I have a value and I want to store that value and a reference to
something inside that value in my own type:
struct Thing {
count: u32,
}
struct Combined<'a>(Thing, &'a u32);
fn make_combined<'a>() -> Combined<'a> {
let thing = Thing { count: 42 };
Combined(thing, &thing.count)
}
Sometimes, I have a value and I want to store that value and a reference to
that value in the same structure:
struct Combined<'a>(Thing, &'a Thing);
fn make_combined<'a>() -> Combined<'a> {
let thing = Thing::new();
Combined(thing, &thing)
}
Sometimes, I'm not even taking a reference of the value and I get the
same error:
struct Combined<'a>(Parent, Child<'a>);
fn make_combined<'a>() -> Combined<'a> {
let parent = Parent::new();
let child = parent.child();
Combined(parent, child)
}
In each of these cases, I get an error that one of the values "does
not live long enough". What does this error mean?
Let's look at a simple implementation of this:
struct Parent {
count: u32,
}
struct Child<'a> {
parent: &'a Parent,
}
struct Combined<'a> {
parent: Parent,
child: Child<'a>,
}
impl<'a> Combined<'a> {
fn new() -> Self {
let parent = Parent { count: 42 };
let child = Child { parent: &parent };
Combined { parent, child }
}
}
fn main() {}
This will fail with the error:
error[E0515]: cannot return value referencing local variable `parent`
--> src/main.rs:19:9
|
17 | let child = Child { parent: &parent };
| ------- `parent` is borrowed here
18 |
19 | Combined { parent, child }
| ^^^^^^^^^^^^^^^^^^^^^^^^^^ returns a value referencing data owned by the current function
error[E0505]: cannot move out of `parent` because it is borrowed
--> src/main.rs:19:20
|
14 | impl<'a> Combined<'a> {
| -- lifetime `'a` defined here
...
17 | let child = Child { parent: &parent };
| ------- borrow of `parent` occurs here
18 |
19 | Combined { parent, child }
| -----------^^^^^^---------
| | |
| | move out of `parent` occurs here
| returning this value requires that `parent` is borrowed for `'a`
To completely understand this error, you have to think about how the
values are represented in memory and what happens when you move
those values. Let's annotate Combined::new with some hypothetical
memory addresses that show where values are located:
let parent = Parent { count: 42 };
// `parent` lives at address 0x1000 and takes up 4 bytes
// The value of `parent` is 42
let child = Child { parent: &parent };
// `child` lives at address 0x1010 and takes up 4 bytes
// The value of `child` is 0x1000
Combined { parent, child }
// The return value lives at address 0x2000 and takes up 8 bytes
// `parent` is moved to 0x2000
// `child` is ... ?
What should happen to child? If the value was just moved like parent
was, then it would refer to memory that no longer is guaranteed to
have a valid value in it. Any other piece of code is allowed to store
values at memory address 0x1000. Accessing that memory assuming it was
an integer could lead to crashes and/or security bugs, and is one of
the main categories of errors that Rust prevents.
This is exactly the problem that lifetimes prevent. A lifetime is a
bit of metadata that allows you and the compiler to know how long a
value will be valid at its current memory location. That's an
important distinction, as it's a common mistake Rust newcomers make.
Rust lifetimes are not the time period between when an object is
created and when it is destroyed!
As an analogy, think of it this way: During a person's life, they will
reside in many different locations, each with a distinct address. A
Rust lifetime is concerned with the address you currently reside at,
not about whenever you will die in the future (although dying also
changes your address). Every time you move it's relevant because your
address is no longer valid.
It's also important to note that lifetimes do not change your code; your
code controls the lifetimes, your lifetimes don't control the code. The
pithy saying is "lifetimes are descriptive, not prescriptive".
Let's annotate Combined::new with some line numbers which we will use
to highlight lifetimes:
{ // 0
let parent = Parent { count: 42 }; // 1
let child = Child { parent: &parent }; // 2
// 3
Combined { parent, child } // 4
} // 5
The concrete lifetime of parent is from 1 to 4, inclusive (which I'll
represent as [1,4]). The concrete lifetime of child is [2,4], and
the concrete lifetime of the return value is [4,5]. It's
possible to have concrete lifetimes that start at zero - that would
represent the lifetime of a parameter to a function or something that
existed outside of the block.
Note that the lifetime of child itself is [2,4], but that it refers
to a value with a lifetime of [1,4]. This is fine as long as the
referring value becomes invalid before the referred-to value does. The
problem occurs when we try to return child from the block. This would
"over-extend" the lifetime beyond its natural length.
This new knowledge should explain the first two examples. The third
one requires looking at the implementation of Parent::child. Chances
are, it will look something like this:
impl Parent {
fn child(&self) -> Child { /* ... */ }
}
This uses lifetime elision to avoid writing explicit generic
lifetime parameters. It is equivalent to:
impl Parent {
fn child<'a>(&'a self) -> Child<'a> { /* ... */ }
}
In both cases, the method says that a Child structure will be
returned that has been parameterized with the concrete lifetime of
self. Said another way, the Child instance contains a reference
to the Parent that created it, and thus cannot live longer than that
Parent instance.
This also lets us recognize that something is really wrong with our
creation function:
fn make_combined<'a>() -> Combined<'a> { /* ... */ }
Although you are more likely to see this written in a different form:
impl<'a> Combined<'a> {
fn new() -> Combined<'a> { /* ... */ }
}
In both cases, there is no lifetime parameter being provided via an
argument. This means that the lifetime that Combined will be
parameterized with isn't constrained by anything - it can be whatever
the caller wants it to be. This is nonsensical, because the caller
could specify the 'static lifetime and there's no way to meet that
condition.
How do I fix it?
The easiest and most recommended solution is to not attempt to put
these items in the same structure together. By doing this, your
structure nesting will mimic the lifetimes of your code. Place types
that own data into a structure together and then provide methods that
allow you to get references or objects containing references as needed.
There is a special case where the lifetime tracking is overzealous:
when you have something placed on the heap. This occurs when you use a
Box<T>, for example. In this case, the structure that is moved
contains a pointer into the heap. The pointed-at value will remain
stable, but the address of the pointer itself will move. In practice,
this doesn't matter, as you always follow the pointer.
Some crates provide ways of representing this case, but they
require that the base address never move. This rules out mutating
vectors, which may cause a reallocation and a move of the
heap-allocated values.
rental (no longer maintained or supported)
owning_ref (has multiple soundness issues)
ouroboros
self_cell
Examples of problems solved with Rental:
Is there an owned version of String::chars?
Returning a RWLockReadGuard independently from a method
How can I return an iterator over a locked struct member in Rust?
How to return a reference to a sub-value of a value that is under a mutex?
How do I store a result using Serde Zero-copy deserialization of a Futures-enabled Hyper Chunk?
How to store a reference without having to deal with lifetimes?
In other cases, you may wish to move to some type of reference-counting, such as by using Rc or Arc.
More information
After moving parent into the struct, why is the compiler not able to get a new reference to parent and assign it to child in the struct?
While it is theoretically possible to do this, doing so would introduce a large amount of complexity and overhead. Every time that the object is moved, the compiler would need to insert code to "fix up" the reference. This would mean that copying a struct is no longer a very cheap operation that just moves some bits around. It could even mean that code like this is expensive, depending on how good a hypothetical optimizer would be:
let a = Object::new();
let b = a;
let c = b;
Instead of forcing this to happen for every move, the programmer gets to choose when this will happen by creating methods that will take the appropriate references only when you call them.
A type with a reference to itself
There's one specific case where you can create a type with a reference to itself. You need to use something like Option to make it in two steps though:
#[derive(Debug)]
struct WhatAboutThis<'a> {
name: String,
nickname: Option<&'a str>,
}
fn main() {
let mut tricky = WhatAboutThis {
name: "Annabelle".to_string(),
nickname: None,
};
tricky.nickname = Some(&tricky.name[..4]);
println!("{:?}", tricky);
}
This does work, in some sense, but the created value is highly restricted - it can never be moved. Notably, this means it cannot be returned from a function or passed by-value to anything. A constructor function shows the same problem with the lifetimes as above:
fn creator<'a>() -> WhatAboutThis<'a> { /* ... */ }
If you try to do this same code with a method, you'll need the alluring but ultimately useless &'a self. When that's involved, this code is even more restricted and you will get borrow-checker errors after the first method call:
#[derive(Debug)]
struct WhatAboutThis<'a> {
name: String,
nickname: Option<&'a str>,
}
impl<'a> WhatAboutThis<'a> {
fn tie_the_knot(&'a mut self) {
self.nickname = Some(&self.name[..4]);
}
}
fn main() {
let mut tricky = WhatAboutThis {
name: "Annabelle".to_string(),
nickname: None,
};
tricky.tie_the_knot();
// cannot borrow `tricky` as immutable because it is also borrowed as mutable
// println!("{:?}", tricky);
}
See also:
Cannot borrow as mutable more than once at a time in one code - but can in another very similar
What about Pin?
Pin, stabilized in Rust 1.33, has this in the module documentation:
A prime example of such a scenario would be building self-referential structs, since moving an object with pointers to itself will invalidate them, which could cause undefined behavior.
It's important to note that "self-referential" doesn't necessarily mean using a reference. Indeed, the example of a self-referential struct specifically says (emphasis mine):
We cannot inform the compiler about that with a normal reference,
since this pattern cannot be described with the usual borrowing rules.
Instead we use a raw pointer, though one which is known to not be null,
since we know it's pointing at the string.
The ability to use a raw pointer for this behavior has existed since Rust 1.0. Indeed, owning-ref and rental use raw pointers under the hood.
The only thing that Pin adds to the table is a common way to state that a given value is guaranteed to not move.
See also:
How to use the Pin struct with self-referential structures?
A slightly different issue which causes very similar compiler messages is object lifetime dependency, rather than storing an explicit reference. An example of that is the ssh2 library. When developing something bigger than a test project, it is tempting to try to put the Session and Channel obtained from that session alongside each other into a struct, hiding the implementation details from the user. However, note that the Channel definition has the 'sess lifetime in its type annotation, while Session doesn't.
This causes similar compiler errors related to lifetimes.
One way to solve it in a very simple way is to declare the Session outside in the caller, and then for annotate the reference within the struct with a lifetime, similar to the answer in this Rust User's Forum post talking about the same issue while encapsulating SFTP. This will not look elegant and may not always apply - because now you have two entities to deal with, rather than one that you wanted!
Turns out the rental crate or the owning_ref crate from the other answer are the solutions for this issue too. Let's consider the owning_ref, which has the special object for this exact purpose:
OwningHandle. To avoid the underlying object moving, we allocate it on the heap using a Box, which gives us the following possible solution:
use ssh2::{Channel, Error, Session};
use std::net::TcpStream;
use owning_ref::OwningHandle;
struct DeviceSSHConnection {
tcp: TcpStream,
channel: OwningHandle<Box<Session>, Box<Channel<'static>>>,
}
impl DeviceSSHConnection {
fn new(targ: &str, c_user: &str, c_pass: &str) -> Self {
use std::net::TcpStream;
let mut session = Session::new().unwrap();
let mut tcp = TcpStream::connect(targ).unwrap();
session.handshake(&tcp).unwrap();
session.set_timeout(5000);
session.userauth_password(c_user, c_pass).unwrap();
let mut sess = Box::new(session);
let mut oref = OwningHandle::new_with_fn(
sess,
unsafe { |x| Box::new((*x).channel_session().unwrap()) },
);
oref.shell().unwrap();
let ret = DeviceSSHConnection {
tcp: tcp,
channel: oref,
};
ret
}
}
The result of this code is that we can not use the Session anymore, but it is stored alongside with the Channel which we will be using. Because the OwningHandle object dereferences to Box, which dereferences to Channel, when storing it in a struct, we name it as such. NOTE: This is just my understanding. I have a suspicion this may not be correct, since it appears to be quite close to discussion of OwningHandle unsafety.
One curious detail here is that the Session logically has a similar relationship with TcpStream as Channel has to Session, yet its ownership is not taken and there are no type annotations around doing so. Instead, it is up to the user to take care of this, as the documentation of handshake method says:
This session does not take ownership of the socket provided, it is
recommended to ensure that the socket persists the lifetime of this
session to ensure that communication is correctly performed.
It is also highly recommended that the stream provided is not used
concurrently elsewhere for the duration of this session as it may
interfere with the protocol.
So with the TcpStream usage, is completely up to the programmer to ensure the correctness of the code. With the OwningHandle, the attention to where the "dangerous magic" happens is drawn using the unsafe {} block.
A further and a more high-level discussion of this issue is in this Rust User's Forum thread - which includes a different example and its solution using the rental crate, which does not contain unsafe blocks.
I've found the Arc (read-only) or Arc<Mutex> (read-write with locking) patterns to be sometimes quite useful tradeoff between performance and code complexity (mostly caused by lifetime-annotation).
Arc:
use std::sync::Arc;
struct Parent {
child: Arc<Child>,
}
struct Child {
value: u32,
}
struct Combined(Parent, Arc<Child>);
fn main() {
let parent = Parent { child: Arc::new(Child { value: 42 }) };
let child = parent.child.clone();
let combined = Combined(parent, child.clone());
assert_eq!(combined.0.child.value, 42);
assert_eq!(child.value, 42);
// combined.0.child.value = 50; // fails, Arc is not DerefMut
}
Arc + Mutex:
use std::sync::{Arc, Mutex};
struct Child {
value: u32,
}
struct Parent {
child: Arc<Mutex<Child>>,
}
struct Combined(Parent, Arc<Mutex<Child>>);
fn main() {
let parent = Parent { child: Arc::new(Mutex::new(Child {value: 42 }))};
let child = parent.child.clone();
let combined = Combined(parent, child.clone());
assert_eq!(combined.0.child.lock().unwrap().value, 42);
assert_eq!(child.lock().unwrap().value, 42);
child.lock().unwrap().value = 50;
assert_eq!(combined.0.child.lock().unwrap().value, 50);
}
See also RwLock (When or why should I use a Mutex over an RwLock?)
As a newcomer to Rust, I had a case similar to your last example:
struct Combined<'a>(Parent, Child<'a>);
fn make_combined<'a>() -> Combined<'a> {
let parent = Parent::new();
let child = parent.child();
Combined(parent, child)
}
In the end, I solved it by using this pattern:
fn make_parent_and_child<'a>(anchor: &'a mut DataAnchorFor1<Parent>) -> Child<'a> {
// construct parent, then store it in anchor object the caller gave us a mut-ref to
*anchor = DataAnchorFor1::holding(Parent::new());
// now retrieve parent from storage-slot we assigned to in the previous line
let parent = anchor.val1.as_mut().unwrap();
// now proceed with regular code, except returning only the child
// (the parent can already be accessed by the caller through the anchor object)
let child = parent.child();
child
}
// this is a generic struct that we can define once, and use whenever we need this pattern
// (it can also be extended to have multiple slots, naturally)
struct DataAnchorFor1<T> {
val1: Option<T>,
}
impl<T> DataAnchorFor1<T> {
fn empty() -> Self {
Self { val1: None }
}
fn holding(val1: T) -> Self {
Self { val1: Some(val1) }
}
}
// for my case, this was all I needed
fn main_simple() {
let anchor = DataAnchorFor1::empty();
let child = make_parent_and_child(&mut anchor);
let child_processing_result = do_some_processing(child);
println!("ChildProcessingResult:{}", child_processing_result);
}
// but if access to parent-data later on is required, you can use this
fn main_complex() {
let anchor = DataAnchorFor1::empty();
// if you want to use the parent object (which is stored in anchor), you must...
// ...wrap the child-related processing in a new scope, so the mut-ref to anchor...
// ...gets dropped at its end, letting us access anchor.val1 (the parent) directly
let child_processing_result = {
let child = make_parent_and_child(&mut anchor);
// do the processing you want with the child here (avoiding ref-chain...
// ...back to anchor-data, if you need to access parent-data afterward)
do_some_processing(child)
};
// now that scope is ended, we can access parent data directly
// so print out the relevant data for both parent and child (adjust to your case)
let parent = anchor.val1.unwrap();
println!("Parent:{} ChildProcessingResult:{}", parent, child_processing_result);
}
This is far from a universal solution! But it worked in my case, and only required usage of the main_simple pattern above (not the main_complex variant), because in my case the "parent" object was just something temporary (a database "Client" object) that I had to construct to pass to the "child" object (a database "Transaction" object) so I could run some database commands.
Anyway, it accomplished the encapsulation/simplification-of-boilerplate that I needed (since I had many functions that needed creation of a Transaction/"child" object, and now all they need is that generic anchor-object creation line), while avoiding the need for using a whole new library.
These are the libraries I'm aware of that may be relevant:
owning-ref
rental
ouroboros
reffers
self_cell
escher
rust-viewbox
However, I scanned through them, and they all seem to have issues of one kind or another (not being updated in years, having multiple unsoundness issues/concerns raised, etc.), so I was hesitant to use them.
So while this isn't as generic of a solution, I figured I would mention it for people with similar use-cases:
Where the caller only needs the "child" object returned.
But the called-function needs to construct a "parent" object to perform its functions.
And the borrowing rules requires that the "parent" object be stored somewhere that persists beyond the "make_parent_and_child" function. (in my case, this was a start_transaction function)
The Rust documentation gives this example where we have an instance of Result<T, E> named some_value:
match some_value {
Ok(value) => println!("got a value: {}", value),
Err(_) => println!("an error occurred"),
}
Is there any way to read from some_value without pattern matching? What about without even checking the type of the contents at runtime? Perhaps we somehow know with absolute certainty what type is contained or perhaps we're just being a bad programmer. In either case, I'm just curious to know if it's at all possible, not if it's a good idea.
It strikes me as a really interesting language feature that this branch is so difficult (or impossible?) to avoid.
At the lowest level, no, you can't read enum fields without a match1.
Methods on an enum can provide more convenient access to data within the enum (e.g. Result::unwrap), but under the hood, they're always implemented with a match.
If you know that a particular case in a match is unreachable, a common practice is to write unreachable!() on that branch (unreachable!() simply expands to a panic!() with a specific message).
1 If you have an enum with only one variant, you could also write a simple let statement to deconstruct the enum. Patterns in let and match statements must be exhaustive, and pattern matching the single variant from an enum is exhaustive. But enums with only one variant are pretty much never used; a struct would do the job just fine. And if you intend to add variants later, you're better off writing a match right away.
enum Single {
S(i32),
}
fn main() {
let a = Single::S(1);
let Single::S(b) = a;
println!("{}", b);
}
On the other hand, if you have an enum with more than one variant, you can also use if let and while let if you're interested in the data from a single variant only. While let and match require exhaustive patterns, if let and while let accept non-exhaustive patterns. You'll often see them used with Option:
fn main() {
if let Some(x) = std::env::args().len().checked_add(1) {
println!("{}", x);
} else {
println!("too many args :(");
}
}
I have code similar to this:
pub trait WorldImpl {
fn new(size: (usize, usize), seed: u32) -> World;
fn three() -> bool;
fn other() -> bool;
fn non_self_methods() -> bool;
}
pub type World = Vec<Vec<UnitOfSpace>>;
// I'm doing this because I want a SPECIAL version of Vec<Vec<UnitOfSpace>>, so I can treat it like a struct but have it be a normal type underneath.
impl WorldImpl for World {
fn new(size: (usize, usize), seed: u32) -> World {
// Code
vec![/* vector stuff */]
}
// Implement other three methods
}
let w = World::new((120, 120), /* seed from UNIX_EPOCH stuff */);
And I get this error, which is clearly wrong:
error[E0061]: this function takes 0 parameters but 2 parameters were supplied
--> src/main.rs:28:28
|
28 | let world = World::new((120 as usize, 120 as usize),
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected 0 parameters
I'm thinking two things:
This is not idiomatic and Rust was never meant to be used this way. In this case, I need to know how to really do this.
It's a stupid error that I'm missing.
When I try similar code to the above on the playground, it works just fine, no errors. I have not found any information on any errors like this anywhere else, so I'll not be surprised to find out I'm just using the language wrong. I have no particular attachment to any of my code, so please tell me what the idiom is for this!
What you are trying to do doesn't quite make sense. You have made World a type alias for Vec<Vec<UnitOfSpace>>, so they are completely interchangeable - the implementations you add for one will apply to the other and vice versa.
If you want to treat this type differently then wrap it in a newtype:
struct World(Vec<Vec<UnitOfSpace>>);
This is now a distinct type from Vec<Vec<UnitOfSpace>>, but with zero runtime overhead.
Your actual error is because you have added a method called new to World as part of its implementation of WorldImpl, but World is a Vec which already has a new method (with zero args!).
Your type World is an alias for Vec<Vec<UnitOfSpace>>. Vec<T> provides an inherent associated function called new that takes no parameters. The compiler prefers selecting inherent associated functions to associated functions defined in traits, thus it selects the inherent new with no parameters instead of your own new that takes 2 parameters.
Here are a few options to solve this:
Invoke the trait's associated function explicitly:
let w = <World as WorldImpl>::new((120, 120), /* seed from UNIX_EPOCH stuff */);
Make World a newtype (struct World(Vec<Vec<UnitOfSpace>>);), which will let you define inherent associated functions (but then Vec's inherent methods won't be available on World).
Rename WorldImpl::new to a name that is not used by an inherent associated function on Vec.
I'm trying to learn Rust, so bear with me if I'm way off :-)
I have a program that inserts enums into a HashMap, and uses Strings as keys. I'm trying to match over the content of the HashMap. Problem is that I can't figure out how to get the correct borrowings, references and types in the eval_output function. How should the eval_output function look to properly handle a reference to a HashMap? Is there any good document that I can read to learn more about this particular subject?
use std::prelude::*;
use std::collections::HashMap;
enum Op {
Not(String),
Value(u16),
}
fn eval_output(output: &str, outputs: &HashMap<String, Op>) -> u16 {
match outputs.get(output) {
Some(&op) => {
match op {
Op::Not(input) => return eval_output(input.as_str(), outputs),
Op::Value(value) => return value,
}
}
None => panic!("Did not find input for wire {}", output),
}
}
fn main() {
let mut outputs = HashMap::new();
outputs.insert(String::from("x"), Op::Value(17));
outputs.insert(String::from("a"), Op::Not(String::from("x")));
println!("Calculated output is {}", eval_output("a", &outputs));
}
Review what the compiler error message is:
error: cannot move out of borrowed content [E0507]
Some(&op) => {
^~~
note: attempting to move value to here
Some(&op) => {
^~
help: to prevent the move, use `ref op` or `ref mut op` to capture value by reference
While technically correct, using Some(ref op) would be a bit silly, as the type of op would then be a double-reference (&&Op). Instead, we simply remove the & and have Some(op).
This is a common mistake that bites people, because to get it right you have to be familiar with both pattern matching and references, plus Rust's strict borrow checker. When you have Some(&op), that says
Match an Option that is the variant Some. The Some must contain a reference to a value. The referred-to thing should be moved out of where it is and placed into op.
When pattern matching, the two keywords ref and mut can come into play. These are not pattern-matched, but instead they control how the value is bound to the variable name. They are analogs of & and mut.
This leads us to the next error:
error: mismatched types:
expected `&Op`,
found `Op`
Op::Not(input) => return eval_output(input.as_str(), outputs),
^~~~~~~~~~~~~~
It's preferred to do match *some_reference, when possible, but in this case you cannot. So we need to update the pattern to match a reference to an Op — &Op. Look at what error comes next...
error: cannot move out of borrowed content [E0507]
&Op::Not(input) => return eval_output(input.as_str(), outputs),
^~~~~~~~~~~~~~~
It's our friend from earlier. This time, we will follow the compilers advice, and change it to ref input. A bit more changes and we have it:
use std::collections::HashMap;
enum Op {
Not(String),
Value(u16),
}
fn eval_output(output: &str, outputs: &HashMap<String, Op>) -> u16 {
match outputs.get(output) {
Some(op) => {
match op {
&Op::Not(ref input) => eval_output(input, outputs),
&Op::Value(value) => value,
}
}
None => panic!("Did not find input for wire {}", output),
}
}
fn main() {
let mut outputs = HashMap::new();
outputs.insert("x".into(), Op::Value(17));
outputs.insert("a".into(), Op::Not("x".into()));
println!("Calculated output is {}", eval_output("a", &outputs));
}
There's no need to use std::prelude::*; — the compiler inserts that automatically.
as_str doesn't exist in stable Rust. A reference to a String (&String) can use deref coercions to act like a string slice (&str).
I used into instead of String::from as it's a bit shorter. No real better reason.
Here's my code:
trait UnaryOperator {
fn apply(&self, expr: Expression) -> Expression;
}
pub enum Expression {
UnaryOp(UnaryOperator, Expression),
Value(i64)
}
Which gives the following errors:
error: the trait 'core::marker::sized' is not implemented for type 'parser::UnaryOperator'
note: 'parser::UnaryOperator' does not have a constant size known at compile-time
I'm not sure how to accomplish what I want. I've tried:
trait UnaryOperator: Sized {
...
}
As well as
pub enum Expression {
UnaryOp(UnaryOperator + Sized, Expression),
...
}
And neither solved the issue.
I've seen ways to possibly accomplish what I want with generics, but then it seems like two expressions with different operators would be different types, but that's not what I want. I want all expressions to be the same type regardless of what the operators are.
Traits do not have a known size - they are unsized. To see why, check this out:
trait AddOne {
fn add_one(&self) -> u8;
}
struct Alpha {
a: u8,
}
struct Beta {
a: [u8; 1024],
}
impl AddOne for Alpha {
fn add_one(&self) -> { 0 }
}
impl AddOne for Beta {
fn add_one(&self) -> { 0 }
}
Both Alpha and Beta implement AddOne, so how big should some arbitrary AddOne be? Oh, and remember that other crates may implement your trait sometime in the future.
That's why you get the first error. There are 3 main solutions (note that none of these solutions immediately fix your problem...):
Use a Box<Trait>. This is kind-of-similar-but-different to languages like Java, where you just accept an interface. This has a known size (a pointer's worth) and owns the trait. This has the downside of requiring an allocation.
trait UnaryOperator {
fn apply(&self, expr: Expression) -> Expression;
}
pub enum Expression {
UnaryOp(Box<UnaryOperator>, Expression),
Value(i64)
}
Use a reference to the trait. This also has a known size (a pointer's worth two pointers' worth, see Matthieu M.'s comment). The downside is that something has to own the object and you need to track the lifetime:
trait UnaryOperator {
fn apply<'a>(&self, expr: Expression<'a>) -> Expression<'a>;
}
pub enum Expression<'a> {
UnaryOp(&'a UnaryOperator, Expression<'a>),
Value(i64)
}
Use a generic. This has a fixed size because each usage of the enum will have been specialized for the specific type. This has the downside of causing code bloat if you have many distinct specializations. Update As you point out, this means that Expression<A> and Expression<B> would have different types. Depending on your usage, this could be a problem. You wouldn't be able to easily create a Vec<Expression<A>> if you had both.
trait UnaryOperator {
fn apply<U>(&self, expr: Expression<U>) -> Expression<U>;
}
pub enum Expression<U>
where U: UnaryOperator
{
UnaryOp(U, Expression<U>),
Value(i64)
}
Now, all of these fail as written because you have a recursive type definition. Let's look at this simplification:
enum Expression {
A(Expression),
B(u8),
}
How big is Expression? Well, it needs to have enough space to hold... an Expression! Which needs to be able to hold an Expression.... you see where this is going.
You need to add some amount of indirection here. Similar concepts to #1 and #2 apply - you can use a Box or a reference to get a fixed size:
enum Expression {
A(Box<Expression>),
B(u8),
}