在 Rust 中实现多个可变(静态分配、静态调度等)回调的等价物的正确方法是什么? [英] What's the correct way to implement the equivalent of multiple mutable (statically allocated, statically dispatched, etc.) callbacks in Rust?
问题描述
我有以下示例代码，它是其他编程语言中事件驱动 API 的标准基础，但在 Rust 中，借用检查器通过不能多次借用 p1 作为可变变量"来阻止它一次":
I have the following example code, which is the standard basis of event-driven APIs in other programming languages, but in Rust the borrow checker blocks it with "cannot borrow p1 as mutable more than once at a time":
struct Pen {
color_cmyk: u32,
ink: usize,
}
impl Pen {
pub fn new() -> Pen {
Pen {
color_cmyk: 0x80800000,
ink: 20000,
}
}
pub fn write(&mut self, text: &str) -> bool {
if self.ink < text.len() {
return false;
}
self.ink -= text.len();
true
}
}
fn main() {
println!("Hello, world !");
let mut p1 = Pen::new();
p1.write("Hello");
println!("ink: {}, color: {}", p1.ink, p1.color_cmyk);
let mut cb = |text| if p1.write(text) {
println!("{}", text);
} else {
println!("Out of ink !");
};
let mut cb2 = |text| {
p1.write(text);
p1.ink
};
cb("Hello");
cb("World");
println!("{}", cb2("Hello"));
}
error[E0499]: cannot borrow `p1` as mutable more than once at a time
--> src/main.rs:37:23
|
31 | let mut cb = |text| if p1.write(text) {
| ------ -- previous borrow occurs due to use of `p1` in closure
| |
| first mutable borrow occurs here
...
37 | let mut cb2 = |text| {
| ^^^^^^ second mutable borrow occurs here
38 | p1.write(text);
| -- borrow occurs due to use of `p1` in closure
...
45 | }
| - first borrow ends here
该代码可用于例如实现对窗口的两个回调:一个用于处理键盘事件，另一个用于处理鼠标事件，两者都更新窗口状态(例如:更改颜色、关闭窗口等.).
The code can be used, for example, to implement two callbacks to a window: one for handling keyboard events and another for handling mouse events, both of which update the window state (ex: changing color, closing the window, etc.).
我知道这个问题出现在 Stack Overflow 和其他论坛的其他地方，但一般来说，答案侧重于描述问题的原因，很少提出完整的通用解决方案:
I know that this question appears elsewhere in Stack Overflow and other forums, but in general, the answers focus on describing the reason of the problem and rarely propose a complete general solution for it:
推荐答案
一种方法是使用 RefCell，它允许你只用 &Pen 改变事物而不是 &mut Pen，代价是将借用检查推送到运行时.它非常便宜:没有分配，只有一个标志测试.主要的缺点是违反规则会导致运行时出现恐慌.一个有用的经验法则是永远不要借用超过必要的时间(将它们视为单线程互斥锁").
One way is to use a RefCell, which allows you to mutate things with only &Pen instead of &mut Pen, at the cost of pushing the borrow-checking to runtime. It’s very cheap: there is no allocation, just a single flag test.
The main downside is that violating the rules will result in a panic at runtime. A useful rule of thumb is to never borrow for any longer than necessary (think of them as "single-threaded mutexes").
use std::cell::RefCell;
fn main() {
println!("Hello, world !");
let p1 = RefCell::new(Pen::new());
{
let mut rp1 = p1.borrow_mut();
rp1.write("Hello");
println!("ink: {}, color: {}", rp1.ink, rp1.color_cmyk);
}
let cb = |text| {
if p1.borrow_mut().write(text) {
println!("{}", text);
}
else {
println!("Out of ink !");
}
};
let cb2 = |text| {
let mut rp1 = p1.borrow_mut();
rp1.write(text);
rp1.ink
};
cb("Hello");
cb("World");
println!("{}", cb2("Hello"));
}
另一种方法是设置回调系统，将您正在修改的对象作为参数传入.权衡是你的回调系统需要知道这个状态.
Another way is to set up the callback system to pass in the object that you’re modifying as an argument. The trade-off is then your callback system needs to be aware of this state.
fn main() {
println!("Hello, world !");
let mut p1 = Pen::new();
p1.write("Hello");
println!("ink: {}, color: {}", p1.ink, p1.color_cmyk);
let cb = |p1: &mut Pen, text| if p1.write(text) {
println!("{}", text);
} else {
println!("Out of ink !");
};
let cb2 = |p1: &mut Pen, text| {
p1.write(text);
p1.ink
};
cb(&mut p1, "Hello");
cb(&mut p1, "World");
println!("{}", cb2(&mut p1, "Hello"));
}
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