Related
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut v = vec![1, 2, 3];
change(&mut v[0], &mut v[1]);
}
When I compile the code above, it has the error:
error[E0499]: cannot borrow `v` as mutable more than once at a time
--> src/main.rs:9:32
|
9 | change(&mut v[0], &mut v[1]);
| - ^ - first borrow ends here
| | |
| | second mutable borrow occurs here
| first mutable borrow occurs here
Why does the compiler prohibit it? v[0] and v[1] occupy different memory positions, so it's not dangerous to use these together. And what should I do if I come across this problem?
You can solve this with split_at_mut():
let mut v = vec![1, 2, 3];
let (a, b) = v.split_at_mut(1); // Returns (&mut [1], &mut [2, 3])
change(&mut a[0], &mut b[0]);
There are uncountably many safe things to do that the compiler unfortunately does not recognize yet. split_at_mut() is just like that, a safe abstraction implemented with an unsafe block internally.
We can do that too, for this problem. The following is something I use in code where I need to separate all three cases anyway (I: Index out of bounds, II: Indices equal, III: Separate indices).
enum Pair<T> {
Both(T, T),
One(T),
None,
}
fn index_twice<T>(slc: &mut [T], a: usize, b: usize) -> Pair<&mut T> {
if a == b {
slc.get_mut(a).map_or(Pair::None, Pair::One)
} else {
if a >= slc.len() || b >= slc.len() {
Pair::None
} else {
// safe because a, b are in bounds and distinct
unsafe {
let ar = &mut *(slc.get_unchecked_mut(a) as *mut _);
let br = &mut *(slc.get_unchecked_mut(b) as *mut _);
Pair::Both(ar, br)
}
}
}
}
Since Rust 1.26, pattern matching can be done on slices. You can use that as long as you don't have huge indices and your indices are known at compile-time.
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut arr = [5, 6, 7, 8];
{
let [ref mut a, _, ref mut b, ..] = arr;
change(a, b);
}
assert_eq!(arr, [7, 6, 5, 8]);
}
The borrow rules of Rust need to be checked at compilation time, that is why something like mutably borrowing a part of a Vec is a very hard problem to solve (if not impossible), and why it is not possible with Rust.
Thus, when you do something like &mut v[i], it will mutably borrow the entire vector.
Imagine I did something like
let guard = something(&mut v[i]);
do_something_else(&mut v[j]);
guard.do_job();
Here, I create an object guard that internally stores a mutable reference to v[i], and will do something with it when I call do_job().
In the meantime, I did something that changed v[j]. guard holds a mutable reference that is supposed to guarantee nothing else can modify v[i]. In this case, all is good, as long as i is different from j; if the two values are equal it is a huge violation of the borrow rules.
As the compiler cannot guarantee that i != j, it is thus forbidden.
This was a simple example, but similar cases are legions, and are why such access mutably borrows the whole container. Plus the fact that the compiler actually does not know enough about the internals of Vec to ensure that this operation is safe even if i != j.
In your precise case, you can have a look at the swap(..) method available on Vec that does the swap you are manually implementing.
On a more generic case, you'll probably need an other container. Possibilities are wrapping all the values of your Vec into a type with interior mutability, such as Cell or RefCell, or even using a completely different container, as #llogiq suggested in his answer with par-vec.
The method [T]::iter_mut() returns an iterator that can yield a mutable reference for each element in the slice. Other collections have an iter_mut method too. These methods often encapsulate unsafe code, but their interface is totally safe.
Here's a general purpose extension trait that adds a method on slices that returns mutable references to two distinct items by index:
pub trait SliceExt {
type Item;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item);
}
impl<T> SliceExt for [T] {
type Item = T;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item) {
match index0.cmp(&index1) {
Ordering::Less => {
let mut iter = self.iter_mut();
let item0 = iter.nth(index0).unwrap();
let item1 = iter.nth(index1 - index0 - 1).unwrap();
(item0, item1)
}
Ordering::Equal => panic!("[T]::get_two_mut(): received same index twice ({})", index0),
Ordering::Greater => {
let mut iter = self.iter_mut();
let item1 = iter.nth(index1).unwrap();
let item0 = iter.nth(index0 - index1 - 1).unwrap();
(item0, item1)
}
}
}
}
On recent nightlies, there is get_many_mut():
#![feature(get_many_mut)]
fn main() {
let mut v = vec![1, 2, 3];
let [a, b] = v
.get_many_mut([0, 1])
.expect("out of bounds or overlapping indices");
change(a, b);
}
Building up on the answer by #bluss you can use split_at_mut() to create a function that can turn mutable borrow of a vector into a vector of mutable borrows of vector elements:
fn borrow_mut_elementwise<'a, T>(v:&'a mut Vec<T>) -> Vec<&'a mut T> {
let mut result:Vec<&mut T> = Vec::new();
let mut current: &mut [T];
let mut rest = &mut v[..];
while rest.len() > 0 {
(current, rest) = rest.split_at_mut(1);
result.push(&mut current[0]);
}
result
}
Then you can use it to get a binding that lets you mutate many items of original Vec at once, even while you are iterating over them (if you access them by index in your loop, not through any iterator):
let mut items = vec![1,2,3];
let mut items_mut = borrow_mut_elementwise(&mut items);
for i in 1..items_mut.len() {
*items_mut[i-1] = *items_mut[i];
}
println!("{:?}", items); // [2, 3, 3]
The problem is that &mut v[…] first mutably borrows v and then gives the mutable reference to the element to the change-function.
This reddit comment has a solution to your problem.
Edit: Thanks for the heads-up, Shepmaster. par-vec is a library that allows to mutably borrow disjunct partitions of a vec.
I publish my daily utils for this to crate.io. Link to the doc.
You may use it like
use arref::array_mut_ref;
let mut arr = vec![1, 2, 3, 4];
let (a, b) = array_mut_ref!(&mut arr, [1, 2]);
assert_eq!(*a, 2);
assert_eq!(*b, 3);
let (a, b, c) = array_mut_ref!(&mut arr, [1, 2, 0]);
assert_eq!(*c, 1);
// ⚠️ The following code will panic. Because we borrow the same element twice.
// let (a, b) = array_mut_ref!(&mut arr, [1, 1]);
It's a simple wrapper around the following code, which is sound. But it requires that the two indexes are different at runtime.
pub fn array_mut_ref<T>(arr: &mut [T], a0: usize, a1: usize) -> (&mut T, &mut T) {
assert!(a0 != a1);
// SAFETY: this is safe because we know a0 != a1
unsafe {
(
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
)
}
}
Alternatively, you may use a method that won't panic with mut_twice
#[inline]
pub fn mut_twice<T>(arr: &mut [T], a0: usize, a1: usize) -> Result<(&mut T, &mut T), &mut T> {
if a0 == a1 {
Err(&mut arr[a0])
} else {
unsafe {
Ok((
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
))
}
}
}
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut v = vec![1, 2, 3];
change(&mut v[0], &mut v[1]);
}
When I compile the code above, it has the error:
error[E0499]: cannot borrow `v` as mutable more than once at a time
--> src/main.rs:9:32
|
9 | change(&mut v[0], &mut v[1]);
| - ^ - first borrow ends here
| | |
| | second mutable borrow occurs here
| first mutable borrow occurs here
Why does the compiler prohibit it? v[0] and v[1] occupy different memory positions, so it's not dangerous to use these together. And what should I do if I come across this problem?
You can solve this with split_at_mut():
let mut v = vec![1, 2, 3];
let (a, b) = v.split_at_mut(1); // Returns (&mut [1], &mut [2, 3])
change(&mut a[0], &mut b[0]);
There are uncountably many safe things to do that the compiler unfortunately does not recognize yet. split_at_mut() is just like that, a safe abstraction implemented with an unsafe block internally.
We can do that too, for this problem. The following is something I use in code where I need to separate all three cases anyway (I: Index out of bounds, II: Indices equal, III: Separate indices).
enum Pair<T> {
Both(T, T),
One(T),
None,
}
fn index_twice<T>(slc: &mut [T], a: usize, b: usize) -> Pair<&mut T> {
if a == b {
slc.get_mut(a).map_or(Pair::None, Pair::One)
} else {
if a >= slc.len() || b >= slc.len() {
Pair::None
} else {
// safe because a, b are in bounds and distinct
unsafe {
let ar = &mut *(slc.get_unchecked_mut(a) as *mut _);
let br = &mut *(slc.get_unchecked_mut(b) as *mut _);
Pair::Both(ar, br)
}
}
}
}
Since Rust 1.26, pattern matching can be done on slices. You can use that as long as you don't have huge indices and your indices are known at compile-time.
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut arr = [5, 6, 7, 8];
{
let [ref mut a, _, ref mut b, ..] = arr;
change(a, b);
}
assert_eq!(arr, [7, 6, 5, 8]);
}
The borrow rules of Rust need to be checked at compilation time, that is why something like mutably borrowing a part of a Vec is a very hard problem to solve (if not impossible), and why it is not possible with Rust.
Thus, when you do something like &mut v[i], it will mutably borrow the entire vector.
Imagine I did something like
let guard = something(&mut v[i]);
do_something_else(&mut v[j]);
guard.do_job();
Here, I create an object guard that internally stores a mutable reference to v[i], and will do something with it when I call do_job().
In the meantime, I did something that changed v[j]. guard holds a mutable reference that is supposed to guarantee nothing else can modify v[i]. In this case, all is good, as long as i is different from j; if the two values are equal it is a huge violation of the borrow rules.
As the compiler cannot guarantee that i != j, it is thus forbidden.
This was a simple example, but similar cases are legions, and are why such access mutably borrows the whole container. Plus the fact that the compiler actually does not know enough about the internals of Vec to ensure that this operation is safe even if i != j.
In your precise case, you can have a look at the swap(..) method available on Vec that does the swap you are manually implementing.
On a more generic case, you'll probably need an other container. Possibilities are wrapping all the values of your Vec into a type with interior mutability, such as Cell or RefCell, or even using a completely different container, as #llogiq suggested in his answer with par-vec.
The method [T]::iter_mut() returns an iterator that can yield a mutable reference for each element in the slice. Other collections have an iter_mut method too. These methods often encapsulate unsafe code, but their interface is totally safe.
Here's a general purpose extension trait that adds a method on slices that returns mutable references to two distinct items by index:
pub trait SliceExt {
type Item;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item);
}
impl<T> SliceExt for [T] {
type Item = T;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item) {
match index0.cmp(&index1) {
Ordering::Less => {
let mut iter = self.iter_mut();
let item0 = iter.nth(index0).unwrap();
let item1 = iter.nth(index1 - index0 - 1).unwrap();
(item0, item1)
}
Ordering::Equal => panic!("[T]::get_two_mut(): received same index twice ({})", index0),
Ordering::Greater => {
let mut iter = self.iter_mut();
let item1 = iter.nth(index1).unwrap();
let item0 = iter.nth(index0 - index1 - 1).unwrap();
(item0, item1)
}
}
}
}
On recent nightlies, there is get_many_mut():
#![feature(get_many_mut)]
fn main() {
let mut v = vec![1, 2, 3];
let [a, b] = v
.get_many_mut([0, 1])
.expect("out of bounds or overlapping indices");
change(a, b);
}
Building up on the answer by #bluss you can use split_at_mut() to create a function that can turn mutable borrow of a vector into a vector of mutable borrows of vector elements:
fn borrow_mut_elementwise<'a, T>(v:&'a mut Vec<T>) -> Vec<&'a mut T> {
let mut result:Vec<&mut T> = Vec::new();
let mut current: &mut [T];
let mut rest = &mut v[..];
while rest.len() > 0 {
(current, rest) = rest.split_at_mut(1);
result.push(&mut current[0]);
}
result
}
Then you can use it to get a binding that lets you mutate many items of original Vec at once, even while you are iterating over them (if you access them by index in your loop, not through any iterator):
let mut items = vec![1,2,3];
let mut items_mut = borrow_mut_elementwise(&mut items);
for i in 1..items_mut.len() {
*items_mut[i-1] = *items_mut[i];
}
println!("{:?}", items); // [2, 3, 3]
The problem is that &mut v[…] first mutably borrows v and then gives the mutable reference to the element to the change-function.
This reddit comment has a solution to your problem.
Edit: Thanks for the heads-up, Shepmaster. par-vec is a library that allows to mutably borrow disjunct partitions of a vec.
I publish my daily utils for this to crate.io. Link to the doc.
You may use it like
use arref::array_mut_ref;
let mut arr = vec![1, 2, 3, 4];
let (a, b) = array_mut_ref!(&mut arr, [1, 2]);
assert_eq!(*a, 2);
assert_eq!(*b, 3);
let (a, b, c) = array_mut_ref!(&mut arr, [1, 2, 0]);
assert_eq!(*c, 1);
// ⚠️ The following code will panic. Because we borrow the same element twice.
// let (a, b) = array_mut_ref!(&mut arr, [1, 1]);
It's a simple wrapper around the following code, which is sound. But it requires that the two indexes are different at runtime.
pub fn array_mut_ref<T>(arr: &mut [T], a0: usize, a1: usize) -> (&mut T, &mut T) {
assert!(a0 != a1);
// SAFETY: this is safe because we know a0 != a1
unsafe {
(
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
)
}
}
Alternatively, you may use a method that won't panic with mut_twice
#[inline]
pub fn mut_twice<T>(arr: &mut [T], a0: usize, a1: usize) -> Result<(&mut T, &mut T), &mut T> {
if a0 == a1 {
Err(&mut arr[a0])
} else {
unsafe {
Ok((
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
))
}
}
}
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut v = vec![1, 2, 3];
change(&mut v[0], &mut v[1]);
}
When I compile the code above, it has the error:
error[E0499]: cannot borrow `v` as mutable more than once at a time
--> src/main.rs:9:32
|
9 | change(&mut v[0], &mut v[1]);
| - ^ - first borrow ends here
| | |
| | second mutable borrow occurs here
| first mutable borrow occurs here
Why does the compiler prohibit it? v[0] and v[1] occupy different memory positions, so it's not dangerous to use these together. And what should I do if I come across this problem?
You can solve this with split_at_mut():
let mut v = vec![1, 2, 3];
let (a, b) = v.split_at_mut(1); // Returns (&mut [1], &mut [2, 3])
change(&mut a[0], &mut b[0]);
There are uncountably many safe things to do that the compiler unfortunately does not recognize yet. split_at_mut() is just like that, a safe abstraction implemented with an unsafe block internally.
We can do that too, for this problem. The following is something I use in code where I need to separate all three cases anyway (I: Index out of bounds, II: Indices equal, III: Separate indices).
enum Pair<T> {
Both(T, T),
One(T),
None,
}
fn index_twice<T>(slc: &mut [T], a: usize, b: usize) -> Pair<&mut T> {
if a == b {
slc.get_mut(a).map_or(Pair::None, Pair::One)
} else {
if a >= slc.len() || b >= slc.len() {
Pair::None
} else {
// safe because a, b are in bounds and distinct
unsafe {
let ar = &mut *(slc.get_unchecked_mut(a) as *mut _);
let br = &mut *(slc.get_unchecked_mut(b) as *mut _);
Pair::Both(ar, br)
}
}
}
}
Since Rust 1.26, pattern matching can be done on slices. You can use that as long as you don't have huge indices and your indices are known at compile-time.
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut arr = [5, 6, 7, 8];
{
let [ref mut a, _, ref mut b, ..] = arr;
change(a, b);
}
assert_eq!(arr, [7, 6, 5, 8]);
}
The borrow rules of Rust need to be checked at compilation time, that is why something like mutably borrowing a part of a Vec is a very hard problem to solve (if not impossible), and why it is not possible with Rust.
Thus, when you do something like &mut v[i], it will mutably borrow the entire vector.
Imagine I did something like
let guard = something(&mut v[i]);
do_something_else(&mut v[j]);
guard.do_job();
Here, I create an object guard that internally stores a mutable reference to v[i], and will do something with it when I call do_job().
In the meantime, I did something that changed v[j]. guard holds a mutable reference that is supposed to guarantee nothing else can modify v[i]. In this case, all is good, as long as i is different from j; if the two values are equal it is a huge violation of the borrow rules.
As the compiler cannot guarantee that i != j, it is thus forbidden.
This was a simple example, but similar cases are legions, and are why such access mutably borrows the whole container. Plus the fact that the compiler actually does not know enough about the internals of Vec to ensure that this operation is safe even if i != j.
In your precise case, you can have a look at the swap(..) method available on Vec that does the swap you are manually implementing.
On a more generic case, you'll probably need an other container. Possibilities are wrapping all the values of your Vec into a type with interior mutability, such as Cell or RefCell, or even using a completely different container, as #llogiq suggested in his answer with par-vec.
The method [T]::iter_mut() returns an iterator that can yield a mutable reference for each element in the slice. Other collections have an iter_mut method too. These methods often encapsulate unsafe code, but their interface is totally safe.
Here's a general purpose extension trait that adds a method on slices that returns mutable references to two distinct items by index:
pub trait SliceExt {
type Item;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item);
}
impl<T> SliceExt for [T] {
type Item = T;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item) {
match index0.cmp(&index1) {
Ordering::Less => {
let mut iter = self.iter_mut();
let item0 = iter.nth(index0).unwrap();
let item1 = iter.nth(index1 - index0 - 1).unwrap();
(item0, item1)
}
Ordering::Equal => panic!("[T]::get_two_mut(): received same index twice ({})", index0),
Ordering::Greater => {
let mut iter = self.iter_mut();
let item1 = iter.nth(index1).unwrap();
let item0 = iter.nth(index0 - index1 - 1).unwrap();
(item0, item1)
}
}
}
}
On recent nightlies, there is get_many_mut():
#![feature(get_many_mut)]
fn main() {
let mut v = vec![1, 2, 3];
let [a, b] = v
.get_many_mut([0, 1])
.expect("out of bounds or overlapping indices");
change(a, b);
}
Building up on the answer by #bluss you can use split_at_mut() to create a function that can turn mutable borrow of a vector into a vector of mutable borrows of vector elements:
fn borrow_mut_elementwise<'a, T>(v:&'a mut Vec<T>) -> Vec<&'a mut T> {
let mut result:Vec<&mut T> = Vec::new();
let mut current: &mut [T];
let mut rest = &mut v[..];
while rest.len() > 0 {
(current, rest) = rest.split_at_mut(1);
result.push(&mut current[0]);
}
result
}
Then you can use it to get a binding that lets you mutate many items of original Vec at once, even while you are iterating over them (if you access them by index in your loop, not through any iterator):
let mut items = vec![1,2,3];
let mut items_mut = borrow_mut_elementwise(&mut items);
for i in 1..items_mut.len() {
*items_mut[i-1] = *items_mut[i];
}
println!("{:?}", items); // [2, 3, 3]
The problem is that &mut v[…] first mutably borrows v and then gives the mutable reference to the element to the change-function.
This reddit comment has a solution to your problem.
Edit: Thanks for the heads-up, Shepmaster. par-vec is a library that allows to mutably borrow disjunct partitions of a vec.
I publish my daily utils for this to crate.io. Link to the doc.
You may use it like
use arref::array_mut_ref;
let mut arr = vec![1, 2, 3, 4];
let (a, b) = array_mut_ref!(&mut arr, [1, 2]);
assert_eq!(*a, 2);
assert_eq!(*b, 3);
let (a, b, c) = array_mut_ref!(&mut arr, [1, 2, 0]);
assert_eq!(*c, 1);
// ⚠️ The following code will panic. Because we borrow the same element twice.
// let (a, b) = array_mut_ref!(&mut arr, [1, 1]);
It's a simple wrapper around the following code, which is sound. But it requires that the two indexes are different at runtime.
pub fn array_mut_ref<T>(arr: &mut [T], a0: usize, a1: usize) -> (&mut T, &mut T) {
assert!(a0 != a1);
// SAFETY: this is safe because we know a0 != a1
unsafe {
(
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
)
}
}
Alternatively, you may use a method that won't panic with mut_twice
#[inline]
pub fn mut_twice<T>(arr: &mut [T], a0: usize, a1: usize) -> Result<(&mut T, &mut T), &mut T> {
if a0 == a1 {
Err(&mut arr[a0])
} else {
unsafe {
Ok((
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
))
}
}
}
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut v = vec![1, 2, 3];
change(&mut v[0], &mut v[1]);
}
When I compile the code above, it has the error:
error[E0499]: cannot borrow `v` as mutable more than once at a time
--> src/main.rs:9:32
|
9 | change(&mut v[0], &mut v[1]);
| - ^ - first borrow ends here
| | |
| | second mutable borrow occurs here
| first mutable borrow occurs here
Why does the compiler prohibit it? v[0] and v[1] occupy different memory positions, so it's not dangerous to use these together. And what should I do if I come across this problem?
You can solve this with split_at_mut():
let mut v = vec![1, 2, 3];
let (a, b) = v.split_at_mut(1); // Returns (&mut [1], &mut [2, 3])
change(&mut a[0], &mut b[0]);
There are uncountably many safe things to do that the compiler unfortunately does not recognize yet. split_at_mut() is just like that, a safe abstraction implemented with an unsafe block internally.
We can do that too, for this problem. The following is something I use in code where I need to separate all three cases anyway (I: Index out of bounds, II: Indices equal, III: Separate indices).
enum Pair<T> {
Both(T, T),
One(T),
None,
}
fn index_twice<T>(slc: &mut [T], a: usize, b: usize) -> Pair<&mut T> {
if a == b {
slc.get_mut(a).map_or(Pair::None, Pair::One)
} else {
if a >= slc.len() || b >= slc.len() {
Pair::None
} else {
// safe because a, b are in bounds and distinct
unsafe {
let ar = &mut *(slc.get_unchecked_mut(a) as *mut _);
let br = &mut *(slc.get_unchecked_mut(b) as *mut _);
Pair::Both(ar, br)
}
}
}
}
Since Rust 1.26, pattern matching can be done on slices. You can use that as long as you don't have huge indices and your indices are known at compile-time.
fn change(a: &mut i32, b: &mut i32) {
let c = *a;
*a = *b;
*b = c;
}
fn main() {
let mut arr = [5, 6, 7, 8];
{
let [ref mut a, _, ref mut b, ..] = arr;
change(a, b);
}
assert_eq!(arr, [7, 6, 5, 8]);
}
The borrow rules of Rust need to be checked at compilation time, that is why something like mutably borrowing a part of a Vec is a very hard problem to solve (if not impossible), and why it is not possible with Rust.
Thus, when you do something like &mut v[i], it will mutably borrow the entire vector.
Imagine I did something like
let guard = something(&mut v[i]);
do_something_else(&mut v[j]);
guard.do_job();
Here, I create an object guard that internally stores a mutable reference to v[i], and will do something with it when I call do_job().
In the meantime, I did something that changed v[j]. guard holds a mutable reference that is supposed to guarantee nothing else can modify v[i]. In this case, all is good, as long as i is different from j; if the two values are equal it is a huge violation of the borrow rules.
As the compiler cannot guarantee that i != j, it is thus forbidden.
This was a simple example, but similar cases are legions, and are why such access mutably borrows the whole container. Plus the fact that the compiler actually does not know enough about the internals of Vec to ensure that this operation is safe even if i != j.
In your precise case, you can have a look at the swap(..) method available on Vec that does the swap you are manually implementing.
On a more generic case, you'll probably need an other container. Possibilities are wrapping all the values of your Vec into a type with interior mutability, such as Cell or RefCell, or even using a completely different container, as #llogiq suggested in his answer with par-vec.
The method [T]::iter_mut() returns an iterator that can yield a mutable reference for each element in the slice. Other collections have an iter_mut method too. These methods often encapsulate unsafe code, but their interface is totally safe.
Here's a general purpose extension trait that adds a method on slices that returns mutable references to two distinct items by index:
pub trait SliceExt {
type Item;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item);
}
impl<T> SliceExt for [T] {
type Item = T;
fn get_two_mut(&mut self, index0: usize, index1: usize) -> (&mut Self::Item, &mut Self::Item) {
match index0.cmp(&index1) {
Ordering::Less => {
let mut iter = self.iter_mut();
let item0 = iter.nth(index0).unwrap();
let item1 = iter.nth(index1 - index0 - 1).unwrap();
(item0, item1)
}
Ordering::Equal => panic!("[T]::get_two_mut(): received same index twice ({})", index0),
Ordering::Greater => {
let mut iter = self.iter_mut();
let item1 = iter.nth(index1).unwrap();
let item0 = iter.nth(index0 - index1 - 1).unwrap();
(item0, item1)
}
}
}
}
On recent nightlies, there is get_many_mut():
#![feature(get_many_mut)]
fn main() {
let mut v = vec![1, 2, 3];
let [a, b] = v
.get_many_mut([0, 1])
.expect("out of bounds or overlapping indices");
change(a, b);
}
Building up on the answer by #bluss you can use split_at_mut() to create a function that can turn mutable borrow of a vector into a vector of mutable borrows of vector elements:
fn borrow_mut_elementwise<'a, T>(v:&'a mut Vec<T>) -> Vec<&'a mut T> {
let mut result:Vec<&mut T> = Vec::new();
let mut current: &mut [T];
let mut rest = &mut v[..];
while rest.len() > 0 {
(current, rest) = rest.split_at_mut(1);
result.push(&mut current[0]);
}
result
}
Then you can use it to get a binding that lets you mutate many items of original Vec at once, even while you are iterating over them (if you access them by index in your loop, not through any iterator):
let mut items = vec![1,2,3];
let mut items_mut = borrow_mut_elementwise(&mut items);
for i in 1..items_mut.len() {
*items_mut[i-1] = *items_mut[i];
}
println!("{:?}", items); // [2, 3, 3]
The problem is that &mut v[…] first mutably borrows v and then gives the mutable reference to the element to the change-function.
This reddit comment has a solution to your problem.
Edit: Thanks for the heads-up, Shepmaster. par-vec is a library that allows to mutably borrow disjunct partitions of a vec.
I publish my daily utils for this to crate.io. Link to the doc.
You may use it like
use arref::array_mut_ref;
let mut arr = vec![1, 2, 3, 4];
let (a, b) = array_mut_ref!(&mut arr, [1, 2]);
assert_eq!(*a, 2);
assert_eq!(*b, 3);
let (a, b, c) = array_mut_ref!(&mut arr, [1, 2, 0]);
assert_eq!(*c, 1);
// ⚠️ The following code will panic. Because we borrow the same element twice.
// let (a, b) = array_mut_ref!(&mut arr, [1, 1]);
It's a simple wrapper around the following code, which is sound. But it requires that the two indexes are different at runtime.
pub fn array_mut_ref<T>(arr: &mut [T], a0: usize, a1: usize) -> (&mut T, &mut T) {
assert!(a0 != a1);
// SAFETY: this is safe because we know a0 != a1
unsafe {
(
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
)
}
}
Alternatively, you may use a method that won't panic with mut_twice
#[inline]
pub fn mut_twice<T>(arr: &mut [T], a0: usize, a1: usize) -> Result<(&mut T, &mut T), &mut T> {
if a0 == a1 {
Err(&mut arr[a0])
} else {
unsafe {
Ok((
&mut *(&mut arr[a0] as *mut _),
&mut *(&mut arr[a1] as *mut _),
))
}
}
}
My code looks like this:
fn swap<T>(mut collection: Vec<T>, a: usize, b: usize) {
let temp = collection[a];
collection[a] = collection[b];
collection[b] = temp;
}
Rust is pretty sure I'm not allowed to "move out of dereference" or "move out of indexed content", whatever that is. How do I convince Rust that this is possible?
There is a swap method defined for &mut [T]. Since a Vec<T> can be mutably dereferenced as a &mut [T], this method can be called directly:
fn main() {
let mut numbers = vec![1, 2, 3];
println!("before = {:?}", numbers);
numbers.swap(0, 2);
println!("after = {:?}", numbers);
}
To implement this yourself, you have to write some unsafe code. Vec::swap is implemented like this:
fn swap(&mut self, a: usize, b: usize) {
unsafe {
// Can't take two mutable loans from one vector, so instead just cast
// them to their raw pointers to do the swap
let pa: *mut T = &mut self[a];
let pb: *mut T = &mut self[b];
ptr::swap(pa, pb);
}
}
It takes two raw pointers from the vector and uses ptr::swap to swap them safely.
There is also a mem::swap(&mut T, &mut T) when you need to swap two distinct variables. That cannot be used here because Rust won't allow taking two mutable borrows from the same vector.
As mentioned by #gsingh2011, the accepted answer is no longer good approach. With current Rust this code works fine:
fn main() {
let mut numbers = vec![1, 2, 3];
println!("before = {:?}", numbers);
numbers.swap(0, 2);
println!("after = {:?}", numbers);
}
try it here