如何对具有动态成员的结构采用发送和同步（未来无法在线程之间安全发送）

Question

考虑下面的代码，它async使用async-traitcrate 的方法声明了一个特征。

use std::{io::Result, sync::Arc};

use async_trait::async_trait;
use tokio;

// A trait that describes a power source.
trait PowerSource {
    fn supply_power(&self) -> Result<()>;
}

// ElectricMotor implements PowerSource.
struct ElectricMotor {}

impl PowerSource for ElectricMotor {
    fn supply_power(&self) -> Result<()> {
        println!("ElectricMotor::supply_power");
        Ok(())
    }
}

// A trait that describes a vehicle
#[async_trait]
trait Vehicle {
    async fn drive(&self) -> Result<()>;
}

// An automobile has some kind of power source and implements Vehicle
struct Automobile {
    power_source: Arc<dyn PowerSource>,
}

#[async_trait]
impl Vehicle for Automobile {
    async fn drive(&self) -> Result<()> {
        self.power_source.supply_power()?;
        println!("Vehicle::Drive");
        Ok(())
    }
}

#[tokio::main]
async fn main() -> std::io::Result<()> {
    let driver = ElectricMotor {};
    let controller = Automobile {
        power_source: Arc::new(driver),
    };

    controller.drive().await?;

    Ok(())
}

这不会编译错误“未来无法在线程之间安全地发送”：

error: future cannot be sent between threads safely
  --> src/main.rs:34:41
   |
34 |       async fn drive(&self) -> Result<()> {
   |  _________________________________________^
35 | |         self.power_source.supply_power()?;
36 | |         println!("Vehicle::Drive");
37 | |         Ok(())
38 | |     }
   | |_____^ future created by async block is not `Send`
   |
   = help: the trait `Sync` is not implemented for `(dyn PowerSource + 'static)`
note: captured value is not `Send`
  --> src/main.rs:34:21
   |
34 |     async fn drive(&self) -> Result<()> {
   |                     ^^^^ has type `&Automobile` which is not `Send`
   = note: required for the cast to the object type `dyn Future<Output = Result<(), std::io::Error>> + Send`

error: future cannot be sent between threads safely
  --> src/main.rs:34:41
   |
34 |       async fn drive(&self) -> Result<()> {
   |  _________________________________________^
35 | |         self.power_source.supply_power()?;
36 | |         println!("Vehicle::Drive");
37 | |         Ok(())
38 | |     }
   | |_____^ future created by async block is not `Send`
   |
   = help: the trait `Send` is not implemented for `(dyn PowerSource + 'static)`
note: captured value is not `Send`
  --> src/main.rs:34:21
   |
34 |     async fn drive(&self) -> Result<()> {
   |                     ^^^^ has type `&Automobile` which is not `Send`
   = note: required for the cast to the object type `dyn Future<Output = Result<(), std::io::Error>> + Send`

如果我正确理解错误，它认为这Automobile不是Send因为该power_source财产不是，所以它不能创造一个适当的未来。我的理解是它Arc是线程安全的并且实现了Sendand Sync，但我对 Rust 并发性仍然很陌生，并且仍然不完全清楚这意味着什么。

我该如何修复这个错误？

Answer 1

您必须修改您的Automobile定义：

struct Automobile {
    power_source: Arc<dyn PowerSource>,
}

在 Rust 中，类型Send当且仅当其所有成员都是Send（除非您手动且不安全地实现Send）。这同样适用于Sync. 因此，鉴于您的power_source不是Send，那么生成的impl Future也将不是Send。

修复它的方法是添加Send + Sync以下要求power_source：

struct Automobile {
    power_source: Arc<dyn PowerSource + Send + Sync>,
}

但Sync你可能会问为什么？Send毕竟编译器只是在抱怨。它需要的原因Sync是因为Arc<T>只有Send当T是Send 并且 Sync

补充阅读：

• 发送和同步
• 关于同一问题的一篇不错的博客文章