why do we need to call take() for Option<T> variable - rust

In this piece of code:
pub struct Post {
state: Option<Box<dyn State>>,
content: String,
}
impl Post {
pub fn new() -> Post {
Post {
state: Some(Box::new(Draft {})),
content: String::new(),
}
}
pub fn add_text(&mut self, text: &str) {
self.content.push_str(text);
}
pub fn content(&self) -> &str {
""
}
pub fn request_review(&mut self) {
if let Some(s) = self.state.take() {
self.state = Some(s.request_review())
}
}
}
trait State {
fn request_review(self: Box<Self>) -> Box<dyn State>;
}
struct Draft {}
impl State for Draft {
fn request_review(self: Box<Self>) -> Box<dyn State> {
Box::new(PendingReview {})
}
}
struct PendingReview {
fn request_review(self: Box<Self>) -> Box<dyn State> {
self
}
}
there is a call to take(); the book says:
To consume the old state, the request_review method needs to take ownership of the state value. This is where the Option in the state field of Post comes in: we call the take method to take the Some value out of the state field and leave a None in its place.
We need to set state to None temporarily rather than setting it directly with code like self.state = self.state.request_review(); to get ownership of the state value. This ensures Post can’t use the old state value after we’ve transformed it into a new state.
How is it possible that Post uses its old state if we set it directly?

If you code like this:
pub struct Post {
state: Box<dyn State>,
content: String,
}
trait State {
fn request_review(self: Box<Self>) -> Box<dyn State>;
}
impl Post {
// ...
pub fn request_review(&mut self) {
self.state = self.state.request_review();
}
// ...
}
You will get a compiler error:
self.state = self.state.request_review();
^^^^^^ move occurs because `self.state` has type `std::boxed::Box<dyn State>`, which does not implement the `Copy` trait'.
This is because calling State::request_review will move Box<self>, which is allocated on heap, and Rust doesn't allow you to just move values away from heap unless you implement Copy, otherwise what's left there? The book uses Option::take() to move ownership out and leave None on the place.

If request_review panic, it will lead to free the Box twice, first in request_review and then when the Option is freed.

Quote from the book; bold font my formatting
we call the take method to take the Some value out of the state field
and leave a None in its place, because Rust doesn’t let us have
unpopulated fields in structs.
As for move and assignment in the same statement, looks like Rust does not detect it as valid.

Related

How can I mutate fields of a struct while referencing other fields?

Here's an example of a problem I ran into:
pub struct Item {
name: String,
value: LockableValue, // another struct that I'd like to mutate
}
impl Item {
pub fn name(&self) -> &str {
&self.name
}
pub fn value_mut(&mut self) -> &mut LockableValue {
&self.value
}
}
pub fn update(item: &mut Item) {
let value = item.value_mut();
value.change(); // how it changes is unimportant
println!("Updated item: {}", item.name());
}
Now, I know why this fails. I have a mutable reference to item through the mutable reference to the value.
If I convert the reference to an owned String, it works fine, but looks strange to me:
pub fn update(item: &mut Item) {
let name = { item.name().to_owned() };
let value = item.value_mut();
value.change(); // how it changes is unimportant
println!("Updated item: {}", name); // It works!
}
If I let value reference drop, then everything is fine.
pub fn update(item: &mut Item) {
{
let value = item.value_mut();
value.change(); // how it changes is unimportant
}
println!("Updated item: {}", item.name()); // It works!
}
The value.change() block is rather large, and accessing other fields in item might be helpful. So while I do have solutions to this issue, I'm wondering if there is a better (code-smell) way to do this. Any suggestions?
My intention behind the above structs was to allow Items to change values, but the name should be immutable. LockableValue is an tool to interface with another memory system, and copying/cloning the struct is not a good idea, as the memory is managed there. (I implement Drop on LockableValue to clean up.)
I was hoping it would be straight-forward to protect members of the struct from modification (even if it were immutable) like this... and I can, but it ends up looking weird to me. Maybe I just need to get used to it?
You could use interior mutability on only the part that you want to mutate by using a RefCell like ths:
use std::cell::{RefCell, RefMut};
pub struct LockableValue;
impl LockableValue {
fn change(&mut self) {}
}
pub struct Item {
name: String,
value: RefCell<LockableValue>, // another struct that I'd like to mutate
}
impl Item {
pub fn name(&self) -> &str {
&self.name
}
pub fn value_mut(&self) -> RefMut<'_, LockableValue> {
self.value.borrow_mut()
}
}
pub fn update(item: &Item) {
let name = item.name();
let mut value = item.value_mut();
value.change(); // how it changes is unimportant
println!("Updated item: {}", name);
}
That way you only need a shared reference to Item and you don't run into an issue with the borrow checker.
Not that this forces the borrow checks on value to be done at runtime though and thus comes with a performance hit.

Make Box<Self> function parameter mutable

I've got this code snippet (playground link) where I'm basically trying to change a trait implementation to allow a state change for the struct it's applied to.
The current method signature is:
fn approve(self: Box<Self>) -> Box<dyn State>
, but I need to make it into something like
fn approve(&mut self: Box<Self>) -> Box<dyn State>
This doesn't compile. I've yet to see a Rust &mut self parameter definition along with a type, though I'm not sure why that would be an issue.
If I try:
fn approve(&mut self) -> Box<dyn State>
I get other compile-time errors when I try to return an unmodified self from the method.
For a little bit more context (before checking the playground), here are some relevant code bits:
trait State {
fn approve(self: Box<Self>) -> Box<dyn State>;
}
struct Draft {
approve_count: u8,
}
impl State for Draft {
fn approve(self: Box<Self>) -> Box<dyn State> {
// self.approve_count += 1;
if self.approve_count == 2 {
Box::new(Approved {})
} else {
self
}
}
}
struct Approved {}
impl State for Approved { ... }
What I would like to do is be able to uncomment the self.approve_count += 1; line and have it work.
The correct syntax for a mutable reference is:
fn approve(self: &mut Box<Self>)
...but then you won't be able to return self as Box<dyn State> because you're only holding a reference to self.
If all you need is for self.approve_count += 1 to work, you don't need to switch to &mut self, you can simply declare mut self: Box<Self>:
impl State for Draft {
fn approve(mut self: Box<Self>) -> Box<dyn State> {
self.approve_count += 1;
if self.approve_count == 2 {
Box::new(Approved {})
} else {
self
}
}
...
}
Playground
That code is a bit suspect in that it modifies self.approved_count even in the case where the function returns a completely new Approved instance (and discards the just incremented value), but that may be fixed by moving the increment into the else clause and adjusting the test as appropriate.

Why RefCell's `into_inner` requires a move?

I have a situation where I have to move a struct from one object to another through a &mut self. Take a look:
pub struct MyStruct {
//no copy trait
device: Device
}
impl MyStruct {
pub fn finalize(&mut self) {
//error: cannot move since I borrowed
let interface = InterfaceBuilder::new(self.device)
}
}
First of all, why I cannot move something out of a borrowed mutable reference? Borrowed mutables are exclusive, there's no chance another code is looking into it.
Well, to address this problem I changed to
pub struct MyStruct {
//no copy trait
device: RefCell<Device>
}
impl MyStruct {
pub fn finalize(&mut self) {
//error on `self.device`: cannot move out of `self.device` which is behind a mutable reference
let interface = InterfaceBuilder::new(self.device.into_inner())
}
}
I know why the error occurs:
pub fn into_inner(self) -> T
calling into_inner makes self.device move. Why RefCell simply does not have an implementation pub fn into_inner(&mut self) -> T? I don't see a problem.
You cannot move out of a mutable reference because that would leave the original object incomplete.
Consider this code:
struct MyStruct {
s: String
}
fn finalize(f: &mut MyStruct) {
let _x = f.s; //error E0507!
}
fn main() {
let mut my = MyStruct {
s: "hi".into()
};
finalize(&mut my);
println!("{}", my.s); //what should this do?
}
Then, RefCell::into_inner(&mut self) -> T has the same problem. You could call it twice in a row and you would get two T values where before there was only one. And that, for a non Copy type is impossible.
If you want this function to consume the inner value, probably it should consume the outer value too:
fn finalize(f: MyStruct) {
let _x = f.s;
}
If you really want to move a value out of a mutable reference, you must leave something valid in its place. The easiest way is to declare an Option and use take() to steal and replace it with a None:
struct MyStruct {
s: Option<String>
}
fn finalize(f: &mut MyStruct) {
let _x = f.s.take();
}
Naturally, Option::take returns an Option so that if you call it twice, the second time you get None. If you are positive you have a value you can do take().uwnrap().
Alternatively, if your field type is Default you can use std::mem::take that replaces it with a default-created value:
struct MyStruct {
s: Vec<i32>
}
fn finalize(f: &mut MyStruct) {
let _x = std::mem::take(&mut f.s);
}
PS #1: there is Cell::take(&self) -> T, but only if T implements Default. It works just like std::mem::take but with a non-mutable reference.
PS #2: there is also unsafe fn ManuallyDrop::take(slot: &mut ManuallyDrop<T>) -> T, that is intented to be used in advanced drop implementations. But it is unsafe so it should never be your first option: if you call it twice you will get undefined behavior.

Can I implement a trait which adds information to an external type in Rust?

I just implemented a simple trait to keep the history of a struct property:
fn main() {
let mut weight = Weight::new(2);
weight.set(3);
weight.set(5);
println!("Current weight: {}. History: {:?}", weight.value, weight.history);
}
trait History<T: Copy> {
fn set(&mut self, value: T);
fn history(&self) -> &Vec<T>;
}
impl History<u32> for Weight {
fn set(&mut self, value: u32) {
self.history.push(self.value);
self.value = value;
}
fn history(&self) -> &Vec<u32> {
&self.history
}
}
pub struct Weight {
value: u32,
history: Vec<u32>,
}
impl Weight {
fn new(value: u32) -> Weight {
Weight {
value,
history: Vec::new(),
}
}
}
I don't expect this is possible, but could you add the History trait (or something equivalent) to something which doesn't already have a history property (like u32 or String), effectively tacking on some information about which values the variable has taken?
No. Traits cannot add data members to the existing structures. Actually, only a programmer can do that by modifying the definition of a structure. Wrapper structures or hash-tables are the ways to go.
No, traits can only contain behavior, not data. But you could make a struct.
If you could implement History for u32, you'd have to keep the entire history of every u32 object indefinitely, in case one day someone decided to call .history() on it. (Also, what would happen when you assign one u32 to another? Does its history come with it, or does the new value just get added to the list?)
Instead, you probably want to be able to mark specific u32 objects to keep a history. A wrapper struct, as red75prime's answer suggests, will work:
mod hist {
use std::mem;
pub struct History<T> {
value: T,
history: Vec<T>,
}
impl<T> History<T> {
pub fn new(value: T) -> Self {
History {
value,
history: Vec::new(),
}
}
pub fn set(&mut self, value: T) {
self.history.push(mem::replace(&mut self.value, value));
}
pub fn get(&self) -> T
where
T: Copy,
{
self.value
}
pub fn history(&self) -> &[T] {
&self.history
}
}
}
It's generic, so you can have a History<u32> or History<String> or whatever you want, but the get() method will only be implemented when the wrapped type is Copy.* Your Weight type could just be an alias for History<u32>. Here it is in the playground.
Wrapping this code in a module is a necessary part of maintaining the abstraction. That means you can't write weight.value, you have to call weight.get(). If value were marked pub, you could assign directly to weight.value (bypassing set) and then history would be inaccurate.
As a side note, you almost never want &Vec<T> when you can use &[T], so I changed the signature of history(). Another thing you might consider is returning an iterator over the previous values (perhaps in reverse order) instead of a slice.
* A better way of getting the T out of a History<T> is to implement Deref and write *foo instead of foo.get().

Lifetime error for returned value of a function

This is a simplified version of a piece of code I am trying to implement:
struct FirstStruct
{
a: i8,
}
impl FirstStruct
{
fn get(&self) -> Option<&str>
{
Some("aaa")
}
}
pub struct SecondStruct<'a>
{
pub name: Option<&'a str>,
}
impl<'a> SecondStruct<'a>
{
fn extract_string(obj: &/*'a*/ FirstStruct) -> Option<&'a str>
{
obj.get() //this is where the error happen
}
pub fn from_string() -> SecondStruct<'a>
{
let obj = FirstStruct{a: 1};
SecondStruct{
name: SecondStruct::extract_string(&obj),
}
}
}
fn main()
{
let g_def_res = SecondStruct::from_string();
}
This code throws the following error :
test2.rs:23:13: 23:18 error: cannot infer an appropriate lifetime for autoref due to conflicting requirements
test2.rs:23 obj.get() //this is where the error happen
^~~~~
test2.rs:21:5: 24:6 help: consider using an explicit lifetime parameter as shown: fn extract_string(obj: &'a FirstStruct) -> Option<&'a str>
test2.rs:21 fn extract_string(obj: &FirstStruct) -> Option<&'a str>
test2.rs:22 {
test2.rs:23 obj.get() //this is where the error happen
test2.rs:24 }
error: aborting due to previous error
Applying the proposed solution throw this error :
test2.rs:30:55: 30:58 error: `obj` does not live long enough
test2.rs:30 name: SecondStruct::extract_string(&obj),
^~~
test2.rs:27:5: 32:6 note: reference must be valid for the lifetime 'a as defined on the block at 27:4...
test2.rs:27 {
test2.rs:28 let obj = FirstStruct{a: 1};
test2.rs:29 SecondStruct{
test2.rs:30 name: SecondStruct::extract_string(&obj),
test2.rs:31 }
test2.rs:32 }
test2.rs:28:37: 32:6 note: ...but borrowed value is only valid for the block suffix following statement 0 at 28:36
test2.rs:28 let obj = FirstStruct{a: 1};
test2.rs:29 SecondStruct{
test2.rs:30 name: SecondStruct::extract_string(&obj),
test2.rs:31 }
test2.rs:32 }
error: aborting due to previous error
To summarise:
How to say that the return value of FirstStruct::get must have the lifetime of either [the return value of SecondStruct::from_str | the struct lifetime 'a]? I think both refer to same thing?
pub fn from_string() -> SecondStruct<'a> {
let obj = FirstStruct { a: 1 };
SecondStruct {
name: SecondStruct::extract_string(&obj),
}
}
This code says "I will return a SecondStruct with the lifetime 'a". The caller of the code gets to determine what the length of the lifetime 'a is. This is almost never what you want!
// Lifetime elision means the method is equivalent to this
// fn get<'a>(&'a self) -> Option<&'a str>
fn get(&self) -> Option<&str> {
Some("aaa")
}
This code uses says that the string returned will live as long as self lives.
Put these two concepts together, and you can understand your error. The variable obj is only defined to live as long as the function call is active. However, you are trying to return a reference to the inner-workings of the struct beyond the call! Actually, you are trying to return it for any arbitrary lifetime the caller decides! This is Rust preventing you from shooting yourself in the foot, hooray for Rust!
So how do you fix your problem? For the provided example code, the easiest thing is to just use the 'static lifetime:
struct FirstStruct { a: i8 }
impl FirstStruct {
fn get(&self) -> Option<&'static str> { Some("aaa") }
}
pub struct SecondStruct<'a> { name: Option<&'a str> }
impl<'a> SecondStruct<'a> {
fn extract_string(obj: &FirstStruct) -> Option<&'static str> { obj.get() }
pub fn from_string() -> SecondStruct<'static> {
let obj = FirstStruct { a: 1 };
SecondStruct { name: SecondStruct::extract_string(&obj) }
}
}
fn main() {
let g_def_res = SecondStruct::from_string();
}
But that's probably not what you really want. The next thing to try would be to embed FirstStruct inside SecondStruct, and simply delegate to it. Another option would be to move from &str to String - String owns the string data, and so you can transfer ownership from First to Second.
Whatever you do, you have to ensure that the source of the string data outlives the function call to from_string.
Either the return value of FirstStruct::get has been allocated on the stack or it has been allocated on the heap.
It's trickier than that. The return value is always on the stack. That is, the Option<&str> takes up space on the stack. The &str may contain a pointer to something that is allocated either on the stack or heap, it's not known by this code. All you know is that the pointed-at value is guaranteed to live for the lifetime of that specific FirstStruct item.
You don't own the string, so you can't transfer ownership around.
I can't move FirstStruct because it is from another library (rustc-serialize
I'm not sure what you mean. If you have an object, then you can embed it into your object. The fact that it comes from another crate doesn't come into play. If you have a reference to something, then you can still capture the reference, but then your object has to live for a shorter period than the reference (so that it never becomes invalid).
Unwrapping Option, updating to a string and rewrapping in Option is a lot of boilerplate.
Have you seen Option::map? It makes this kind of thing very succinct. Combined with From, you can write a very short thing to convert an Option<&str> to Option<String>:
// fn is just to establish some types, you'd just use the `.map` call in real code
fn foo(a: Option<&str>) -> Option<String> {
a.map(From::from)
}

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