Conditional compilation with a if let enum matching which consists of one item - enums

I have the following enum:
pub enum Game {
Match(GameWorker),
#[cfg(feature = "cups")]
Cup(CupWorker),
}
So, this enum consists of one item if cups feature is disabled. The code below with match compiles okay but in place where I use if lets on matching this enum there is a error:
Working match:
fn clear(&mut self, silent: bool) {
match *self {
Game::Match(ref mut gm) => gm.clear(silent),
#[cfg(feature = "cups")]
Game::Cup(ref mut c) => c.clear(silent),
}
}
if let which leads to a compile error:
let m: &mut Game = Game::Match(...);
if let Game::Match(ref mut gamematch) = *m {
// ...
}
Error:
error[E0162]: irrefutable if-let pattern
--> src/game.rs:436:32
|
436 | if let Game::Match(ref mut gamematch) = *m {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ irrefutable pattern
Minimal example
Is there a way to allow such if lets ? I like this construction but somewhy it is not allowed to use it, I don't understand why. As shown above, match construction works okay in the same case. In my personal opinion here should be a silenceable warning instead of error.

if let expects a refutable pattern, similar to how if expects a bool. You can't write if () { something }, even though () is "valid" in some sense. If you had if () {} else { something_else } it would be statically known that the else cannot occur.
Arguably if true { something } is also statically known, but there's a difference: The condition is a bool, which has two values, so even if you statically know the value, the type still offers multiple variants.
With if let it's the same, but you can use user defined types instead of just bool. If your enum has multiple variants, you can't statically decide that the if let is always taken. If the enum has a single variant, you know for a fact that the if condition is always true, so even if you had an else branch, it would not make any sense at all to exist.

Related

generic callback with data

There is already a very popular question about this topic but I don;t fully understand the answer.
The goal is:
I need a list (read a Vec) of "function pointers" that modify data stored elsewhere in a program. The simplest example I can come up with are callbacks to be called when a key is pressed. So when any key is pressed, all functions passed to the object will be called in some order.
Reading the answer, it is not clear to me how I would be able to make such a list. It sounds like I would need to restrict the type of the callback to something known, else I don't know how you would be able to make an array of it.
It's also not clear to me how to store the data pointers/references.
Say I have
struct Processor<CB>
where
CB: FnMut(),
{
callback: CB,
}
Like the answer suggests, I can't make an array of processors, can I? since each Processor is technically a different type depending on the generic isntantiation.
Indeed, you can't make a vector of processors. Usually, closures all have different, innominable types. What you want instead are trait objects, which allow you to have dynamic dispatch of callback calls. Since those are not Sized, you'd probably want to put them in a Box. The final type is Vec<Box<dyn FnMut()>>.
fn add_callback(list: &mut Vec<Box<dyn FnMut()>>, cb: impl FnMut() + 'static) {
list.push(Box::new(cb))
}
fn run_callback(list: &mut [Box<dyn FnMut()>]) {
for cb in list {
cb()
}
}
see the playground
If you do like that, however, you might have some issues with the lifetimes (because your either force to move-in everything, or only modify values that life for 'static, which isn't very convenient. Instead, the following might be better
#[derive(Default)]
struct Producer<'a> {
list: Vec<Box<dyn FnMut() + 'a>>,
}
impl<'a> Producer<'a> {
fn add_callback(&mut self, cb: impl FnMut() + 'a) {
self.list.push(Box::new(cb))
}
fn run_callbacks(&mut self) {
for cb in &mut self.list {
cb()
}
}
}
fn callback_1() {
println!("Hello!");
}
fn main() {
let mut modified = 0;
let mut prod = Producer::default();
prod.add_callback(callback_1);
prod.add_callback(
|| {
modified += 1;
println!("World!");
}
);
prod.run_callbacks();
drop(prod);
println!("{}", modified);
}
see the playground
Just a few things to note:
You manually have to drop the producer, otherwise Rust will complain that it will be dropped at the end of the scope, but it contains (through the closure) an exclusive reference to modified, which is not ok since I try to read it.
Current, run_callbacks take a &mut self, because we only require for a FnMut. If you wanted it to be only a &self, you'd need to replace FnMut with Fn, which means the callbacks can still modify things outside of them, but not inside.
Yes, all closures are differents type, so if you want to have a vec of different closure you will need to make them trait objects. This can be archieve with Box<dyn Trait> (or any smart pointer). Box<dyn FnMut()> implements FnMut(), so you can have Processor<Box<dyn FnMut()>> and can make a vec of them, and call the callbacks on them: playground

Why do we use the Option enum?

I don't get what the Option enum is for. I read that Rust doesn't have null values. The Option enum is defined like this:
enum Option<T> {
Some(T),
None,
}
I read its implementation and I came across this example:
fn main() {
fn divide(numerator: f64, denominator: f64) -> Option<f64> {
if denominator == 0.0 {
None
} else {
Some(numerator / denominator)
}
}
// The return value of the function is an option
let result = divide(2.0, 3.0);
// Pattern match to retrieve the value
match result {
// The division was valid
Some(x) => println!("Result: {}", x),
// The division was invalid
None => println!("Cannot divide by 0"),
}
}
When they could also do it like this:
fn main() {
fn divide(numerator: f64, denominator: f64) -> String {
if denominator == 0.0 {
format!("Can't divide")
} else {
let x = numerator / denominator;
format!("{}", x)
}
}
let result = divide(2.0, 3.0);
println!("{}", result);
}
Both programs output:
0.6666666666666666
Maybe the above example is not a very good example of Option, but the following example shows Option at its very best:
fn main() {
let name = String::from("naufil");
println!(
"Character at index 6: {}",
match name.chars().nth(6) {
Some(c) => c.to_string(),
None => "No character at index 6!".to_string(),
}
)
}
When we are not sure whether there is a character at 6th element and you don't want your program to crash, Option comes to the rescue. Here is another example from The Rust Programming Language:
fn plus_one(x: Option<i32>) -> Option<i32> {
match x {
None => None,
Some(i) => Some(i + 1),
}
}
let five = Some(5);
let six = plus_one(five);
let none = plus_one(None);
Listing 6-5: A function that uses a match expression on
an Option<i32>
Let’s examine the first execution of plus_one in more detail. When we call
plus_one(five), the variable x in the body of plus_one will have the
value Some(5). We then compare that against each match arm.
None => None,
The Some(5) value doesn’t match the pattern None, so we continue to the
next arm.
Some(i) => Some(i + 1),
Does Some(5) match Some(i)? Why yes it does! We have the same variant. The
i binds to the value contained in Some, so i takes the value 5. The
code in the match arm is then executed, so we add 1 to the value of i and
create a new Some value with our total 6 inside.
Now let’s consider the second call of plus_one in Listing 6-5, where x is
None. We enter the match and compare to the first arm.
None => None,
It matches! There’s no value to add to, so the program stops and returns the
None value on the right side of =>. Because the first arm matched, no other
arms are compared.
Combining match and enums is useful in many situations. You’ll see this
pattern a lot in Rust code: match against an enum, bind a variable to the
data inside, and then execute code based on it. It’s a bit tricky at first, but
once you get used to it, you’ll wish you had it in all languages. It’s
consistently a user favorite.
The reason the Option enum is used for the same reason the Result enum is used. It allows the programmer to see the breadth of returning values they might receive, but without having to dig through code you don't remember all the details about, or have never seen.
Option isn't a special value, it's just an enum, like Result. You could also use something like:
enum Division_Result {
Successful(f64),
DividedByZero,
}
fn divide(numerator: f64, denominator: f64) -> Division_Result {
if denominator == 0.0 {
Division_Result::DividedByZero
} else {
Division_Result::Successful(numerator / denominator)
}
}
It just so happens that Optional values are some of the most common types of values that you have to deal with in a program. They're baked the Optional enum into standard because otherwise you would have to deal with everyone coming up with their own enum for the simple concept of an Optional value.
Returning an enum is an improvement over returning unwrapped magic values because it is more explicit to the programmer that the return value might diverge from what they wanted from the function.

Is there a way to make an immutable reference mutable?

I want to solve a leetcode question in Rust (Remove Nth Node From End of List). My solution uses two pointers to find the Node to remove:
#[derive(PartialEq, Eq, Debug)]
pub struct ListNode {
pub val: i32,
pub next: Option<Box<ListNode>>,
}
impl ListNode {
#[inline]
fn new(val: i32) -> Self {
ListNode { next: None, val }
}
}
// two-pointer sliding window
impl Solution {
pub fn remove_nth_from_end(head: Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> {
let mut dummy_head = Some(Box::new(ListNode { val: 0, next: head }));
let mut start = dummy_head.as_ref();
let mut end = dummy_head.as_ref();
for _ in 0..n {
end = end.unwrap().next.as_ref();
}
while end.as_ref().unwrap().next.is_some() {
end = end.unwrap().next.as_ref();
start = start.unwrap().next.as_ref();
}
// TODO: fix the borrow problem
// ERROR!
// start.unwrap().next = start.unwrap().next.unwrap().next.take();
dummy_head.unwrap().next
}
}
I borrow two immutable references of the linked-list. After I find the target node to remove, I want to drop one and make the other mutable. Each of the following code examples leads to a compiler error:
// ERROR
drop(end);
let next = start.as_mut().unwrap.next.take();
// ERROR
let mut node = *start.unwrap()
I don't know if this solution is possible to be written in Rust. If I can make an immutable reference mutable, how do I do it? If not, is there anyway to implement the same logic while making the borrow checker happy?
The correct answer is that you should not be doing this. This is undefined behavior, and breaks many assumptions made by the compiler when compiling your program.
However, it is possible to do this. Other people have also mentioned why this is not a good idea, but they haven't actually shown what the code to do something like this would look like. Even though you should not do this, this is what it would look like:
unsafe fn very_bad_function<T>(reference: &T) -> &mut T {
let const_ptr = reference as *const T;
let mut_ptr = const_ptr as *mut T;
&mut *mut_ptr
}
Essentially, you convert a constant pointer into a mutable one, and then make the mutable pointer into a reference.
Here's one example why this is very unsafe and unpredictable:
fn main() {
static THIS_IS_IMMUTABLE: i32 = 0;
unsafe {
let mut bad_reference = very_bad_function(&THIS_IS_IMMUTABLE);
*bad_reference = 5;
}
}
If you run this... you get a segfault. What happened? Essentially, you invalidated memory rules by trying to write to an area of memory that had been marked as immutable. Essentially, when you use a function like this, you break the trust the compiler has made with you to not mess with constant memory.
Which is why you should never use this, especially in a public API, because if someone passes an innocent immutable reference to your function, and your function mutates it, and the reference is to an area of memory not meant to be written to, you'll get a segfault.
In short: don't try to cheat the borrow checker. It's there for a reason.
EDIT: In addition to the reasons I just mentioned on why this is undefined behavior, another reason is breaking reference aliasing rules. That is, since you can have both a mutable and immutable reference to a variable at the same time with this, it causes loads of problems when you pass them in separately to the same function, which assumes the immutable and mutable references are unique. Read this page from the Rust docs for more information about this.
Is there a way to make an immutable reference mutable?
No.
You could write unsafe Rust code to force the types to line up, but the code would actually be unsafe and lead to undefined behavior. You do not want this.
For your specific problem, see:
How to remove the Nth node from the end of a linked list?
How to use two pointers to iterate a linked list in Rust?

Sharing a common value in all enum values

I have the following code where every variant of the enum Message has a Term value associated with it:
type Term = usize;
pub enum Message {
AppendRequest(Term),
AppendResponse(Term),
VoteRequest(Term),
VoteResponse(Term),
}
impl Message {
pub fn term(&self) -> Term {
match *self {
Message::AppendRequest(term) => term,
Message::AppendResponse(term) => term,
Message::VoteRequest(term) => term,
Message::VoteResponse(term) =>term,
}
}
}
I want to, given a Message be able to get its term without having to deconstruct the actual Message value I have. The best I could come up with was creating a public function that unpacked the value for me, but this feels clunky. If I ever add a new enum value, I'm going to have to remember to update match statement in the term function.
Is there a more succinct/ergonomic way to express the code above? Is there some way to say "hey, every value for this enum will have also have a Term value associated with it.
Is there some way to say "hey, every value for this enum will have also have a Term value associated with it.
No. This is usually handled by splitting the enum into two parts, with a struct containing all the common parts:
pub struct Message {
term: Term,
kind: MessageKind,
}
pub enum MessageKind {
AppendRequest,
AppendResponse,
VoteRequest,
VoteResponse,
}
One option is to implement the Deref (and/or DerefMut) trait to convert to the common part.
You still have to update that implementation each time you add to the Enum, but there is less boilerplate at the point of use.
E.g., an example below, note that main accesses the field number on the Enum.
use std::ops::Deref;
use std::string::String;
enum JudgedNumber {
GoodNumber(Number),
BadNumber(Number, String),
}
struct Number { number: i32 }
fn main() {
let nice = JudgedNumber::GoodNumber(Number{number: 42});
let naughty = JudgedNumber::BadNumber(
Number{number: 666}, "Damn you to hell".to_string());
println!("j1 = {}", j1.number);
println!("j2 = {}", j2.number);
}
impl Deref for JudgedNumber {
type Target = Number;
fn deref(&self) -> &Number {
match self {
JudgedNumber::GoodNumber(n) => n,
JudgedNumber::BadNumber(n, _) => n,
}
}
}
I learnt this from https://github.com/rust-embedded/svd/blob/master/src/svd/cluster.rs

Rust: How to specify lifetimes in closure arguments?

I'm writing a parser generator as a project to learn rust, and I'm running into something I can't figure out with lifetimes and closures. Here's my simplified case (sorry it's as complex as it is, but I need to have the custom iterator in the real version and it seems to make a difference in the compiler's behavior):
Playpen link: http://is.gd/rRm2aa
struct MyIter<'stat, T:Iterator<&'stat str>>{
source: T
}
impl<'stat, T:Iterator<&'stat str>> Iterator<&'stat str> for MyIter<'stat, T>{
fn next(&mut self) -> Option<&'stat str>{
self.source.next()
}
}
struct Scanner<'stat,T:Iterator<&'stat str>>{
input: T
}
impl<'main> Scanner<'main, MyIter<'main,::std::str::Graphemes<'main>>>{
fn scan_literal(&'main mut self) -> Option<String>{
let mut token = String::from_str("");
fn get_chunk<'scan_literal,'main>(result:&'scan_literal mut String,
input: &'main mut MyIter<'main,::std::str::Graphemes<'main>>)
-> Option<&'scan_literal mut String>{
Some(input.take_while(|&chr| chr != "\"")
.fold(result, |&mut acc, chr|{
acc.push_str(chr);
&mut acc
}))
}
get_chunk(&mut token,&mut self.input);
println!("token is {}", token);
Some(token)
}
}
fn main(){
let mut scanner = Scanner{input:MyIter{source:"\"foo\"".graphemes(true)}};
scanner.scan_literal();
}
There are two problems I know of here. First, I have to shadow the 'main lifetime in the get_chunk function (I tried using the one in the impl, but the compiler complains that 'main is undefined inside get_chunk). I think it will still work out because the call to get_chunk later will match the 'main from the impl with the 'main from get_chunk, but I'm not sure that's right.
The second problem is that the &mut acc inside the closure needs to have a lifetime of 'scan_literal in order to work like I want it to (accumulating characters until the first " is encountered for this example). I can't add an explicit lifetime to &mut acc though, and the compiler says its lifetime is limited to the closure itself, and thus I can't return the reference to use in the next iteration of fold. I've gotten the function to compile and run in various other ways, but I don't understand what the problem is here.
My main question is: Is there any way to explicitly specify the lifetime of an argument to a closure? If not, is there a better way to accumulate the string using fold without doing multiple copies?
First, about lifetimes. Functions defined inside other functions are static, they are not connected with their outside code in any way. Consequently, their lifetime parameters are completely independent. You don't want to use 'main as a lifetime parameter for get_chunk() because it will shadow the outer 'main lifetime and give nothing but confusion.
Next, about closures. This expression:
|&mut acc, chr| ...
very likely does not what you really think it does. Closure/function arguments allow irrefutable patterns in them, and & have special meaning in patterns. Namely, it dereferences the value it is matched against, and assigns its identifier to this dereferenced value:
let x: int = 10i;
let p: &int = &x;
match p {
&y => println!("{}", y) // prints 10
}
You can think of & in a pattern as an opposite to & in an expression: in an expression it means "take a reference", in a pattern it means "remove the reference".
mut, however, does not belong to & in patterns; it belongs to the identifier and means that the variable with this identifier is mutable, i.e. you should write not
|&mut acc, chr| ...
but
|& mut acc, chr| ...
You may be interested in this RFC which is exactly about this quirk in the language syntax.
It looks like that you want to do a very strange thing, I'm not sure I understand where you're getting at. It is very likely that you are confusing different string kinds. First of all, you should read the official guide which explains ownership and borrowing and when to use them (you may also want to read the unfinished ownership guide; it will soon get into the main documentation tree), and then you should read strings guide.
Anyway, your problem can be solved in much simpler and generic way:
#[deriving(Clone)]
struct MyIter<'s, T: Iterator<&'s str>> {
source: T
}
impl<'s, T: Iterator<&'s str>> Iterator<&'s str> for MyIter<'s, T>{
fn next(&mut self) -> Option<&'s str>{ // '
self.source.next()
}
}
#[deriving(Clone)]
struct Scanner<'s, T: Iterator<&'s str>> {
input: T
}
impl<'m, T: Iterator<&'m str>> Scanner<'m, T> { // '
fn scan_literal(&mut self) -> Option<String>{
fn get_chunk<'a, T: Iterator<&'a str>>(input: T) -> Option<String> {
Some(
input.take_while(|&chr| chr != "\"")
.fold(String::new(), |mut acc, chr| {
acc.push_str(chr);
acc
})
)
}
let token = get_chunk(self.input.by_ref());
println!("token is {}", token);
token
}
}
fn main(){
let mut scanner = Scanner{
input: MyIter {
source: "\"foo\"".graphemes(true)
}
};
scanner.scan_literal();
}
You don't need to pass external references into the closure; you can generate a String directly in fold() operation. I also generified your code and made it more idiomatic.
Note that now impl for Scanner also works with arbitrary iterators returning &str. It is very likely that you want to write this instead of specializing Scanner to work only with MyIter with Graphemes inside it. by_ref() operation turns &mut I where I is an Iterator<T> into J, where J is an Iterator<T>. It allows further chaining of iterators even if you only have a mutable reference to the original iterator.
By the way, your code is also incomplete; it will only return Some("") because the take_while() will stop at the first quote and won't scan further. You should rewrite it to take initial quote into account.

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