I have this piece of code:
#[derive(Debug)]
struct Foo<'a> {
x: &'a i32,
}
impl<'a> Foo<'a> {
fn set(&mut self, r: &'a i32) {
self.x = r;
}
}
fn main() {
let v = 5;
let w = 7;
let mut f = Foo { x: &v };
println!("f is {:?}", f);
f.set(&w);
println!("now f is {:?}", f);
}
My understanding is that in the first borrow of the value of v, the generic lifetime parameter 'a on the struct declaration is filled in with the lifetime of the value of v. This means that the resulting Foo object must not live longer than this 'a lifetime or that the value of v must live at least as long as the Foo object.
In the call to the method set, the lifetime parameter on the impl block is used and the lifetime of the value of w is filled in for 'a in the method signature. &mut self is assigned a different lifetime by the compiler, which is the lifetime of f (the Foo object). If I switched the order of the bindings of w and f in the main function, this would result in an error.
I wondered what would happen if I annotated the &mut self reference with the same lifetime parameter 'a as r in the set method:
impl<'a> Foo<'a> {
fn set(&'a mut self, r: &'a i32) {
self.x = r;
}
}
Which results in the following error:
error[E0502]: cannot borrow `f` as immutable because it is also borrowed as mutable
--> src/main.rs:21:31
|
19 | f.set(&w);
| - mutable borrow occurs here
20 |
21 | println!("now f is {:?}", f);
| ^ immutable borrow occurs here
22 | }
| - mutable borrow ends here
In contrast to the example above, f is still considered mutably borrowed by the time the second println! is called, so it cannot be borrowed simultaneously as immutable.
How did this come to be?
By not leaving off the lifetime annotation the compiler filled one in for me for &mut self in the first example. This happens by the rules of lifetime elision. However by explicitly setting it to 'a in the second example I linked the lifetimes of the value of f and the value of w.
Is f considered borrowed by itself somehow?
And if so, what is the scope of the borrow? Is it min(lifetime of f, lifetime of w) -> lifetime of f?
I assume I haven't fully understood the &mut self reference in the function call yet. I mean, the function returns, but f is still considered to be borrowed.
I am trying to fully understand lifetimes. I am primarily looking for corrective feedback on my understanding of the concepts. I am grateful for every bit of advice and further clarification.
In the call to the method set the lifetime parameter on the impl block is used and the lifetime of the value of w is filled in for 'a in the method signature.
No. The value of the lifetime parameter 'a is fixed at the creation of the Foo struct, and will never change as it is part of its type.
In your case, the compiler actually choses for 'a a value that is compatible with both the lifetimes of v and w. If that was not possible, it would fail, such as in this example:
fn main() {
let v = 5;
let mut f = Foo { x: &v };
println!("f is {:?}", f);
let w = 7;
f.set(&w);
println!("now f is {:?}", f);
}
which outputs:
error[E0597]: `w` does not live long enough
--> src/main.rs:21:1
|
18 | f.set(&w);
| - borrow occurs here
...
21 | }
| ^ `w` dropped here while still borrowed
|
= note: values in a scope are dropped in the opposite order they are created
Exactly because the 'a lifetime imposed by v is not compatible with the shorter lifetime of w.
In the second example, by forcing the lifetime of self to be 'a as well, you are tying the mutable borrow to the lifetime 'a as well, and thus the borrow ends when all items of lifetime 'a goes out of scope, namely v and w.
Related
I'm having some trouble understanding the compiler error message for this code:
struct A {
b: B,
c: C,
}
struct B {
f: u32,
}
struct C {
f: u32,
}
fn foo(b: &mut B) -> &u32 { &b.f }
fn bar(c: &mut C) -> &u32 { &c.f }
fn quux(z: &mut A) -> u32 { z.b.f }
fn baz(a: &mut A) {
let x = foo(&mut a.b);
let y = bar(&mut a.c);
let z = quux(a); // equivalent to: quux(&mut *a)
println!("{} {}", x, y);
}
error[E0499]: cannot borrow `*a` as mutable more than once at a time
--> src/main.rs:22:18
|
20 | let x = foo(&mut a.b);
| -------- first mutable borrow occurs here
21 | let y = bar(&mut a.c);
22 | let z = quux(a);
| ^ second mutable borrow occurs here
23 | println!("{} {}", x, y);
| - first borrow later used here
For more information about this error, try `rustc --explain E0499`.
error: could not compile `playground` due to previous error
If line 20 is mutably borrowing *a (as the compilation error asserts), then line 21 must also mutably borrow *a (How couldn't it? It's nearly identical to line 20 -- only difference is which field is borrowed), so by my reasoning the code should still fail to compile after deleting line 22; however, the code actually compiles just fine if I delete line 22. The fact that this would then compile seems to contradict the compiler's claim that &mut a.b results in a mutable borrow of *a.
x and y are immutable borrows, so if anything I would expect a compilation error regarding the overlapping scopes of the immutable x/y and the mutable borrow of *a in line 22 (otherwise quux could conceivably mutate a while we're borrowing its fields).
What knowledge am I likely missing that makes this error message confusing to me? If you're not surprised by this message, could you break down how you reason about this?
Some more thinking before I submitted this question:
I suppose the immutable references returned from foo and bar somehow extend the original mutable borrows of a.b and a.c (why that would be necessary, I don't know). If that's the case, maybe the way to see this is that &mut a.b and &mut a.c in turn extend the mutable borrow of *a. The error message says the first mutable borrow of *a occurs on line 20, but maybe the way to see this is:
The true first borrow of *a is when when the caller gives us the &mut A.
While the compiler suggests there's a mutable borrow of *a on line 20, this isn't exactly the case: the mut &a.b expression borrows (*a).b, but merely extends the existing borrow given to us (as opposed to being a new mutable borrow of *a).
The last point, but with respect to mut &a.c.
Insert the aforementioned reasoning of "extending borrows" wrt *a.
The call to quux then mutably borrows the *a a second time.
So maybe what the compilation error calls the initial borrow of *a is actually reporting the location of the expression responsible for extending the original *a borrow?
Line 20 does not conflict with 21 since they borrow different fields within a and therefore cause no collision.
The issue is that foo and bar return references that inherit the lifetime of the mutable reference. Even though they return immutable references, their lifetimes rely on the lifetime of the mutable borrow. So by requiring they live for a period of time, by extension the mutable borrow must do so as well. The compiler always takes the function signature on face value so it can not downgrade the initial mutable reference. If it did not, it could lead to issues if the function was modified later or if it contained unsafe code that relied on consuming a mutable reference for safety.
Here is a counter example where all of the function signatures remain the same. Since the compiler can not make assumptions on how quux and foo will use their values quux may make the references returned by foo or bar invalid.
struct A {
b: B,
c: C,
}
struct B {
f: Option<u32>,
}
struct C {
f: u32,
}
fn foo(b: &mut B) -> &u32 {
b.f.as_ref().unwrap()
}
fn bar(c: &mut C) -> &u32 { &c.f }
fn quux(z: &mut A) -> u32 {
z.b.f = None;
z.c.f
}
fn baz(a: &mut A) {
let x = foo(&mut a.b);
let y = bar(&mut a.c);
let z = quux(a);
println!("{} {}", x, y);
}
There are two issues here. First: why does let x = foo(&mut a.b); prevent you from doing let z = quux(a);, and second why does let x = foo(&mut a.b); not prevent you from doing let y = bar(&mut a.c);?
Why does let x = foo(&mut a.b); prevent you from doing let z = quux(a);?
This is due to the lifetime elision rules. You declared foo as:
fn foo(b: &mut B) -> &u32
Since you use references there must be a lifetime, but since you didn't specify one yourself the compiler assumes that all references have the same lifetime. In other words, the compiler rewrites your code as:
fn foo<'a>(b: &'a mut B) -> &'a u32
Which means that the returned value has the same lifetime as the mutable borrow, so the mutable borrow lasts until the println where you use the return value x, which prevents other borrows of a.b or a as a whole.
But then why does let x = foo(&mut a.b); not prevent you from doing let y = bar(&mut a.c);?
That's because the compiler is smart enough to realize that a.b and a.c are disjoint parts of a, which can be borrowed without conflict (in the same way that you can borrow different local variables simultaneously).
I was playing around with Rust closures when I hit this interesting scenario:
fn main() {
let mut y = 10;
let f = || &mut y;
f();
}
This gives an error:
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
--> src/main.rs:4:16
|
4 | let f = || &mut y;
| ^^^^^^
|
note: first, the lifetime cannot outlive the lifetime as defined on the body at 4:13...
--> src/main.rs:4:13
|
4 | let f = || &mut y;
| ^^^^^^^^^
note: ...so that closure can access `y`
--> src/main.rs:4:16
|
4 | let f = || &mut y;
| ^^^^^^
note: but, the lifetime must be valid for the call at 6:5...
--> src/main.rs:6:5
|
6 | f();
| ^^^
note: ...so type `&mut i32` of expression is valid during the expression
--> src/main.rs:6:5
|
6 | f();
| ^^^
Even though the compiler is trying to explain it line by line, I still haven't understood what exactly it is complaining about.
Is it trying to say that the mutable reference cannot outlive the enclosing closure?
The compiler does not complain if I remove the call f().
Short version
The closure f stores a mutable reference to y. If it were allowed to return a copy of this reference, you would end up with two simultaneous mutable references to y (one in the closure, one returned), which is forbidden by Rust's memory safety rules.
Long version
The closure can be thought of as
struct __Closure<'a> {
y: &'a mut i32,
}
Since it contains a mutable reference, the closure is called as FnMut, essentially with the definition
fn call_mut(&mut self, args: ()) -> &'a mut i32 { self.y }
Since we only have a mutable reference to the closure itself, we can't move the field y out, neither are we able to copy it, since mutable references aren't Copy.
We can trick the compiler into accepting the code by forcing the closure to be called as FnOnce instead of FnMut. This code works fine:
fn main() {
let x = String::new();
let mut y: u32 = 10;
let f = || {
drop(x);
&mut y
};
f();
}
Since we are consuming x inside the scope of the closure and x is not Copy, the compiler detects that the closure can only be FnOnce. Calling an FnOnce closure passes the closure itself by value, so we are allowed to move the mutable reference out.
Another more explicit way to force the closure to be FnOnce is to pass it to a generic function with a trait bound. This code works fine as well:
fn make_fn_once<'a, T, F: FnOnce() -> T>(f: F) -> F {
f
}
fn main() {
let mut y: u32 = 10;
let f = make_fn_once(|| {
&mut y
});
f();
}
There are two main things at play here:
Closures cannot return references to their environment
A mutable reference to a mutable reference can only use the lifetime of the outer reference (unlike with immutable references)
Closures returning references to environment
Closures cannot return any references with the lifetime of self (the closure object). Why is that? Every closure can be called as FnOnce, since that's the super-trait of FnMut which in turn is the super-trait of Fn. FnOnce has this method:
fn call_once(self, args: Args) -> Self::Output;
Note that self is passed by value. So since self is consumed (and now lives within the call_once function`) we cannot return references to it -- that would be equivalent to returning references to a local function variable.
In theory, the call_mut would allow to return references to self (since it receives &mut self). But since call_once, call_mut and call are all implemented with the same body, closures in general cannot return references to self (that is: to their captured environment).
Just to be sure: closures can capture references and return those! And they can capture by reference and return that reference. Those things are something different. It's just about what is stored in the closure type. If there is a reference stored within the type, it can be returned. But we can't return references to anything stored within the closure type.
Nested mutable references
Consider this function (note that the argument type implies 'inner: 'outer; 'outer being shorter than 'inner):
fn foo<'outer, 'inner>(x: &'outer mut &'inner mut i32) -> &'inner mut i32 {
*x
}
This won't compile. On the first glance, it seems like it should compile, since we're just peeling one layer of references. And it does work for immutable references! But mutable references are different here to preserve soundness.
It's OK to return &'outer mut i32, though. But it's impossible to get a direct reference with the longer (inner) lifetime.
Manually writing the closure
Let's try to hand code the closure you were trying to write:
let mut y = 10;
struct Foo<'a>(&'a mut i32);
impl<'a> Foo<'a> {
fn call<'s>(&'s mut self) -> &'??? mut i32 { self.0 }
}
let mut f = Foo(&mut y);
f.call();
What lifetime should the returned reference have?
It can't be 'a, because we basically have a &'s mut &'a mut i32. And as discussed above, in such a nested mutable reference situation, we can't extract the longer lifetime!
But it also can't be 's since that would mean the closure returns something with the lifetime of 'self ("borrowed from self"). And as discussed above, closures can't do that.
So the compiler can't generate the closure impls for us.
Consider this code:
fn main() {
let mut y: u32 = 10;
let ry = &mut y;
let f = || ry;
f();
}
It works because the compiler is able to infer ry's lifetime: the reference ry lives in the same scope of y.
Now, the equivalent version of your code:
fn main() {
let mut y: u32 = 10;
let f = || {
let ry = &mut y;
ry
};
f();
}
Now the compiler assigns to ry a lifetime associated to the scope of the closure body, not to the lifetime associated with the main body.
Also note that the immutable reference case works:
fn main() {
let mut y: u32 = 10;
let f = || {
let ry = &y;
ry
};
f();
}
This is because &T has copy semantics and &mut T has move semantics, see Copy/move semantics documentation of &T/&mut T types itself for more details.
The missing piece
The compiler throws an error related to a lifetime:
cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
but as pointed out by Sven Marnach there is also a problem related to the error
cannot move out of borrowed content
But why doesn't the compiler throw this error?
The short answer is that the compiler first executes type checking and then borrow checking.
the long answer
A closure is made up of two pieces:
the state of the closure: a struct containing all the variables captured by the closure
the logic of the closure: an implementation of the FnOnce, FnMut or Fn trait
In this case the state of the closure is the mutable reference y and the logic is the body of the closure { &mut y } that simply returns a mutable reference.
When a reference is encountered, Rust controls two aspects:
the state: if the reference points to a valid memory slice, (i.e. the read-only part of lifetime validity);
the logic: if the memory slice is aliased, in other words if it is pointed from more than one reference simultaneously;
Note the move out from borrowed content is forbidden for avoiding memory aliasing.
The Rust compiler executes its job through several stages, here's a simplified workflow:
.rs input -> AST -> HIR -> HIR postprocessing -> MIR -> HIR postprocessing -> LLVM IR -> binary
The compiler reports a lifetime problem because it first executes the type checking phase in HIR postprocessing (which comprises lifetime analysis) and after that, if successful, executes borrow checking in the MIR postprocessing phase.
I was playing around with Rust closures when I hit this interesting scenario:
fn main() {
let mut y = 10;
let f = || &mut y;
f();
}
This gives an error:
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
--> src/main.rs:4:16
|
4 | let f = || &mut y;
| ^^^^^^
|
note: first, the lifetime cannot outlive the lifetime as defined on the body at 4:13...
--> src/main.rs:4:13
|
4 | let f = || &mut y;
| ^^^^^^^^^
note: ...so that closure can access `y`
--> src/main.rs:4:16
|
4 | let f = || &mut y;
| ^^^^^^
note: but, the lifetime must be valid for the call at 6:5...
--> src/main.rs:6:5
|
6 | f();
| ^^^
note: ...so type `&mut i32` of expression is valid during the expression
--> src/main.rs:6:5
|
6 | f();
| ^^^
Even though the compiler is trying to explain it line by line, I still haven't understood what exactly it is complaining about.
Is it trying to say that the mutable reference cannot outlive the enclosing closure?
The compiler does not complain if I remove the call f().
Short version
The closure f stores a mutable reference to y. If it were allowed to return a copy of this reference, you would end up with two simultaneous mutable references to y (one in the closure, one returned), which is forbidden by Rust's memory safety rules.
Long version
The closure can be thought of as
struct __Closure<'a> {
y: &'a mut i32,
}
Since it contains a mutable reference, the closure is called as FnMut, essentially with the definition
fn call_mut(&mut self, args: ()) -> &'a mut i32 { self.y }
Since we only have a mutable reference to the closure itself, we can't move the field y out, neither are we able to copy it, since mutable references aren't Copy.
We can trick the compiler into accepting the code by forcing the closure to be called as FnOnce instead of FnMut. This code works fine:
fn main() {
let x = String::new();
let mut y: u32 = 10;
let f = || {
drop(x);
&mut y
};
f();
}
Since we are consuming x inside the scope of the closure and x is not Copy, the compiler detects that the closure can only be FnOnce. Calling an FnOnce closure passes the closure itself by value, so we are allowed to move the mutable reference out.
Another more explicit way to force the closure to be FnOnce is to pass it to a generic function with a trait bound. This code works fine as well:
fn make_fn_once<'a, T, F: FnOnce() -> T>(f: F) -> F {
f
}
fn main() {
let mut y: u32 = 10;
let f = make_fn_once(|| {
&mut y
});
f();
}
There are two main things at play here:
Closures cannot return references to their environment
A mutable reference to a mutable reference can only use the lifetime of the outer reference (unlike with immutable references)
Closures returning references to environment
Closures cannot return any references with the lifetime of self (the closure object). Why is that? Every closure can be called as FnOnce, since that's the super-trait of FnMut which in turn is the super-trait of Fn. FnOnce has this method:
fn call_once(self, args: Args) -> Self::Output;
Note that self is passed by value. So since self is consumed (and now lives within the call_once function`) we cannot return references to it -- that would be equivalent to returning references to a local function variable.
In theory, the call_mut would allow to return references to self (since it receives &mut self). But since call_once, call_mut and call are all implemented with the same body, closures in general cannot return references to self (that is: to their captured environment).
Just to be sure: closures can capture references and return those! And they can capture by reference and return that reference. Those things are something different. It's just about what is stored in the closure type. If there is a reference stored within the type, it can be returned. But we can't return references to anything stored within the closure type.
Nested mutable references
Consider this function (note that the argument type implies 'inner: 'outer; 'outer being shorter than 'inner):
fn foo<'outer, 'inner>(x: &'outer mut &'inner mut i32) -> &'inner mut i32 {
*x
}
This won't compile. On the first glance, it seems like it should compile, since we're just peeling one layer of references. And it does work for immutable references! But mutable references are different here to preserve soundness.
It's OK to return &'outer mut i32, though. But it's impossible to get a direct reference with the longer (inner) lifetime.
Manually writing the closure
Let's try to hand code the closure you were trying to write:
let mut y = 10;
struct Foo<'a>(&'a mut i32);
impl<'a> Foo<'a> {
fn call<'s>(&'s mut self) -> &'??? mut i32 { self.0 }
}
let mut f = Foo(&mut y);
f.call();
What lifetime should the returned reference have?
It can't be 'a, because we basically have a &'s mut &'a mut i32. And as discussed above, in such a nested mutable reference situation, we can't extract the longer lifetime!
But it also can't be 's since that would mean the closure returns something with the lifetime of 'self ("borrowed from self"). And as discussed above, closures can't do that.
So the compiler can't generate the closure impls for us.
Consider this code:
fn main() {
let mut y: u32 = 10;
let ry = &mut y;
let f = || ry;
f();
}
It works because the compiler is able to infer ry's lifetime: the reference ry lives in the same scope of y.
Now, the equivalent version of your code:
fn main() {
let mut y: u32 = 10;
let f = || {
let ry = &mut y;
ry
};
f();
}
Now the compiler assigns to ry a lifetime associated to the scope of the closure body, not to the lifetime associated with the main body.
Also note that the immutable reference case works:
fn main() {
let mut y: u32 = 10;
let f = || {
let ry = &y;
ry
};
f();
}
This is because &T has copy semantics and &mut T has move semantics, see Copy/move semantics documentation of &T/&mut T types itself for more details.
The missing piece
The compiler throws an error related to a lifetime:
cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
but as pointed out by Sven Marnach there is also a problem related to the error
cannot move out of borrowed content
But why doesn't the compiler throw this error?
The short answer is that the compiler first executes type checking and then borrow checking.
the long answer
A closure is made up of two pieces:
the state of the closure: a struct containing all the variables captured by the closure
the logic of the closure: an implementation of the FnOnce, FnMut or Fn trait
In this case the state of the closure is the mutable reference y and the logic is the body of the closure { &mut y } that simply returns a mutable reference.
When a reference is encountered, Rust controls two aspects:
the state: if the reference points to a valid memory slice, (i.e. the read-only part of lifetime validity);
the logic: if the memory slice is aliased, in other words if it is pointed from more than one reference simultaneously;
Note the move out from borrowed content is forbidden for avoiding memory aliasing.
The Rust compiler executes its job through several stages, here's a simplified workflow:
.rs input -> AST -> HIR -> HIR postprocessing -> MIR -> HIR postprocessing -> LLVM IR -> binary
The compiler reports a lifetime problem because it first executes the type checking phase in HIR postprocessing (which comprises lifetime analysis) and after that, if successful, executes borrow checking in the MIR postprocessing phase.
In section 3.2 of the Nomicon, under the heading "liveness", it says
However it's often the case that Rust isn't sufficiently smart to
prove that multiple borrows are disjoint.
What is an example where the Rust compiler cannot prove that they are disjoint? Will this ever occur in a tuple struct?
The key is in the previous sentence:
Rust explicitly enables [reborrowing into multiple mutable references] to be done with disjoint struct fields, because disjointness can be statically proven
Outside of this case, the compiler cannot tell that two borrows are disjoint. In practice, this means that the compiler cannot tell that borrows resulting from a function call will be disjoint.
struct Thing {
a: i32,
b: i32,
}
fn example_works(thing: &mut Thing) {
let a = &mut thing.a;
let b = &mut thing.b;
}
fn get_a(thing: &mut Thing) -> &mut i32 {
&mut thing.a
}
fn get_b(thing: &mut Thing) -> &mut i32 {
&mut thing.b
}
fn example_doesnt_work(thing: &mut Thing) {
let a = get_a(thing);
let b = get_b(thing);
println!("{}, {}", a, b);
}
error[E0499]: cannot borrow `*thing` as mutable more than once at a time
--> src/lib.rs:26:19
|
25 | let a = get_a(thing);
| ----- first mutable borrow occurs here
26 | let b = get_b(thing); // cannot borrow `*thing` as mutable more than once at a time
| ^^^^^ second mutable borrow occurs here
27 | println!("{}, {}", a, b);
| - first borrow later used here
Will this ever occur in a tuple struct?
Not specifically because it's a tuple struct, but yes, it can happen for the same reasons. If you obtain a borrow from a function call, you will get the same problem as a "traditional" struct.
fn works<'a>(foo: &Option<&'a mut String>, s: &'a mut String) {}
fn error<'a>(foo: &RefCell<Option<&'a mut String>>, s: &'a mut String) {}
let mut s = "hi".to_string();
let foo = None;
works(&foo, &mut s);
// with this, it errors
// let bar = RefCell::new(None);
// error(&bar, &mut s);
s.len();
If I put in the two lines with the comment, the following error occurs:
error[E0502]: cannot borrow `s` as immutable because it is also borrowed as mutable
--> <anon>:16:5
|
14 | error(&bar, &mut s);
| - mutable borrow occurs here
15 |
16 | s.len();
| ^ immutable borrow occurs here
17 | }
| - mutable borrow ends here
The signatures of works() and errors() look fairly similar. But apparently the compiler knows that you can cheat on it with a RefCell, because the borrow checker behaves differently.
I can even "hide" the RefCell in another type of my own, but the compiler still always does the right thing (errors in case a RefCell could be used). How does the compiler know all that stuff and how does it work? Does the compiler mark types as "interior mutability container" or something like that?
RefCell<T> contains an UnsafeCell<T> which is a special lang item. It is UnsafeCell that causes the error. You could check with:
fn error<'a>(foo: &UnsafeCell<Option<&'a mut String>>, s: &'a mut String) {}
...
let bar = UnsafeCell::new(None);
error(&bar, &mut s);
But the error is not due to compiler recognizing an UnsafeCell introduces interior mutability, but that an UnsafeCell is invariant in T. In fact, we could reproduce the error using PhantomData:
struct Contravariant<T>(PhantomData<fn(T)>);
fn error<'a>(foo: Contravariant<&'a i32>, s: &'a mut String) {}
...
let bar = Contravariant(PhantomData);
error(bar, &mut s);
or even just anything that is contravariant or invariant in the lifetime 'a:
fn error<'a>(foo: Option<fn(&'a i32)>, s: &'a mut String) {}
let bar = None;
error(bar, &mut s);
The reason you can't hide a RefCell is because variance is derived through the fields of the structure. Once you used RefCell<T> somewhere, no matter how deep, the compiler will figure out T is invariant.
Now let's see how the compiler determine the E0502 error. First, it's important to remember that the compiler has to choose two specific lifetimes here: the lifetime in the type of the expression &mut s ('a) and the lifetime in the type of bar (let's call it 'x). Both are restricted: the former lifetime 'a has to be shorter than the scope of s, otherwise we would end up with a reference living longer than the original string. 'x has to be larger than the scope of bar, otherwise we could access an dangling pointer through bar (if a type has a lifetime parameter the compiler assume the type can access a value with that lifetime).
With these two basic restriction, the compiler goes through the following steps:
The type bar is Contravariant<&'x i32>.
The error function accepts any subtype of Contravariant<&'a i32>, where 'a is the lifetime of that &mut s expression.
Thus bar should be a subtype of Contravariant<&'a i32>
Contravariant<T> is contravariant over T, i.e. if U <: T, then Contravariant<T> <: Contravariant<U>.
So the subtyping relation can be satisfied when &'x i32 is a supertype of &'a i32.
Thus 'x should be shorter than 'a, i.e. 'a should outlive 'x.
Similarly, for an invariant type, the derived relation is 'a == 'x, and for convariant, 'x outlives 'a.
Now, the problem here is that the lifetime in the type of bar lives until the end of scope (as per restriction mentioned above):
let bar = Contravariant(PhantomData); // <--- 'x starts here -----+
error(bar, // |
&mut s); // <- 'a starts here ---+ |
s.len(); // | |
// <--- 'x ends here¹ --+---+
// |
// <--- 'a ends here² --+
}
// ¹ when `bar` goes out of scope
// ² 'a has to outlive 'x
In both contravariant and invariant cases, 'a outlives (or equals to) 'x means the statement s.len() must be included in the range, causing borrowck error.
Only in the covariant case we could make the range of 'a shorter than 'x, allowing the temporary object &mut s be dropped before s.len() is called (meaning: at s.len(), s is not considered borrowed anymore):
let bar = Covariant(PhantomData); // <--- 'x starts here -----+
// |
error(bar, // |
&mut s); // <- 'a starts here --+ |
// | |
// <- 'a ends here ----+ |
s.len(); // |
} // <--- 'x ends here -------+