如何将通用 T 发送到另一个线程？

Question

如何发送通用 T？

我尝试将通用 T 发送到另一个线程，但我得到：

error[E0308]: mismatched types
  --> src/main.rs:23:22
   |
23 |             t1.merge(Element(vec![3]));
   |                      ^^^^^^^^^^^^^^^^ expected associated type, found struct `Element`
   |
   = note: expected associated type `<T as Join>::Item`
                       found struct `Element`
   = help: consider constraining the associated type `<T as Join>::Item` to `Element`

完整代码：

trait Join {
    type Item;
    fn merge(&mut self, other: Self::Item);
}

#[derive(Debug, Default)]
struct Element(Vec<u8>);

impl Join for Element {
    type Item = Element;
    fn merge(&mut self, mut other: Self::Item) {
        self.0.append(&mut other.0);
    }
}

fn work<T>()
where
    T: Default + Join + Send + Sync + 'static,
{
    let (sender, receiver) = std::sync::mpsc::channel::<(T)>();
    std::thread::spawn(move || {
        while let (mut t1) = receiver.recv().unwrap() {
            t1.merge(Element(vec![3]));
        }
    });

    loop {
        let mut t1 = T::default();
        sender.send(t1);
        std::thread::sleep(std::time::Duration::from_secs(5));
    }
}

fn main() {
    // works!
    let mut e = Element(vec![1]);
    e.merge(Element(vec![2]));

    // bad!
    work::<Element>();
}

Playground link

Answer 1

当您使用泛型时，您让调用者决定您的泛型函数必须使用哪些类型。

示例中的这一行 t1.merge(Element(vec![3])); 无效，因为它假定 T = Element 但调用者可以从无限多种可能的 T 类型中进行选择，其中 T != Element 这就是为什么编译器在抱怨。

为了让你的函数完全通用，你必须做一些事情，比如在函数签名中添加一个绑定到 <T as Join>::Item 的 Default，然后将有问题的行更改为 t1.merge(<T as Join>::Item::default());.

更新了工作评论示例：

use std::fmt::Debug;

trait Join {
    type Item;
    fn merge(&mut self, other: Self::Item);
}

#[derive(Debug)]
struct Element(Vec<u8>);

// updated Default impl so we can observe merges
impl Default for Element {
    fn default() -> Self {
        Element(vec![1])
    }
}

impl Join for Element {
    type Item = Element;
    fn merge(&mut self, mut other: Self::Item) {
        self.0.append(&mut other.0);
    }
}

fn work<T>() -> Result<(), Box<dyn std::error::Error>>
where
    T: Default + Join + Send + Sync + Debug + 'static,
    <T as Join>::Item: Default, // added Default bound here
{
    let (sender, receiver) = std::sync::mpsc::channel::<T>();
    std::thread::spawn(move || {
        while let Ok(mut t1) = receiver.recv() {
            // changed this to use Default impl
            t1.merge(<T as Join>::Item::default());

            // prints "Element([1, 1])" three times
            println!("{:?}", t1);
        }
    });

    let mut iterations = 3;
    loop {
        let t1 = T::default();
        sender.send(t1)?;
        std::thread::sleep(std::time::Duration::from_millis(100));
        iterations -= 1;
        if iterations == 0 {
            break;
        }
    }

    Ok(())
}

fn main() -> Result<(), Box<dyn std::error::Error>> {
    // works!
    let mut e = Element(vec![1]);
    e.merge(Element(vec![2]));

    // now also works!
    work::<Element>()?;

    Ok(())
}

playground

如何将通用 T 发送到另一个线程？

How to send generic T to another thread?

generics

multithreading

traits

rust

parametric-polymorphism