Box<dyn GameObj<'a> + 'a>>,
}
impl<'a> container<'a> {
fn get_obj<'b>(&'b self) -> Option<&'b Box<dyn GameObj<'a> + 'a>> {
return Some(&self.data[0]);
}
}
pub fn main() {
let a = String::from("hallo");
let b = String::from("asdf");
{
let abc = object { content: &a };
let def = object { content: &b };
let mut cont = container { data: Vec::new() };
cont.data.push(Box::new(abc));
cont.data.push(Box::new(def));
loop {
for i in 0..cont.data.len() {
let mut obj = cont.data.remove(0);
obj.update(&mut cont);
cont.data.insert(i, obj);
}
}
}
}
playground
When I try to build the code, it results in the following error message.
If I comment out/delete let any = val.as_any(); in the update function it compiles fine.
error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'a` due to conflicting requirements
--> src/main.rs:18:24
|
18 | let val = cont.get_obj().unwrap();
| ^^^^^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #3 defined on the method body at 17:5...
--> src/main.rs:17:5
|
17 | / fn update(&mut self, cont: &mut container) {
18 | | let val = cont.get_obj().unwrap();
19 | | let any = val.as_any();
20 | | }
| |_____^
= note: ...so that the types are compatible:
expected &container<'_>
found &container<'_>
= note: but, the lifetime must be valid for the static lifetime...
note: ...so that the declared lifetime parameter bounds are satisfied
--> src/main.rs:19:23
|
19 | let any = val.as_any();
| ^^^^^^
How I can make this work without using 'static, or why is this impossible?
Any is declared trait Any: 'static and can only store 'static types. So in order to make dyn Any + 'a a well-formed type, your as_any method was given an implicit 'a: 'static bound, leading to the lifetime error you showed.
If not for this restriction, you would be able to break safety by putting in an 'a type into an Any and getting out a 'static type, because TypeId can’t tell the difference—lifetimes are erased during compilation. See the discussion on RFC 1849 for more information.
You should think more carefully about why you want to use Any. It’s almost never what you actually want. Perhaps something as simple as an enum type of all the different object types you might want to store would satisfy your use case better?
If you really want to use Any, then you’ll need to find a way to make your types 'static. Rc (or Arc, if threads are involved) is often helpful for this purpose; for example, you could have your object store Rc<String> (or better, Rc<str>) instead of &'a String.