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I don't understand the following code written for LLVM IR. I hope, you can give me a hint.
%struct.foo_struct = type {[3 x i32], i16*, i32}
;struct foo_struct {
; [3 x i32] f0;
; i16* f1;
; i32 f2;
; };
define i32 #foo(%struct.foo_struct* %P) {
entry:
; &P[0].f1
%tmp0 = getelementptr inbounds %struct.foo_struct, %struct.foo_struct* %P, i64 0, i32 1
; P[0].f1
%tmp1 = load i16*, i16** %tmp0
; &P[0].f1[0]
%tmp2 = getelementptr inbounds i16, i16* %tmp1, i64 0
Specifically, in the first code line in entry, we have at the end i32 1. Why i32? Since we want to jump to the next field, namely f1, we have to jump over an array (f0), which 3xi32. So what is this i32? What would be there if we want to have e.g. &P[0].f2?
Thank you for any help
Since we want to jump to the next field, namely f1, we have to jump over an array (f0), which 3xi32. So what is this i32?
1 is the index of the member in the struct and i32 is the type of this index.
The fact that we're jumping over a 3xi32 to get to f1 or that we're thus jumping by 12 bytes is not directly encoded in the instruction. All we're specifying is that we want the second member (i.e. the member at index 1) and how that translates to an offset in bytes is calculated by LLVM based on the types involved. We don't spell this out in the instruction.
What would be there if we want to have e.g. &P[0].f2?
i32 2
Why i32?
Struct indices always have the type i32. Since struct indices must always be constants, there wasn't really a need to allow different types and i32 is large enough for all practical purposes (i.e. you won't have a struct with more than 2^32 members).
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 _),
))
}
}
}
I find it impossible to do a subtraction between an immutable array reference and a mutable one using the ndarray crate:
#[macro_use]
extern crate ndarray;
use ndarray::Array1;
fn main() {
let a: &Array1<f64> = &array![3.0, 2.0, 1.0];
let b: &mut Array1<f64> = &mut array![1.0, 1.0, 1.0];
let c = a - &b.view(); // This compiles
let d = a - b; // This fails to compile
}
The error message I get is:
let d = a - b;
^ no implementation for `f64 - &mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>`
I don't understand what these two types mean, but is there any special reason why this is not implemented?
The Sub trait in ndarray::ArrayBase is not implemented for &mut arguments (it is for immutable references to other arrays). It is not needed because the right-handed value should not be modified. The second operand is a &mut Array<A, D>, and it ends up being one of the cases where type weakening into an immutable reference does not happen automatically. Still, you can explicitly reborrow the value:
let a: &Array1<f64> = &array![3.0, 2.0, 1.0];
let b: &mut Array1<f64> = &mut array![1.0, 1.0, 1.0];
let c = a - &b.view();
let d = a - &*b;
This assumes that a and b were obtained elsewhere. In fact, you can make these variables own the arrays instead:
let a: Array1<f64> = array![3.0, 2.0, 1.0];
let mut b: Array1<f64> = array![1.0, 1.0, 1.0];
let c = &a - &b.view();
let d = &a - &b;
The full error message is:
error[E0277]: the trait bound `f64: std::ops::Sub<&mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>>` is not satisfied
--> src/main.rs:10:15
|
10 | let d = a - b;
| ^ no implementation for `f64 - &mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>`
|
= help: the trait `std::ops::Sub<&mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>>` is not implemented for `f64`
= note: required because of the requirements on the impl of `std::ops::Sub<&mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>>` for `&ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>`
error[E0277]: the trait bound `&mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>: ndarray::ScalarOperand` is not satisfied
--> src/main.rs:10:15
|
10 | let d = a - b;
| ^ the trait `ndarray::ScalarOperand` is not implemented for `&mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>`
|
= note: required because of the requirements on the impl of `std::ops::Sub<&mut ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>>` for `&ndarray::ArrayBase<ndarray::OwnedRepr<f64>, ndarray::Dim<[usize; 1]>>`
The first error message is quite obscure for me but the second is more enlightening:
the trait ndarray::ScalarOperand is not implemented for `&mut ndarray::ArrayBase
The reason of this error is that Rust does not perform coercions when matching traits: If there is an impl for some type U and T coerces to U, that does not constitute an implementation for T (copied directly from the Nomicon)
Without entering into ndarray internals, this is a minimal example reproducing the same problem:
trait MyShape {}
fn foo<T: MyShape>(x: T) -> T {
x
}
impl<'a> MyShape for &'a i32 {}
fn main() {
let mut num = 1;
let arg: &mut i32 = &mut num;
foo(arg);
}
Result:
error[E0277]: the trait bound `&mut i32: MyShape` is not satisfied
--> src/main.rs:12:5
|
12 | foo(arg);
| ^^^ the trait `MyShape` is not implemented for `&mut i32`
|
= help: the following implementations were found:
<&'a i32 as MyShape>
= note: required by `foo`
Why this doesn't compile:
fn main() {
let mut b = Box::new(Vec::new());
b.push(Vec::new());
b.get_mut(0).unwrap().push(1);
}
While this does:
fn main() {
let a = Box::new(Vec::new());
let mut b = *a;
b.push(Vec::new());
b.get_mut(0).unwrap().push(1);
}
And also this does:
fn main() {
let mut b = Vec::new();
b.push(Vec::new());
b.get_mut(0).unwrap().push(Vec::new());
b.get_mut(0).unwrap().get_mut(0).unwrap().push(1)
}
The first and third one for me are the conceptually the same - Box of a Vector of Vectors of integers and a Vector of Vector of Vectors of integers, but the last one results in each vector being mutable, whereas the first one makes the inner vector immutable.
In at least Rust 1.25.0, all three original examples work. This was a bug in some previous version of Rust.
You need to unbox your value before accessing it as a mutable:
fn main() {
let mut b = Box::new(Vec::new());
b.push(Vec::new());
(*b).get_mut(0).unwrap().push(1);
}
This is because the . operator uses the Deref trait instead of DerefMut.
The best way to achieve this would be:
fn main() {
let mut b = Box::new(Vec::new());
b.push(Vec::new());
b[0].push(1);
}