I am on my way of learning Rust. In practical purpose I decided to make my own linked list collection. But I faced with some problems soon. I tried but can't find any way to fix this problem. Can I do this using this type of structure for linked list? The main problem I faced is implementation of deleting function.This function should delete element from the list with replacing address of the current node with address of the next node. But I can't change the value because it was borrowed already.
#[derive(Debug, PartialEq)]
enum AlmostList {
Cons(i32, Box<AlmostList>),
Nil,
}
The base of linked list is enumeration.
#[derive(Debug, PartialEq)]
struct List {
length: u32,
data: AlmostList,
}
But I want the instance of list knows its own length. Because of it I decided to use structure to store the length of the list in it.
use crate::AlmostList::{Cons, Nil};
//Implementetion of functions
impl List {
fn new() -> List {
List{ length: 0_u32, data: AlmostList::Nil }
}
fn append(&mut self, elem: i32) {
let mut alist = &mut self.data;
loop {
match alist {
Cons(_, ref mut ptr) => alist = ptr,
Nil => {
let node = Cons(elem, Box::new(Nil));
self.length += 1;
*alist = node;
break;
},
}
}
}
fn show(&self) {
let mut alist = &self.data;
if let Nil = alist {
panic!("List is empty");
}
loop {
match alist {
Cons(value, ptr) => {
print!("{} ", value);
alist = &*ptr;
},
Nil => break,
}
}
}
}
Everything was fine until I started to write function of deleting elements.This function should delete element from the list with replacing address of the current node with address of the next node.
fn delete(&mut self, number: i32) -> Option<i32> {
let mut alist = &mut self.data;
let result = loop {
match alist {
Cons(value, ptr) => {
if *value != number {
alist = ptr;
}
self.length -= 1;
*alist = *ptr.clone();
break Some(number);
},
Nil => break None,
}
};
result
}
}
Got this error:
error[E0506]: cannot assign to `*alist` because it is borrowed
--> main.rs:79:25
|
73 | Cons(value, ptr) => {
| --- borrow of `*alist` occurs here
...
79 | *alist = *ptr.clone();
| ^^^^^^
| |
| assignment to borrowed `*alist` occurs here
| borrow later used here
error[E0502]: cannot borrow `*ptr` as immutable because it is also borrowed as mutable
--> main.rs:79:35
|
75 | alist = ptr;
| --- mutable borrow occurs here
...
79 | *alist = *ptr.clone();
| ------ ^^^ immutable borrow occurs here
| |
| mutable borrow later used here
error: aborting due to 2 previous errors
Some errors have detailed explanations: E0502, E0506.
For more information about an error, try `rustc --explain E0502`.
Can I solve this problem somehow?
Yes, you can solve this problem. I fixed it in that way:
pub fn delete(&mut self, number: i32) -> Option<i32> {
let mut alist = &mut self.data;
let result = loop {
match alist {
AlmostList::Cons(value, ref mut ptr) => {
if *value != number {
alist = ptr;
} else {
break Some(ptr);
}
}
AlmostList::Nil => break None,
}
};
if let Some(ptr) = result {
*ptr = Box::new(AlmostList::Nil);
Some(number)
} else {
None
}
}
At least the code above was compiled successfully. Full rust playground example: https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=5bd28014bdec30e5c2e851827f15dad5
Additional hints:
write better namings. Examples: AlmostLeast -> ListNode (or just Node), Nil -> None, Cons -> Some. I have feeling that you have experience with Lisp in the past :)
format your code with cargo fmt and check with cargo clippy
any data structure implementations should not panic. Use Result type
Update:
Code above is wrong. I've implemented it using recursion: https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=ea6dbd3a627a59014df823be29e296b2
Related
I have the following code:
use std::collections::HashMap;
type Elem = i64;
enum Op {
Num(Elem),
Add((String, String)),
}
type Statements = HashMap<String, Op>;
fn main() {
let mut statements = Statements::new();
statements.insert("lhs".to_string(), Op::Num(64));
statements.insert("op".to_string(), Op::Add(("lhs".to_string(), "rhs".to_string())));
statements.insert("rhs".to_string(), Op::Num(64));
let mut stack = Vec::new();
let mut op_name = "op";
loop {
let result: Elem;
match statements.get(op_name).unwrap() {
Op::Num(_) => {
panic!("{}: Did not expect Num", op_name);
},
Op::Add((lhs, rhs)) => {
if let Op::Num(lhs_value) = statements.get(lhs).unwrap() {
if let Op::Num(rhs_value) = statements.get(rhs).unwrap() {
result = lhs_value + rhs_value;
} else {
stack.push(op_name);
op_name = rhs;
continue;
}
} else {
stack.push(op_name);
op_name = lhs;
continue;
}
},
};
if let Some(new_op_name) = stack.pop() {
*(statements.get_mut(op_name).unwrap()) = Op::Num(result);
op_name = new_op_name;
} else {
println!("Result: {}", result);
return;
}
}
}
which gives me the following compiler error:
error[E0502]: cannot borrow `statements` as mutable because it is also borrowed as immutable
--> minimal.rs:43:15
|
22 | match statements.get(op_name).unwrap() {
| ----------------------- immutable borrow occurs here
...
43 | *(statements.get_mut(op_name).unwrap()) = Op::Num(result);
| ^^^^^^^^^^^-------^^^^^^^^^
| | |
| | immutable borrow later used by call
| mutable borrow occurs here
error: aborting due to previous error
For more information about this error, try `rustc --explain E0502`.
I was expecting the borrowing to end with the match statement, since I do not see how I could still access the result of the borrowing after this statement. Why is the value still being borrowed after the match statement?
Is there something I could do to make my code above work?
In my opinion you'll find it difficult to work with strings and strings references. I would rather switch to some type for your HashMap keys that would be small, and copyable, like an integer number. This way you'll not be required to deal with references and mutability - you can just copy your integers with little to no cost.
I resolved your compilation errors, but the overall code becomes more convoluted. Basically now each insert into the hash map will give you a unique integer ID of the operation. You can only query operations from the hash map using this provided ID. It also requires you to push statements in your hash map in the order of appearance. You can't insert add(lhs, rhs) unless you already know IDs of both lhs and rhs
use std::collections::HashMap;
type Elem = i64;
type StatementId = usize;
enum Op {
Num(Elem),
Add((StatementId, StatementId)),
}
#[derive(Default)]
struct StatementsRecord {
record: HashMap<StatementId, (String, Op)>,
next_statement_id: StatementId,
}
impl StatementsRecord {
fn insert(&mut self, name: String, operation: Op) -> StatementId {
self.record.insert(self.next_statement_id, (name, operation));
self.next_statement_id += 1;
self.next_statement_id - 1 // return id of newly stored operation
}
fn get(&self, id: &StatementId) -> Option<&(String, Op)> {
self.record.get(id)
}
fn get_mut(&mut self, id: &StatementId) -> Option<&mut (String, Op)> {
self.record.get_mut(id)
}
fn update(&mut self, id: &StatementId, value: Op) {
self.get_mut(id).unwrap().1 = value;
}
}
fn main() {
let mut statements = StatementsRecord::default();
let lhs_id = statements.insert("lhs".to_string(), Op::Num(64));
let rhs_id = statements.insert("rhs".to_string(), Op::Num(64));
// 64+64 = 128
let mut op_id = statements.insert("op".to_string(), Op::Add((lhs_id, rhs_id)));
// (64+64) + (64+64) = 256
let mut op_id = statements.insert("op".to_string(), Op::Add((op_id, op_id)));
let mut stack = Vec::new();
loop {
let result: Elem;
match statements.get(&op_id).unwrap() {
(op_name, Op::Num(_)) => {
panic!("{}: Did not expect Num", op_name);
},
(op_name, Op::Add((lhs, rhs))) => {
if let (_, Op::Num(lhs_value)) = statements.get(lhs).unwrap() {
if let (_, Op::Num(rhs_value)) = statements.get(rhs).unwrap() {
result = lhs_value + rhs_value;
} else {
stack.push(op_id);
op_id = *rhs;
continue;
}
} else {
stack.push(op_id);
op_id = *lhs;
continue;
}
},
};
if let Some(new_op_id) = stack.pop() {
statements.update(&op_id, Op::Num(result));
op_id = new_op_id;
} else {
println!("Result: {}", result);
return;
}
}
}
Prints
Result: 256
I'm trying to learn Rust after many years of C++. I have a situation where the compiler is complaining about a borrow, and it doesn't seem to matter whether it is mutable or immutable. I don't seem to be able to use self as a parameter inside a loop that start with: for item in self.func.drain(..).I've tried calling appropriate() as a function:
Self::appropriate(&self,&item,index)
and I have tried it as a method:
self.appropriate(&item,index)
but I get the same message in either case:
The function or method appropriate() is intended imply examine the relationship among its parameters and return a bool without modifying anything. How can I call either a function or method on self without violating borrowing rules?This program is a learning exercise from exercism.org and doesn't include a main() so it won't run but should almost compile except for the error in question. Here's the code I have:
use std::collections::HashMap;
pub type Value = i32;
pub type Result = std::result::Result<(), Error>;
pub struct Forth {
v: Vec<Value>,
f: HashMap<String,usize>,
s: Vec<Vec<String>>,
func: Vec<String>
}
#[derive(Debug, PartialEq)]
pub enum Error {
DivisionByZero,
StackUnderflow,
UnknownWord,
InvalidWord,
}
impl Forth {
pub fn new() -> Forth {
let mut temp: Vec<Vec<String>> = Vec::new();
temp.push(Vec::new());
Forth{v: Vec::<Value>::new(), f: HashMap::new(), s: temp, func: Vec::new()}
}
pub fn stack(&self) -> &[Value] {
&self.v
}
pub fn eval(&mut self, input: &str) -> Result {
self.v.clear();
self.s[0].clear();
let mut count = 0;
{
let temp: Vec<&str> = input.split(' ').collect();
let n = temp.len() as i32;
for x in 0..n as usize {
self.s[0].push(String::from(temp[x]));
}
}
let mut collecting = false;
let mut xlist: Vec<(usize,usize)> = Vec::new();
let mut sx: usize = 0;
let mut z: i32 = -1;
let mut x: usize;
let mut n: usize = self.s[0].len();
loop {
count += 1;
if count > 20 {break;}
z += 1;
x = z as usize;
if x >= n {break;}
z = x as i32;
let word = &self.s[sx][x];
if word == ";" {
if collecting {
collecting = false;
let index: usize = self.s.len();
self.s.push(Vec::<String>::new());
for item in self.func.drain(..) {
if self.s[index].len() > 0 &&
Self::appropriate(&self,&item,index)
{
let sx = *self.f.get(&self.s[index][0]).unwrap();
let n = self.s[sx].len();
for x in 1..n as usize {
let symbol = self.s[sx][x].clone();
self.s[index].push(symbol);
}
}
else {
self.s[index].push(item);
}
}
self.f.insert(self.s[index][0].clone(), index);
self.func.clear();
continue;
}
if 0 < xlist.len() {
(x, n) = xlist.pop().unwrap();
continue;
}
return Err(Error::InvalidWord);
}
if collecting {
self.func.push(String::from(word));
continue;
}
if Self::is_op(word) {
if self.v.len() < 2 {
return Err(Error::StackUnderflow);
}
let b = self.v.pop().unwrap();
let a = self.v.pop().unwrap();
let c = match word.as_str() {
"+" => a + b,
"-" => a - b,
"*" => a * b,
"/" => {if b == 0 {return Err(Error::DivisionByZero);} a / b},
_ => 0
};
self.v.push(c);
continue;
}
match word.parse::<Value>() {
Ok(value) => { self.v.push(value); continue;},
_ => {}
}
if word == ":" {
collecting = true;
self.func.clear();
continue;
}
if word == "drop" {
if self.v.len() < 1 {
return Err(Error::StackUnderflow);
}
self.v.pop();
continue;
}
if word == "dup" {
if self.v.len() < 1 {
return Err(Error::StackUnderflow);
}
let temp = self.v[self.v.len() - 1];
self.v.push(temp);
continue;
}
if !self.f.contains_key(word) {
return Err(Error::UnknownWord);
}
xlist.push((sx,n));
sx = *self.f.get(word).unwrap();
n = self.s[sx].len();
z = 0;
}
Ok(())
}
fn is_op(input: &str) -> bool {
match input {"+"|"-"|"*"|"/" => true, _ => false}
}
fn appropriate(&self, item:&str, index:usize) -> bool
{
false
}
fn prev_def_is_short(&self, index: usize) -> bool {
if index >= self.s.len() {
false
}
else {
if let Some(&sx) = self.f.get(&self.func[0]) {
self.s[sx].len() == 2
}
else {
false
}
}
}
}
The error message relates to the call to appropriate(). I haven't even written the body of that function yet; I'd like to get the parameters right first. The compiler's complaint is:
As a subroutine call
error[E0502]: cannot borrow `self` as immutable because it is also borrowed as mutable
--> src/lib.rs:85:47
|
81 | for item in self.func.drain(..) {
| -------------------
| |
| mutable borrow occurs here
| mutable borrow later used here
...
85 | Self::appropriate(&self,&item,index)
| ^^^^^ immutable borrow occurs here
For more information about this error, try `rustc --explain E0502`.
as a method call
error[E0502]: cannot borrow `*self` as immutable because it is also borrowed as mutable
--> src/lib.rs:85:29
|
81 | for item in self.func.drain(..) {
| -------------------
| |
| mutable borrow occurs here
| mutable borrow later used here
...
85 | self.appropriate(&item,index)
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ immutable borrow occurs here
For more information about this error, try `rustc --explain E0502`.
Is there any canonical way to deal with this situation?
The problem is that self.func.drain() will consume the elements contained in self.func, thus an exclusive (&mut) access is needed on self.func for the entire for loop.
If during the iteration you need to pass a reference to self globally, then its func member is potentially accessible while the loop holds an exclusive access to it: Rust forbids that.
Since you use drain() in order to consume all the elements inside self.func, I suggest you swap this vector with an empty one just before the loop, then iterate on this other vector that is not anymore part of self.
No copy of the content of the vector is involved here; swap() only deals with pointers.
Here is an over-simplified version of your code, adapted consequently.
struct Forth {
func: Vec<String>,
}
impl Forth {
fn eval(&mut self) {
/*
for item in self.func.drain(..) {
self.appropriate(&self);
}
*/
let mut func = Vec::new();
std::mem::swap(&mut self.func, &mut func);
for item in func.drain(..) {
let b = self.appropriate();
println!("{:?} {:?}", item, b);
}
}
fn appropriate(&self) -> bool {
false
}
}
fn main() {
let mut f = Forth {
func: vec!["aaa".into(), "bbb".into()],
};
f.eval();
}
I've never understood why I have received the Rust error "cannot move out of borrowed content".
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
use std::vec::Vec;
pub struct user_type {
pub name: String,
pub ilist: Vec<i32>,
pub user_type_list: VecDeque<Option<Rc<RefCell<user_type>>>>,
pub parent: Option<Rc<RefCell<user_type>>>,
}
impl user_type {
pub fn new(name: String) -> Self {
user_type {
name: name.clone(),
ilist: Vec::new(),
user_type_list: VecDeque::new(),
parent: Option::None,
}
}
pub fn to_string(&self) -> String {
let mut result: String = String::new();
result += "name is ";
result += &self.name;
let n = self.user_type_list.len();
for iter in &self.user_type_list {
match iter {
Some(ref x) => {
let temp = x.into_inner();
let temp2 = temp.to_string();
result += &temp2[..];
}
None => panic!("to_string"),
}
result += "\n";
}
result
}
}
The full error message is:
error[E0507]: cannot move out of borrowed content
--> src/main.rs:34:32
|
34 | let temp = x.into_inner();
| ^ cannot move out of borrowed content
What is the origin of this kind of error?
Look carefully at this code:
for iter in &self.user_type_list {
match iter {
Some(ref x) => {
let temp = x.into_inner();
let temp2 = temp.to_string();
result += &temp2[..];
}
None => panic!("to_string"),
}
result += "\n";
}
Here, you are iterating &self.user_type_list so the type of iter is actually a reference to the contained value: &Option<Rc<RefCell<user_type>>>. That is nice, because you do not want to take ownership of the container or its values.
Then you match iter to Some(ref x). Older compiler versions would fail because you are matching a reference to a non-reference, but new compilers will do as if you are matching a Option<&T> instead of a &Option<T>, if needed. That is handy, and means that you can write just Some(x) => and x will be of type &Rc<RefCell<user_type>> instead of &&Rc<..> (not that it really matters, automatic dereferencing will make those equivalent).
Now you are calling x.into_inner() with a &Rc<RefCell<..>> and that will never work. It looks like you want to get the RefCell into temp that is not needed, Rc implements Deref so you get that for free. Instead the compiler thinks you are calling RefCell::into_inner(self) -> T, but this function consumes the self to get to the contained value. And you do not own it, you just borrowed it. That is what the error message means: you are trying to consume (move out) and object you do not own (borrowd).
What you really want is just to borrow the user_type enough to call to_string():
Some(x) => {
let temp = x.borrow().to_string();
result += &temp;
}
I'm trying to recurse down a structure of nodes, modifying them, and then returning the last Node that I get to. I solved the problems with mutable references in the loop using an example in the non-lexical lifetimes RFC. If I try to return the mutable reference to the last Node, I get a use of moved value error:
#[derive(Debug)]
struct Node {
children: Vec<Node>,
}
impl Node {
fn new(children: Vec<Self>) -> Self {
Self { children }
}
fn get_last(&mut self) -> Option<&mut Node> {
self.children.last_mut()
}
}
fn main() {
let mut root = Node::new(vec![Node::new(vec![])]);
let current = &mut root;
println!("Final: {:?}", get_last(current));
}
fn get_last(mut current: &mut Node) -> &mut Node {
loop {
let temp = current;
println!("{:?}", temp);
match temp.get_last() {
Some(child) => { current = child },
None => break,
}
}
current
}
Gives this error
error[E0382]: use of moved value: `*current`
--> test.rs:51:5
|
40 | let temp = current;
| ---- value moved here
...
51 | current
| ^^^^^^^ value used here after move
|
= note: move occurs because `current` has type `&mut Node`, which does not implement the `Copy` trait
If I return the temporary value instead of breaking, I get the error cannot borrow as mutable more than once.
fn get_last(mut current: &mut Node) -> &mut Node {
loop {
let temp = current;
println!("{:?}", temp);
match temp.get_last() {
Some(child) => { current = child },
None => return temp,
}
}
}
error[E0499]: cannot borrow `*temp` as mutable more than once at a time
--> test.rs:47:28
|
43 | match temp.get_last() {
| ---- first mutable borrow occurs here
...
47 | None => return temp,
| ^^^^ second mutable borrow occurs here
48 | }
49 | }
| - first borrow ends here
How can I iterate through the structure with mutable references and return the last Node? I've searched, but I haven't found any solutions for this specific problem.
I can't use Obtaining a mutable reference by iterating a recursive structure because it gives me a borrowing more than once error:
fn get_last(mut current: &mut Node) -> &mut Node {
loop {
let temp = current;
println!("{:?}", temp);
match temp.get_last() {
Some(child) => current = child,
None => current = temp,
}
}
current
}
This is indeed different from Cannot obtain a mutable reference when iterating a recursive structure: cannot borrow as mutable more than once at a time. If we look at the answer there, modified a bit, we can see that it matches on a value and is able to return the value that was matched on in the terminal case. That is, the return value is an Option:
fn back(&mut self) -> &mut Option<Box<Node>> {
let mut anchor = &mut self.root;
loop {
match {anchor} {
&mut Some(ref mut node) => anchor = &mut node.next,
other => return other, // transferred ownership to here
}
}
}
Your case is complicated by two aspects:
The lack of non-lexical lifetimes.
The fact that you want to take a mutable reference and "give it up" in one case (there are children) and not in the other (no children). This is conceptually the same as this:
fn maybe_identity<T>(_: T) -> Option<T> { None }
fn main() {
let name = String::from("vivian");
match maybe_identity(name) {
Some(x) => println!("{}", x),
None => println!("{}", name),
}
}
The compiler cannot tell that the None case could (very theoretically) continue to use name.
The straight-forward solution is to encode this "get it back" action explicitly. We create an enum that returns the &mut self in the case of no children, a helper method that returns that enum, and rewrite the primary method to use the helper:
enum LastOrNot<'a> {
Last(&'a mut Node),
NotLast(&'a mut Node),
}
impl Node {
fn get_last_or_self(&mut self) -> LastOrNot<'_> {
match self.children.is_empty() {
false => LastOrNot::Last(self.children.last_mut().unwrap()),
true => LastOrNot::NotLast(self),
}
}
fn get_last(mut current: &mut Node) -> &mut Node {
loop {
match { current }.get_last_or_self() {
LastOrNot::Last(child) => current = child,
LastOrNot::NotLast(end) => return end,
}
}
}
}
Note that we are using all of the techniques exposed in both Returning a reference from a HashMap or Vec causes a borrow to last beyond the scope it's in? and Cannot obtain a mutable reference when iterating a recursive structure: cannot borrow as mutable more than once at a time.
With an in-progress reimplementation of NLL, we can simplify get_last_or_self a bit to avoid the boolean:
fn get_last_or_self(&mut self) -> LastOrNot<'_> {
match self.children.last_mut() {
Some(l) => LastOrNot::Last(l),
None => LastOrNot::NotLast(self),
}
}
The final version of Polonius should allow reducing the entire problem to a very simple form:
fn get_last(mut current: &mut Node) -> &mut Node {
while let Some(child) = current.get_last() {
current = child;
}
current
}
See also:
Returning a reference from a HashMap or Vec causes a borrow to last beyond the scope it's in?
Cannot obtain a mutable reference when iterating a recursive structure: cannot borrow as mutable more than once at a time
I'm trying to implement a function that steps though two iterators at the same time, calling a function for each pair. This callback can control which of the iterators is advanced in each step by returning a (bool, bool) tuple. Since the iterators take a reference to a buffer in my use case, they can't implement the Iterator trait from the stdlib, but instead are used though a next_ref function, which is identical to Iterator::next, but takes an additional lifetime parameter.
// An iterator-like type, that returns references to itself
// in next_ref
struct RefIter {
value: u64
}
impl RefIter {
fn next_ref<'a>(&'a mut self) -> Option<&'a u64> {
self.value += 1;
Some(&self.value)
}
}
// Iterate over two RefIter simultaneously and call a callback
// for each pair. The callback returns a tuple of bools
// that indicate which iterators should be advanced.
fn each_zipped<F>(mut iter1: RefIter, mut iter2: RefIter, callback: F)
where F: Fn(&Option<&u64>, &Option<&u64>) -> (bool, bool)
{
let mut current1 = iter1.next_ref();
let mut current2 = iter2.next_ref();
loop {
let advance_flags = callback(¤t1, ¤t2);
match advance_flags {
(true, true) => {
current1 = iter1.next_ref();
current2 = iter2.next_ref();
},
(true, false) => {
current1 = iter1.next_ref();
},
(false, true) => {
current2 = iter1.next_ref();
},
(false, false) => {
return
}
}
}
}
fn main() {
let mut iter1 = RefIter { value: 3 };
let mut iter2 = RefIter { value: 4 };
each_zipped(iter1, iter2, |val1, val2| {
let val1 = *val1.unwrap();
let val2 = *val2.unwrap();
println!("{}, {}", val1, val2);
(val1 < 10, val2 < 10)
});
}
error[E0499]: cannot borrow `iter1` as mutable more than once at a time
--> src/main.rs:28:28
|
22 | let mut current1 = iter1.next_ref();
| ----- first mutable borrow occurs here
...
28 | current1 = iter1.next_ref();
| ^^^^^ second mutable borrow occurs here
...
42 | }
| - first borrow ends here
error[E0499]: cannot borrow `iter2` as mutable more than once at a time
--> src/main.rs:29:28
|
23 | let mut current2 = iter2.next_ref();
| ----- first mutable borrow occurs here
...
29 | current2 = iter2.next_ref();
| ^^^^^ second mutable borrow occurs here
...
42 | }
| - first borrow ends here
error[E0499]: cannot borrow `iter1` as mutable more than once at a time
--> src/main.rs:32:28
|
22 | let mut current1 = iter1.next_ref();
| ----- first mutable borrow occurs here
...
32 | current1 = iter1.next_ref();
| ^^^^^ second mutable borrow occurs here
...
42 | }
| - first borrow ends here
error[E0499]: cannot borrow `iter1` as mutable more than once at a time
--> src/main.rs:35:28
|
22 | let mut current1 = iter1.next_ref();
| ----- first mutable borrow occurs here
...
35 | current2 = iter1.next_ref();
| ^^^^^ second mutable borrow occurs here
...
42 | }
| - first borrow ends here
I understand why it complains, but can't find a way around it. I'd appreciate any help on the subject.
Link to this snippet in the playground.
Since the iterators take a reference to a buffer in my use case, they can't implement the Iterator trait from the stdlib, but instead are used though a next_ref function, which is identical to Iterator::next, but takes an additional lifetime parameter.
You are describing a streaming iterator. There is a crate for this, aptly called streaming_iterator. The documentation describes your problem (emphasis mine):
While the standard Iterator trait's functionality is based off of
the next method, StreamingIterator's functionality is based off of
a pair of methods: advance and get. This essentially splits the
logic of next in half (in fact, StreamingIterator's next method
does nothing but call advance followed by get).
This is required because of Rust's lexical handling of borrows (more
specifically a lack of single entry, multiple exit borrows). If
StreamingIterator was defined like Iterator with just a required
next method, operations like filter would be impossible to define.
The crate does not currently have a zip function, and certainly not the variant you have described. However, it's easy enough to implement:
extern crate streaming_iterator;
use streaming_iterator::StreamingIterator;
fn each_zipped<A, B, F>(mut iter1: A, mut iter2: B, callback: F)
where
A: StreamingIterator,
B: StreamingIterator,
F: for<'a> Fn(Option<&'a A::Item>, Option<&'a B::Item>) -> (bool, bool),
{
iter1.advance();
iter2.advance();
loop {
let advance_flags = callback(iter1.get(), iter2.get());
match advance_flags {
(true, true) => {
iter1.advance();
iter2.advance();
}
(true, false) => {
iter1.advance();
}
(false, true) => {
iter1.advance();
}
(false, false) => return,
}
}
}
struct RefIter {
value: u64
}
impl StreamingIterator for RefIter {
type Item = u64;
fn advance(&mut self) {
self.value += 1;
}
fn get(&self) -> Option<&Self::Item> {
Some(&self.value)
}
}
fn main() {
let iter1 = RefIter { value: 3 };
let iter2 = RefIter { value: 4 };
each_zipped(iter1, iter2, |val1, val2| {
let val1 = *val1.unwrap();
let val2 = *val2.unwrap();
println!("{}, {}", val1, val2);
(val1 < 10, val2 < 10)
});
}
The issue with this code is that RefIter is being used in two ways, which are fundamentally at odds with one another:
Callers of next_ref recieve a reference to the stored value, which is tied to the lifetime of RefIter
RefIter's value needs to be mutable, so that it can be incremented on each call
This perfectly describes mutable aliasing (you're trying to modify 'value' while a reference to it is being held) - something which Rust is explicitly designed to prevent.
In order to make each_zipped work, you'll need to modify RefIter to avoid handing out references to data that you wish to mutate.
I've implemented one possibility below using a combination of RefCell and Rc:
use std::cell::RefCell;
use std::rc::Rc;
// An iterator-like type, that returns references to itself
// in next_ref
struct RefIter {
value: RefCell<Rc<u64>>
}
impl RefIter {
fn next_ref(&self) -> Option<Rc<u64>> {
let new_val = Rc::new(**self.value.borrow() + 1);
*self.value.borrow_mut() = new_val;
Some(Rc::clone(&*self.value.borrow()))
}
}
// Iterate over two RefIter simultaniously and call a callback
// for each pair. The callback returns a tuple of bools
// that indicate which iterators should be advanced.
fn each_zipped<F>(iter1: RefIter, iter2: RefIter, callback: F)
where F: Fn(&Option<Rc<u64>>, &Option<Rc<u64>>) -> (bool, bool)
{
let mut current1 = iter1.next_ref();
let mut current2 = iter2.next_ref();
loop {
let advance_flags = callback(¤t1, ¤t2);
match advance_flags {
(true, true) => {
current1 = iter1.next_ref();
current2 = iter2.next_ref();
},
(true, false) => {
current1 = iter1.next_ref();
},
(false, true) => {
current2 = iter1.next_ref();
},
(false, false) => {
return
}
}
}
}
fn main() {
let iter1 = RefIter { value: RefCell::new(Rc::new(3)) };
let iter2 = RefIter { value: RefCell::new(Rc::new(4)) };
each_zipped(iter1, iter2, |val1, val2| {
// We can't use unwrap() directly, since we're only passed a reference to an Option
let val1 = **val1.iter().next().unwrap();
let val2 = **val2.iter().next().unwrap();
println!("{}, {}", val1, val2);
(val1 < 10, val2 < 10)
});
}
This version of RefIter hands out Rcs to consumers, instead of references. This avoids the issue of mutable aliasing - updating value is done by placing
a new Rc into the outer RefCell. A side effect of this is that consumers are able to hold onto an 'old' reference to the buffer (through the returned Rc), even after RefIter has advanced.