Lifetimes issues when placing a match with mutable references into a function - rust

Let's say I have this code (it's contrived, since it's heavily simplified):
enum Foo<'a, T> where T: 'a {
Bar(&'a mut Vec<T>),
}
fn main() {
let mut v = vec![1, 2];
let foo: Foo<isize> = Foo::Bar(&mut v);
let a = match foo {
Foo::Bar(ref mut v) => &mut v[..],
};
a[1] = 10;
println!("a = {:?}", a);
}
This works and everyone is happy. However, if I try to clean it up a bit by placing the match statement inside a function, I can't even get the function to compile, let alone put it to work for me. One of my many attempts looks like this:
fn unpack<'a, T>(foo: Foo<'a, T>) -> &'a mut [T] {
match foo {
Foo::Bar(ref mut v) => &mut v[..],
}
}
I just get the error:
error: `foo.0` does not live long enough
I have been trying to change to multiple lifetimes and adding a where clause (just like I had to do to get the Enum to work), but I can't seem to get it to work. My question is simply if it works (and how) or if the only option here is to use a macro.

Foo::Bar(ref mut v) borrows &mut Vec<T> from foo so it can't outlive foo (which only lives until the end of the function). What you actually want to do is take it (by value) so you should match on Foo::Bar(v):
fn unpack<'a, T>(foo: Foo<'a, T>) -> &'a mut [T] {
match foo {
Foo::Bar(v) => &mut v[..],
}
}
Note: Rust will deref v as needed so the following will also work:
fn unpack<'a, T>(foo: Foo<'a, T>) -> &'a mut [T] {
match foo {
Foo::Bar(v) => v,
}
}

Related

ref to 'static does not live long enough?

Consider next code:
fn get_ref<'a, R>(slice: &'a Vec<i32>, f: fn(&'a Vec<i32>) -> R) -> R
where
R: 'a,
{
f(slice)
}
fn main() {
let v = [1,2,3,4,5,6];
let iter = get_ref(&v, |x| x.iter().skip(1).take(2));
println!("{:?}", iter.collect::<Vec<_>>());
}
I create some static variable, then apply some function to its reference and get a result.
It seems to work totally fine. At least it successfully compiles.
Now I am trying to add next level of abstraction. And things are getting weird...
fn owned<'a, R>(owner: Vec<i32>, f: fn(&'a Vec<i32>) -> R)
where
R: 'a,
{
let _ = get_ref(&owner, f); // error occurs here
// `owner` does not live long enough.
}
// get_ref is the same as in the first example
fn get_ref<'a, R>(slice: &'a Vec<i32>, f: fn(&'a Vec<i32>) -> R) -> R
where
R: 'a,
{
f(slice)
}
fn main() {
let v = [1,2,3,4,5,6];
owned(v, |x| x.iter().skip(1).take(2));
}
For me it looks like pretty the same code. But Rust fails to compile it. I really don't understand why this is happening and how should I rewrite my code to compile.
Imagine if I decide to call the function owned<'static, i32>(Vec<i32>, foo) with foo defined as:
fn foo(vec: &'static Vec<i32>) -> i32 { ... }
This satisfies the bounds for owned, since i32: 'static. However this means that you must have a static reference to call f, but owner does not live forever, since it is destroyed at the end of owned.
One way to fix it is to use the following:
fn owned<R>(owner: Vec<i32>, f: for<'a> fn(&'a Vec<i32>) -> R) {
let _ = get_ref(&owner, f);
}
It says that f must be callable with any lifetime, rather than just some specific lifetime. However, this has the consequence that R cannot borrow from the argument, since R is declared in a larger scope than 'a. There isn't any way to fix that while keeping the generics as is.
This answer was taken from my response to this thread on URLO.

Why Rust can't coerce mutable reference to immutable reference in a type constructor?

It is possible to coerce &mut T into &T but it doesn't work if the type mismatch happens within a type constructor.
playground
use ndarray::*; // 0.13.0
fn print(a: &ArrayView1<i32>) {
println!("{:?}", a);
}
pub fn test() {
let mut x = array![1i32, 2, 3];
print(&x.view_mut());
}
For the above code I get following error:
|
9 | print(&x.view_mut());
| ^^^^^^^^^^^^^ types differ in mutability
|
= note: expected reference `&ndarray::ArrayBase<ndarray::ViewRepr<&i32>, ndarray::dimension::dim::Dim<[usize; 1]>>`
found reference `&ndarray::ArrayBase<ndarray::ViewRepr<&mut i32>, ndarray::dimension::dim::Dim<[usize; 1]>>`
It is safe to coerce &mut i32 to &i32 so why it is not applied in this situation? Could you provide some examples on how could it possibly backfire?
In general, it's not safe to coerce Type<&mut T> into Type<&T>.
For example, consider this wrapper type, which is implemented without any unsafe code and is therefore sound:
#[derive(Copy, Clone)]
struct Wrapper<T>(T);
impl<T: Deref> Deref for Wrapper<T> {
type Target = T::Target;
fn deref(&self) -> &T::Target { &self.0 }
}
impl<T: DerefMut> DerefMut for Wrapper<T> {
fn deref_mut(&mut self) -> &mut T::Target { &mut self.0 }
}
This type has the property that &Wrapper<&T> automatically dereferences to &T, and &mut Wrapper<&mut T> automatically dereferences to &mut T. In addition, Wrapper<T> is copyable if T is.
Assume that there exists a function that can take a &Wrapper<&mut T> and coerce it into a &Wrapper<&T>:
fn downgrade_wrapper_ref<'a, 'b, T: ?Sized>(w: &'a Wrapper<&'b mut T>) -> &'a Wrapper<&'b T> {
unsafe {
// the internals of this function is not important
}
}
By using this function, it is possible to get a mutable and immutable reference to the same value at the same time:
fn main() {
let mut value: i32 = 0;
let mut x: Wrapper<&mut i32> = Wrapper(&mut value);
let x_ref: &Wrapper<&mut i32> = &x;
let y_ref: &Wrapper<&i32> = downgrade_wrapper_ref(x_ref);
let y: Wrapper<&i32> = *y_ref;
let a: &mut i32 = &mut *x;
let b: &i32 = &*y;
// these two lines will print the same addresses
// meaning the references point to the same value!
println!("a = {:p}", a as &mut i32); // "a = 0x7ffe56ca6ba4"
println!("b = {:p}", b as &i32); // "b = 0x7ffe56ca6ba4"
}
Full playground example
This is not allowed in Rust, leads to undefined behavior and means that the function downgrade_wrapper_ref is unsound in this case. There may be other specific cases where you, as the programmer, can guarantee that this won't happen, but it still requires you to implement it specifically for those case, using unsafe code, to ensure that you take the responsibility of making those guarantees.
Consider this check for an empty string that relies on content staying unchanged for the runtime of the is_empty function (for illustration purposes only, don't use this in production code):
struct Container<T> {
content: T
}
impl<T> Container<T> {
fn new(content: T) -> Self
{
Self { content }
}
}
impl<'a> Container<&'a String> {
fn is_empty(&self, s: &str) -> bool
{
let str = format!("{}{}", self.content, s);
&str == s
}
}
fn main() {
let mut foo : String = "foo".to_owned();
let container : Container<&mut String> = Container::new(&mut foo);
std::thread::spawn(|| {
container.content.replace_range(1..2, "");
});
println!("an empty str is actually empty: {}", container.is_empty(""))
}
(Playground)
This code does not compile since &mut String does not coerce into &String. If it did, however, it would be possible that the newly created thread changed the content after the format! call but before the equal comparison in the is_empty function, thereby invalidating the assumption that the container's content was immutable, which is required for the empty check.
It seems type coercions don't apply to array elements when array is the function parameter type.
playground

Is it possible to create a wrapper around an &mut that acts like an &mut

The following code fails to compile because MutRef is not Copy. It can not be made copy because &'a mut i32 is not Copy. Is there any way give MutRef similar semantics to &'a mut i32?
The motivation for this is being able to package up a large set of function parameters into a struct so that they can be passed as a group instead of needing to be passed individually.
struct MutRef<'a> {
v: &'a mut i32
}
fn wrapper_use(s: MutRef) {
}
fn raw_use(s: &mut i32) {
}
fn raw_ref() {
let mut s: i32 = 9;
let q = &mut s;
raw_use(q);
raw_use(q);
}
fn wrapper() {
let mut s: i32 = 9;
let q = MutRef{ v: &mut s };
wrapper_use(q);
wrapper_use(q);
}
No.
The name for this feature is "implicit reborrowing" and it happens when you pass a &mut reference where the compiler expects a &mut reference of a possibly different lifetime. The compiler only implicitly reborrows when the actual type and the expected type are both &mut references. It does not work with generic arguments or structs that contain &mut references. There is no way in current Rust to make a custom type that can be implicitly reborrowed. There is an open issue about this limitation dating from 2015, but so far nobody has proposed any way to lift it.
You can always implement your own method to explicitly reborrow:
impl<'a> MutRef<'a> {
// equivalent to fn reborrow(&mut self) -> MutRef<'_>
fn reborrow<'b>(&'b mut self) -> MutRef<'b> {
MutRef {v: self.v}
}
}
fn wrapper() {
let mut s: i32 = 9;
let mut q = MutRef{ v: &mut s };
wrapper_use(q.reborrow()); // does not move q
wrapper_use(q); // moves q
}
See also
Why is the mutable reference not moved here?
Type inference and borrowing vs ownership transfer

IntoIterator as a function argument doesn't accept adapter struct

I want to have a function, that accepts &IntoIterator<Item=u32>, so I could pass to it both &Vec<u32> and iterators' adapter structs (like Map, Filter and any other, which I believe all implement IntoIterator)
So I have a function like
pub fn f<'a, T>(it_src: &'a T) -> u32
where &'a T: IntoIterator<Item = u32> {
let it = it_src.into_iter();
let result: u32;
// more more usage
result
}
And this is how I tried to use it (same signature, but different name)
pub fn f_with_feature()<'a, T>(it_src: &'a T) -> u32
where &'a T: IntoIterator<Item = u32> {
let adjusted_values = it_src.into_iter()
.map(|e| adjust(e));
f(&adjusted_values)
}
What I've got is an error
error[E0308]: mismatched types
--> src\main.rs:14:7
|
14 | f(&adjusted_values)
| ^^^^^^^^^^^^^^^^ expected type parameter, found struct `std::iter::Map`
|
= note: expected type `&T`
found type `&std::iter::Map<<&T as std::iter::IntoIterator>::IntoIter, [closure#src\main.rs:13:14: 13:27]>`
How is it that Map doesn't match as T?
Also, I've come up with an idea, that passing iterators' adaptors with static dispatch isn't a good idea since each other closure used to generate a Map will create a new function specialization. Though I've seen that static dispatch approach for most of the times is idiomatic in Rust. How to manage this situation?
I think you want to have trait bounds on T (and not on &'a T). So I guess you actually want the following:
pub fn f<'a, T>(it_src: &'a T) -> u32
where T: IntoIterator<Item = u32> {
let it = it_src.into_iter();
let result: u32 = 1;
// more more usage
result
}
pub fn f_with_feature<'a, T>(it_src: &'a T) -> u32
where T: IntoIterator<Item = u32> {
let adjusted_values = it_src.into_iter()
.map(|e| adjust(e));
f(&adjusted_values)
}
Which brings us to the next problem: IntoIterator's into_iter consumes self, which means that you cannot call it_src.into_iter if you only borrow it_src.
So if you really want to use into_iter, you can try this:
pub fn f<T>(it_src: T) -> u32
where T: IntoIterator<Item = u32> {
let it = it_src.into_iter();
let result: u32 = 1;
// more more usage
result
}
pub fn f_with_feature<T>(it_src: T) -> u32
where T: IntoIterator<Item = u32> {
let adjusted_values = it_src.into_iter()
.map(|e| adjust(e));
f(adjusted_values)
}
The above, however, requires you to move the values into f resp. f_with_feature.
In my experience, just taking an iterator (and doing the conversion at call site if necessary), leads to simple, straightforward solutions:
pub fn f<T>(it_src: T) -> u32
where T: Iterator<Item = u32> {
let it = it_src.into_iter();
let result: u32 = 1;
// more more usage
result
}
pub fn f_with_feature<T>(it_src: T) -> u32
where T: Iterator<Item = u32> {
let adjusted_values = it_src.into_iter()
.map(|e| adjust(e));
f(adjusted_values)
}

borrowed value does not live long enough in this case ( Vec<&Fn(i32) -> i32> )

I am having this error, other times I had something similar and I have been able to solve, in different ways but now is not how to solve in this case:
borrowed value does not live long enough in
I moved the code that fails one more simple, but I can not find the error:
fn main(){
let mut v: Vec<&Fn(i32) -> i32> = Vec::new();
v.push(&ops_code1);
//v.push(&ops_code2);
//v.push(&ops_code3);
}
fn ops_code1(value: i32) -> i32 {
..//
error: borrowed value does not live long enough
v.push(&ops_code1);
play.rust
What you are doing here is creating a Vec of closures. In Rust static functions are treated slightly differently from closures, so when we create the reference a closure is actually created. If we do that after creating the Vec the resulting closure gets a shorter lifetime than the Vec, which is an error. We can instead use a let to create the closure before the Vec, giving a long enough lifetime, outliving the Vec:
fn main() {
let extended = &ops_code1;
let mut v: Vec<&Fn(i32) -> i32> = Vec::new();
// Note that placing it here does not work:
// let extended = &ops_code1;
v.push(extended);
//v.push(&ops_code2);
//v.push(&ops_code3);
}
fn ops_code1(value: i32) -> i32 {
println!("ops_code1 {}", value);
value
}
Rust Playground
However, if you only use static functions - and not closures - the following also works fine, and lets you avoid the extra let:
fn main() {
let mut v: Vec<fn(i32) -> i32> = Vec::new();
v.push(ops_code1);
v.push(ops_code2);
}
fn ops_code1(value: i32) -> i32 {
println!("ops_code1 {}", value);
value
}
fn ops_code2(value: i32) -> i32 {
println!("ops_code2 {}", value);
value
}
Rust Playground
A third option is to use boxed closures, which let's you use both closures and static functions without the extra lets, but with its own trade-offs:
fn main() {
let mut v: Vec<Box<Fn(i32) -> i32>> = Vec::new();
v.push(Box::new(ops_code1));
v.push(Box::new(ops_code2));
for f in v {
f(1);
}
}
fn ops_code1(value: i32) -> i32 {
println!("ops_code1 {}", value);
value
}
fn ops_code2(value: i32) -> i32 {
println!("ops_code2 {}", value);
value
}
Rust Playground

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