[英]How can I use serde to serialize a struct to another Rust data structure?
[英]How Can I Get Two Structs in Rust to Use and Share a reference to Another Struct
我一直在尝试使用更复杂的结构,当值包含在另一个结构中时,我在尝试使用一个结构编辑值时遇到了麻烦。 这样做的目的是能够从潜在用户的角度编写一个简单的抽象,这样他们就只有一个结构,他们必须改变和使用。
示例代码:
#[derive(Debug)]
pub struct Widget {
counter: u16,
}
impl Widget{
pub fn new() -> Widget {
let nw = Widget {
counter: 0
};
return nw;
}
}
pub struct Market {
widgets: Vec<Widget>
}
impl Market {
pub fn new() -> Market {
let market_vec = Vec::new();
let market = Market {
widgets: market_vec
};
return market;
}
pub fn new_user(&mut self) -> User {
let user_widget = Widget::new();
let user = User::new(user_widget);
self.widgets.push(user_widget);
return user;
}
}
pub struct User {
name: String,
widget: Widget
}
impl User {
pub fn new(user_widget: Widget) -> User {
let user = User {
name: "User1".to_string(),
widget: user_widget
};
return user;
}
pub fn update_count(&mut self) {
self.widget.counter +=1;
}
}
pub fn main() {
let mut market = Market::new();
let mut user1 = market.new_user();
println!("{:?}", market.widgets);
user1.update_count();
println!("{:?}", market.widgets);
}
示例输出:
Compiling playground v0.0.1 (/playground)
error[E0382]: use of moved value: `user_widget`
--> src/main.rs:31:27
|
29 | let user_widget = Widget::new();
| ----------- move occurs because `user_widget` has type `Widget`, which does not implement the `Copy` trait
30 | let user = User::new(user_widget);
| ----------- value moved here
31 | self.widgets.push(user_widget);
| ^^^^^^^^^^^ value used here after move
For more information about this error, try `rustc --explain E0382`.
error: could not compile `playground` due to previous error
理论上,我希望用户中的小部件是对小部件的引用,但我无法使用引用初始化用户,然后修改该引用。 我已经研究过尝试使用Arc<T>
或RC<T>
但我不确定是否需要包装存储具有这些类型的小部件的向量和引用它的用户。 我可以在 User 结构中只使用一次吗?
您实际上是在通过所有这些实例修改值,这使问题变得更加困难。
rust 中的所有权基础说明了三件事:
这也适用于Rc
和Arc
,这意味着,虽然它们可以访问多个“所有者”,但它们只是不可变地这样做。
要实际修改这些值,您需要创建内部 mutability 。 这通常在单线程情况下使用RefCell
或在多线程情况下使用Mutex
来完成。
这是您的代码Rc<RefCell>
:
use std::{cell::RefCell, rc::Rc};
#[derive(Debug)]
pub struct Widget {
counter: u16,
}
impl Widget {
pub fn new() -> Widget {
let nw = Widget { counter: 0 };
return nw;
}
}
pub struct Market {
widgets: Vec<Rc<RefCell<Widget>>>,
}
impl Market {
pub fn new() -> Market {
let market_vec = Vec::new();
let market = Market {
widgets: market_vec,
};
return market;
}
pub fn new_user(&mut self) -> User {
let user_widget = Rc::new(RefCell::new(Widget::new()));
let user = User::new(user_widget.clone());
self.widgets.push(user_widget);
return user;
}
}
pub struct User {
name: String,
widget: Rc<RefCell<Widget>>,
}
impl User {
pub fn new(user_widget: Rc<RefCell<Widget>>) -> User {
let user = User {
name: "User1".to_string(),
widget: user_widget,
};
return user;
}
pub fn update_count(&mut self) {
self.widget.borrow_mut().counter += 1;
}
}
pub fn main() {
let mut market = Market::new();
println!("{:?}", market.widgets);
let mut user1 = market.new_user();
user1.update_count();
println!("{:?}", market.widgets);
}
[]
[RefCell { value: Widget { counter: 1 } }]
在您的具体情况下,我注意到您实际更新的唯一内容是counter
。
因此,您实际上不需要使整个Widget
可变,而是可以只使计数器可变。 计数器比Widget
类更简单,因此我们可以对其进行一些优化。
在单线程情况下,我们可以使用Cell
。 Cell
与RefCell
相同,但不能失败。 但Cell
只存在于可复制的对象。
在多线程的情况下,我们可以使用AtomicU16
。 它比Mutex
效率高得多; 实际上,在大多数情况下,与普通的u16
相比,它的开销为零。
这是Cell<u16>
的解决方案:
use std::{cell::Cell, rc::Rc};
#[derive(Debug)]
pub struct Widget {
counter: Cell<u16>,
}
impl Widget {
pub fn new() -> Widget {
let nw = Widget { counter: 0.into() };
return nw;
}
}
pub struct Market {
widgets: Vec<Rc<Widget>>,
}
impl Market {
pub fn new() -> Market {
let market_vec = Vec::new();
let market = Market {
widgets: market_vec,
};
return market;
}
pub fn new_user(&mut self) -> User {
let user_widget = Rc::new(Widget::new());
let user = User::new(user_widget.clone());
self.widgets.push(user_widget);
return user;
}
}
pub struct User {
name: String,
widget: Rc<Widget>,
}
impl User {
pub fn new(user_widget: Rc<Widget>) -> User {
let user = User {
name: "User1".to_string(),
widget: user_widget,
};
return user;
}
pub fn update_count(&mut self) {
let prev = self.widget.counter.get();
self.widget.counter.set(prev + 1);
}
}
pub fn main() {
let mut market = Market::new();
println!("{:?}", market.widgets);
let mut user1 = market.new_user();
user1.update_count();
println!("{:?}", market.widgets);
}
[]
[Widget { counter: Cell { value: 1 } }]
为了完整起见,这里是多线程上下文中的相同解决方案。
使用Arc<Mutex>
:
use std::sync::{Arc, Mutex};
#[derive(Debug)]
pub struct Widget {
counter: u16,
}
impl Widget {
pub fn new() -> Widget {
let nw = Widget { counter: 0 };
return nw;
}
}
pub struct Market {
widgets: Vec<Arc<Mutex<Widget>>>,
}
impl Market {
pub fn new() -> Market {
let market_vec = Vec::new();
let market = Market {
widgets: market_vec,
};
return market;
}
pub fn new_user(&mut self) -> User {
let user_widget = Arc::new(Mutex::new(Widget::new()));
let user = User::new(user_widget.clone());
self.widgets.push(user_widget);
return user;
}
}
pub struct User {
name: String,
widget: Arc<Mutex<Widget>>,
}
impl User {
pub fn new(user_widget: Arc<Mutex<Widget>>) -> User {
let user = User {
name: "User1".to_string(),
widget: user_widget,
};
return user;
}
pub fn update_count(&mut self) {
self.widget.lock().unwrap().counter += 1;
}
}
pub fn main() {
let mut market = Market::new();
println!("{:?}", market.widgets);
let mut user1 = market.new_user();
user1.update_count();
println!("{:?}", market.widgets);
}
[]
[Mutex { data: Widget { counter: 1 }, poisoned: false, .. }]
使用AtomicU16
:
use std::{
sync::atomic::{AtomicU16, Ordering},
sync::Arc,
};
#[derive(Debug)]
pub struct Widget {
counter: AtomicU16,
}
impl Widget {
pub fn new() -> Widget {
let nw = Widget { counter: 0.into() };
return nw;
}
}
pub struct Market {
widgets: Vec<Arc<Widget>>,
}
impl Market {
pub fn new() -> Market {
let market_vec = Vec::new();
let market = Market {
widgets: market_vec,
};
return market;
}
pub fn new_user(&mut self) -> User {
let user_widget = Arc::new(Widget::new());
let user = User::new(user_widget.clone());
self.widgets.push(user_widget);
return user;
}
}
pub struct User {
name: String,
widget: Arc<Widget>,
}
impl User {
pub fn new(user_widget: Arc<Widget>) -> User {
let user = User {
name: "User1".to_string(),
widget: user_widget,
};
return user;
}
pub fn update_count(&mut self) {
self.widget.counter.fetch_add(1, Ordering::SeqCst);
}
}
pub fn main() {
let mut market = Market::new();
println!("{:?}", market.widgets);
let mut user1 = market.new_user();
user1.update_count();
println!("{:?}", market.widgets);
}
[]
[Widget { counter: 1 }]
使用Rc
和Mutex
,您可以包装每个Widget
,以便传递对单个结构的可变引用。
我也对您的代码进行了一些修改,因此对我来说更容易工作,我希望仍然具有足够的可读性:
use std::rc::Rc;
use std::sync::Mutex;
#[derive(Debug)]
pub struct Widget {
counter: u16,
}
impl Widget {
pub fn new() -> Widget {
Widget { counter: 0 }
}
}
type MutWidget = Rc<Mutex<Widget>>;
pub struct Market {
widgets: Vec<MutWidget>,
}
impl Market {
pub fn new() -> Market {
Market {
widgets: Vec::new(),
}
}
pub fn new_user(&mut self) -> User {
let user_widget = Rc::new(Mutex::new(Widget::new()));
let user = User::new(user_widget.clone());
self.widgets.push(user_widget.clone());
return user;
}
}
pub struct User {
name: String,
widget: MutWidget,
}
impl User {
pub fn new(user_widget: MutWidget) -> User {
User {
name: "User1".to_string(),
widget: user_widget,
}
}
pub fn update_count(&mut self) {
self.widget.lock().unwrap().counter += 1;
}
}
pub fn main() {
let mut market = Market::new();
println!("{:?}", market.widgets);
let mut user1 = market.new_user();
user1.update_count();
println!("{:?}", market.widgets);
}
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