Borrow a struct field mutably and pass it to mutable methods - rust

Because rust doesn't track individual field borrows across method calls - it assumes you've borrowed from the whole struct, it's not possible to borrow a field from the struct, return it, and then pass that into other mutable methods. That sounds contrived but I have this exact issue right now.
struct Foo {
x: i32,
y: i32
}
impl Foo {
pub fn borrow_x(&mut self) -> &mut i32 {
&mut self.x
}
pub fn cross_product(&mut self, other: &mut i32) {
self.y *= *other;
*other *= self.y;
}
}
fn main() {
let mut foo = Foo{x: 2, y: 4};
let x_ref = foo.borrow_x();
foo.cross_product(x_ref);
println!("x={} y={}", foo.x, foo.y);
}
This of course isn't permitted in rust, the error from the compiler is:
error[E0499]: cannot borrow `foo` as mutable more than once at a time
--> src/main.rs:20:5
|
19 | let x_ref = foo.borrow_x();
| --- first mutable borrow occurs here
20 | foo.cross_product(x_ref);
| ^^^ ----- first borrow later used here
| |
| second mutable borrow occurs here
My first reaction is to use unsafe to tell the compiler that it doesn't really borrow from self:
pub fn borrow_x<'a>(&mut self) -> &'a mut i32 {
unsafe {
std::mem::transmute::<&mut i32, &'a mut i32>(&mut self.x)
}
}
It works, but it requires that you don't actually break rust's aliasing rules and reference foo.x elsewhere by accident.
Then I thought maybe I could split the struct so that the borrow and cross_product methods are on different sibling structs. This does work:
struct Bar {
x: X,
y: Y,
}
struct X {
inner: i32
}
impl X {
pub fn borrow_x(&mut self) -> &mut i32 {
&mut self.inner
}
}
struct Y {
inner: i32
}
impl Y {
pub fn cross_product(&mut self, other: &mut i32) {
self.inner *= *other;
*other *= self.inner;
}
}
fn main() {
let mut bar = Bar{x: X{inner: 2}, y: Y{inner: 4}};
let x_ref = bar.x.borrow_x();
bar.y.cross_product(x_ref);
println!("x={} y={}", bar.x.inner, bar.y.inner);
}
I think one could ditch the method all-together and pass references to both fields into cross_product as a free function. That should be equivalent to the last solution. I find that kind of ugly though, especially as the number of fields needed from the struct grow.
Are there other solutions to this problem?

The thing to keep in mind is that rust is working at a per-function level.
pub fn cross_product(&mut self, other: &mut i32) {
self.y *= *other;
*other *= self.y;
}
The mutability rules ensure that self.y and other are not the same thing.
If you're not careful you can run into this kind of bug in C++/C/Java quite easily.
i.e. imagine if instead of
let x_ref = foo.borrow_x();
foo.cross_product(x_ref);
you were doing
let y_ref = foo.borrow_y();
foo.cross_product(y_ref);
What value should foo.y end up with? (Allowing these to refer to the same object can cause certain performance optimisations to not be applicable too)
As you mentioned, you could go with a free function
fn cross_product(a: &mut i32, b: &mut i32) {
a *= b;
b *= a;
}
but I'd consider ditching mutability altogether
fn cross_product(a:i32, b:i32) -> (i32,i32) {
(a*b, a*b*b)
}
ASIDE: If the return value seems odd to you (it did to me initially too) and you expected (a*b, a*b) you need to think a little harder... and I'd say that alone is a good reason to avoid the mutable version.
Then I can use it like this
let (_x,_y) = cross_product(foo.x, foo.y);
foo.x = _x;
foo.y = _y;
This is a bit verbose, but when RFC-2909 lands we can instead write:
(foo.y, foo.x) = cross_product(foo.x, foo.y);

Related

Mutable borrow into two parts with cleanup

I have some object that I want to split into two parts via a mutable borrow, then combine those back together into the original object when the split references go out of scope.
The simplified example below is for a Count struct that holds a single i32, which we want to split into two &mut i32s, who are both incorporated back into the original Count when the two mutable references go out of scope.
The approach I am taking below is to use an intermediate object CountSplit which holds a mutable reference to the original Count object and has the Drop trait implemented to do the re-combination logic.
This approach feels kludgy. In particular, this is awkward:
let mut ms = c.make_split();
let (x, y) = ms.split();
Doing this in one line like let (x, y) = c.make_split().split(); is not allowed because the intermediate object must have a longer lifetime. Ideally I would be able to do something like let (x, y) = c.magic_split(); and avoid exposing the intermediate object altogether.
Is there a way to do this which doesn't require doing two let's every time, or some other way to tackle this pattern that would be more idiomatic?
#[derive(Debug)]
struct Count {
val: i32,
}
trait MakeSplit<'a> {
type S: Split<'a>;
fn make_split(&'a mut self) -> Self::S;
}
impl<'a> MakeSplit<'a> for Count {
type S = CountSplit<'a>;
fn make_split(&mut self) -> CountSplit {
CountSplit {
top: self,
second: 0,
}
}
}
struct CountSplit<'a> {
top: &'a mut Count,
second: i32,
}
trait Split<'a> {
fn split(&'a mut self) -> (&'a mut i32, &'a mut i32);
}
impl<'a, 'b> Split<'a> for CountSplit<'b> {
fn split(&mut self) -> (&mut i32, &mut i32) {
(&mut self.top.val, &mut self.second)
}
}
impl<'a> Drop for CountSplit<'a> {
fn drop(&mut self) {
println!("custom drop occurs here");
self.top.val += self.second;
}
}
fn main() {
let mut c = Count { val: 2 };
println!("{:?}", c); // Count { val: 2 }
{
let mut ms = c.make_split();
let (x, y) = ms.split();
println!("split: {} {}", x, y); // split: 2 0
// each of these lines correctly gives a compile-time error
// c.make_split(); // can't borrow c as mutable
// println!("{:?}", c); // or immutable
// ms.split(); // also can't borrow ms
*x += 100;
*y += 5000;
println!("split: {} {}", x, y); // split: 102 5000
} // custom drop occurs here
println!("{:?}", c); // Count { val: 5102 }
}
playground:
I don't think a reference to a temporary value like yours can be made to work in today's Rust.
If it's any help, if you specifically want to call a function with two &mut i32 parameters like you mentioned in the comments, e.g.
fn foo(a: &mut i32, b: &mut i32) {
*a += 1;
*b += 2;
println!("split: {} {}", a, b);
}
you can already do that with the same number of lines as you'd have if your chaining worked.
With the chaining, you'd call
let (x, y) = c.make_split().split();
foo(x, y);
And if you just leave out the conversion to a tuple, it looks like this:
let mut ms = c.make_split();
foo(&mut ms.top.val, &mut ms.second);
You can make it a little prettier by e.g. storing the mutable reference to val directly in CountSplit as first, so that it becomes foo(&mut ms.first, &mut ms.second);. If you want it to feel even more like a tuple, I think you can use DerefMut to be able to write foo(&mut ms.0, &mut ms.1);.
Alternatively, you can of course formulate this as a function taking a function
impl Count {
fn as_split<F: FnMut(&mut i32, &mut i32)>(&mut self, mut f: F) {
let mut second = 0;
f(&mut self.val, &mut second);
self.val += second;
}
}
and then just call
c.as_split(foo);

Can I borrow self immutably for self.callbacks: Vec<Box<FnMut>>?

pub struct Notifier<'a, T> {
callbacks: Vec<Box<'a + FnMut(&T)>>
}
impl<'a, T> Notifier<'a, T>{
fn add_callback<F: 'a + FnMut(&T)>(&mut self, callback: F) {
self.callbacks.push(Box::new(callback));
}
fn trigger(&mut self, payload: T) {
for callback in &mut self.callbacks {
callback(&payload);
}
}
}
struct A {
x: i64
}
impl A {
fn foo(&mut self, x: &i64) {
self.x = x + 1;
}
}
fn test() {
let mut bar = A {x: 3};
let mut notifier = Notifier{callbacks: Vec::new()};
notifier.add_callback(|x| bar.foo(x));
}
Playground
This is a simple observer pattern implemented using callbacks. It works.
However, the fact that trigger(&mut self... causes much trouble in my later coding (How to update self based on reference of value from hashmap from self). Is it possible to make trigger(&self ... instead?
I'm using rustc 1.19.0-nightly.
If you want to change the interior of a struct without having the struct mutable, you should use a Cell:
Values of the Cell<T> and RefCell<T> types may be mutated through shared references (i.e. the common &T type), whereas most Rust types can only be mutated through unique (&mut T) references. We say that Cell<T> and RefCell<T> provide 'interior mutability', in contrast with typical Rust types that exhibit 'inherited mutability'.
Playground

How to resolve lifetime error for mutable reference in Rust?

I am not sure why the following code does not compile.
use std::cmp::Ordering;
struct MyItr<'a> {
cur: &'a i32,
}
impl<'a> Ord for MyItr<'a> {
fn cmp(&self, other: &MyItr) -> Ordering {
self.cur.cmp(&other.cur)
}
}
impl<'a> PartialOrd for MyItr<'a> {
fn partial_cmp(&self, other: &MyItr<'a>) -> Option<Ordering> {
Some(self.cmp(other))
}
}
impl<'a> PartialEq for MyItr<'a> {
fn eq(&self, other: &MyItr) -> bool {
self.cur == other.cur
}
}
impl<'a> Eq for MyItr<'a> {}
fn f0<'a>(t0: &'a mut MyItr<'a>, t1: &'a mut MyItr<'a>, i: &'a i32) {
let t = std::cmp::max(t0, t1);
t.cur = i;
}
fn f1() {
let i0 = 1;
let i1 = 2;
let mut z0 = MyItr { cur: &i0 };
let mut z1 = MyItr { cur: &i1 };
let i2 = 3;
f0(&mut z0, &mut z1, &i2);
}
$ cargo build
Compiling foo v0.1.0 (file:///private/tmp/foo)
error: `z1` does not live long enough
--> lib.rs:40:1
|
39 | f0(&mut z0, &mut z1, &i2);
| -- borrow occurs here
40 | }
| ^ `z1` dropped here while still borrowed
|
= note: values in a scope are dropped in the opposite order they are created
My understanding is the borrowed references to z0 and z1 are backed once the f0 invocation ends. However, The compiler seems to assume the borrowed references are not backed.
$ cargo --version
cargo 0.20.0-nightly (41e490480 2017-05-16)
There are two problems here. The first is that you've over-specified lifetimes, creating a situation the compiler just can't deal with.
fn f0<'a>(t0: &'a mut MyItr<'a>, t1: &'a mut MyItr<'a>, i: &'a i32)
You've told the compiler that all the arguments must be pointers with the same lifetime. The compiler can narrow overlapping lifetimes, but in this case that doesn't help. You've specified that the pointer to the MyItrs has the same lifetime as the thing they point to, and the outer pointers are mutable.
The second problem is that (even after fixing that), what you're trying to do is just outright unsafe and will lead to dangling pointers.
Here's a more minimal example:
struct S<'a> {
ptr: &'a i32,
}
fn f<'b>(t: &'b mut S<'b>, new_ptr: &'b i32) {}
fn main() {
let i0 = 1;
let mut s = S { ptr: &i0 };
let i1 = 2;
f(&mut s, &i1);
}
What is 'b? Well, the compiler can only narrow lifetimes, so usually you'd just take the lifetimes of everything you're trying to pass and pick the shortest one. In this case, that would be the lifetime of i1. So, it has to narrow the lifetime of &s. The lifetime on the pointer to s itself isn't a problem (you can narrow how long you take a borrow for), but narrowing the inner lifetime (the one used for the ptr field) is a problem.
If the compiler narrowed the lifetime of s.ptr, you would then be able to store &i1 in that field. s expects s.ptr to outlive itself, but that will no longer be true: i1 will be destroyed before s is, meaning s.ptr will contain a dangling pointer. And Rust will not allow that to happen.
As a result, Rust can't narrow s's inner 'a lifetime... but if it can't narrow it, then that means 'b must be the full, un-narrowed 'a. But wait, that means that 'b is longer than the lifetime of s itself and i1. And that's impossible.
Hence the failure.
The solution requires two things. First, you need to not over-specify lifetimes. Secondly, you need to ensure that some valid lifetime exists at all; in the case of your original code, that means moving i2 above z0 and z1 so that it outlives them. Like so:
fn f0<'a>(t0: &mut MyItr<'a>, t1: &mut MyItr<'a>, i: &'a i32) {
let t: &mut MyItr<'a> = std::cmp::max(t0, t1);
t.cur = i;
}
fn f1() {
let i0 = 1;
let i1 = 2;
let i2 = 3;
let mut z0 = MyItr { cur: &i0 };
let mut z1 = MyItr { cur: &i1 };
f0(&mut z0, &mut z1, &i2);
}
A rule of thumb: don't just spam a single lifetime everywhere. Only use the same lifetime for things that should be the same.

When is it useful to define multiple lifetimes in a struct?

In Rust, when we want a struct to contain references, we typically define their lifetimes as such:
struct Foo<'a> {
x: &'a i32,
y: &'a i32,
}
But it's also possible to define multiple lifetimes for different references in the same struct:
struct Foo<'a, 'b> {
x: &'a i32,
y: &'b i32,
}
When is it ever useful to do this? Can someone provide some example code that doesn't compile when both lifetimes are 'a but does compile when the lifetimes are 'a and 'b (or vice versa)?
After staying up way too late, I was able to come up with an example case where the lifetimes matter. Here is the code:
static ZERO: i32 = 0;
struct Foo<'a, 'b> {
x: &'a i32,
y: &'b i32,
}
fn get_x_or_zero_ref<'a, 'b>(x: &'a i32, y: &'b i32) -> &'a i32 {
if *x > *y {
return x
} else {
return &ZERO
}
}
fn main() {
let x = 1;
let v;
{
let y = 2;
let f = Foo { x: &x, y: &y };
v = get_x_or_zero_ref(&f.x, &f.y);
}
println!("{}", *v);
}
If you were to change the definition of Foo to this:
struct Foo<'a> {
x: &'a i32,
y: &'a i32,
}
Then the code won't compile.
Basically, if you want to use the fields of the struct on any function that requires it's parameters to have different lifetimes, then the fields of the struct must have different lifetimes as well.
I want to re-answer my question here since it's still showing up high in search results and I feel I can explain better. Consider this code:
Rust Playground
struct Foo<'a> {
x: &'a i32,
y: &'a i32,
}
fn main() {
let x = 1;
let v;
{
let y = 2;
let f = Foo { x: &x, y: &y };
v = f.x;
}
println!("{}", *v);
}
And the error:
error[E0597]: `y` does not live long enough
--> src/main.rs:11:33
|
11 | let f = Foo { x: &x, y: &y };
| ^^ borrowed value does not live long enough
12 | v = f.x;
13 | }
| - `y` dropped here while still borrowed
14 | println!("{}", *v);
| -- borrow later used here
What's going on here?
The lifetime of f.x has the requirement of being at least large enough to encompass the scope of x up until the println! statement (since it's initialized with &x and then assigned to v).
The definition of Foo specifies that both f.x and f.y use the same generic lifetime 'a, so the lifetime of f.y must be at least as large as f.x.
But, that can't work, because we assign &y to f.y, and y goes out of scope before the println!. Error!
The solution here is to allow Foo to use separate lifetimes for f.x and f.y, which we do using multiple generic lifetime parameters:
Rust Playground
struct Foo<'a, 'b> {
x: &'a i32,
y: &'b i32,
}
Now the lifetimes of f.x and f.y aren't tied together. The compiler will still use a lifetime that's valid until the println! statement for f.x. But there's no longer a requirement that f.y uses the same lifetime, so the compiler is free to choose a smaller lifetime for f.y, such as one that is valid only for the scope of y.
Here is another simple example where the struct definition has to use two lifetimes in order to operate as expected. It does not split the aggregate into fields of different lifetimes, but nests the struct with another struct.
struct X<'a>(&'a i32);
struct Y<'a, 'b>(&'a X<'b>);
fn main() {
let z = 100;
//taking the inner field out of a temporary
let z1 = ((Y(&X(&z))).0).0;
assert!(*z1 == z);
}
The struct Y has two lifetime parameters, one for its contained field &X, and one for X's contained field &z.
In the operation ((Y(&X(&z))).0).0, X(&z) is created as a temporary and is borrowed. Its lifetime is only in the scope of this operation, expiring at the statement end. But since X(&z)'s lifetime is different from the its contained field &z, the operation is fine to return &z, whose value can be accessed later in the function.
If using single lifetime for Y struct. This operation won't work, because the lifetime of &z is the same as its containing struct X(&z), expiring at the statement end; therefore the returned &z is no longer valid to be accessed afterwards.
See code in the playground.

HashMap borrow issue when trying to implement find or insert

I tried to implement own analogue of find_or_insert method that looks like this:
use std::collections::HashMap;
pub struct SomeManager {
next: i32,
types: HashMap<i32, i32>,
}
impl SomeManager {
pub fn get_type<'a>(&'a mut self, k: i32) -> &'a i32 {
match self.types.get(&k) {
Some(ref x) => return *x,
None => {
self.types.insert(k, self.next);
self.next += 1;
return self.types.get(&k).unwrap();
}
}
}
}
fn main() {}
Error:
error[E0502]: cannot borrow `self.types` as mutable because it is also borrowed as immutable
--> src/main.rs:13:17
|
10 | match self.types.get(&k) {
| ---------- immutable borrow occurs here
...
13 | self.types.insert(k, self.next);
| ^^^^^^^^^^ mutable borrow occurs here
...
18 | }
| - immutable borrow ends here
I know that there are some standard methods that implement this functionality, but I want this method to be as light as possible - it will be called very-very often and almost all of the time the values will already exist.
As I understand it, when we call self.types.get we borrow it to scope of match statement, so we can't call self.types.insert here. I have tried to move methods from None branch out of match statement, but it also fails.
The only working solution that I found requires invoking get twice:
pub fn get_type<'a>(&'a mut self, k: i32) -> &'a i32 {
let is_none = match self.types.get(&k) {
Some(ref x) => false,
None => true,
};
if is_none {
self.types.insert(k, self.next);
self.next += 1;
}
self.types.get(&k).unwrap()
}
How can I work around such situations?
There are a handful of methods on HashMap to achieve these sorts of complex cases. Most notably, for your case, HashMap::entry and Entry::or_insert_with:
pub fn get_type<'a>(&'a mut self, k: i32) -> &'a i32 {
self.types.entry(k).or_insert_with(|| {
let value = self.next;
self.next += 1;
value
})
}
In your case, however, there’s the borrow of self inside, so that won’t do.
We thus shift the borrow of self.next outside of the closure so the compiler can reason about it as disjoint from self.types. Problem solved with only one lookup, as it should be.
pub fn get_type<'a>(&'a mut self, k: i32) -> &'a i32 {
let next = &mut self.next;
self.types.entry(k).or_insert_with(|| {
let value = *next;
*next += 1;
value
})
}
Note that in your first case you're doing one lookup when the key exists in the map and three when it does not exist. Your last attempt does two lookups in either case. This is somewhat prettified version of the latter:
pub fn get_type<'a>(&'a mut self, k: i32) -> &'a i32 {
let contains = self.types.contains_key(&k);
if !contains {
self.types.insert(k, self.next);
self.next += 1;
}
self.types.get(&k).unwrap()
}
I don't think it is possible to avoid the second lookup without some support from the map's implementation because of borrowing restrictions.
In any case, using the solution by Chris Morgan is superior to the above one (for example, it may be more efficient and in fact require less lookups), so I suggest sticking with it.

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