Rust是否有一种使用默认关联类型的方法？

Question

在https://stackoverflow.com/a/65924807/5884503之后，我将添加关联类型，以便从我的特性方法返回期货：

pub trait RtspClient: Runnable {
    type PlayResult: Future<Output = std::result::Result<(), ()>>;
    type ConnectResult: Future<Output = std::result::Result<(), RtspConnectionError>>;
    type DisconnectResult: Future<Output = std::result::Result<(), RtspDisconnectionError>>;

    fn connect(&self) -> RtspClient::ConnectResult {
        Ok(())
    }

    fn disconnect(&self) ->  RtspClient::DisconnectResult {
        Ok(())
    }
}

但是，我添加的方法越多，得到的关联类型就越多。

这不是一个问题，只是当我需要使用dyn RtspClient时，我需要指定所有这些

  |
12 |     pub rtsp_client: Option<Arc<Mutex<dyn RtspClient>>>,
   |                                           ^^^^^^^^^^ help: specify the associated types: `RtspClient<PlayResult = Type, ConnectResult = Type, DisconnectResult = Type>`
   | 

有没有办法强迫铁锈使用我在特征定义中已经写过的默认方法？

Answer 1

这里有很多误解。关联类型的Future<...>部分不是默认的，而是一个约束。Future不是一种类型，而是一种特征。例如，它的意思是实现的PlayResult必须满足Future<...>约束。

默认关联类型如下所示：

pub trait RtspClient  {
    type PlayResult = Ready<Result<(), ()>>; // using std::future::Ready as an example
}

但是这个语法还没有稳定下来：

error[E0658]: associated type defaults are unstable
 --> src/lib.rs:4:5
  |
4 |     type PlayResult = Ready<Result<(), ()>>; // using std::future::Ready as an example
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
  = note: see issue #29661 <https://github.com/rust-lang/rust/issues/29661> for more information

而且，即使您使用夜间编译器并启用该特性，默认情况下也无法实现connect()和disconnect()，因为它必须对所有可能的实现都有效(在某些情况下可以这样做，但我认为这不是其中之一)：

error[E0308]: mismatched types
  --> src/lib.rs:13:9
   |
12 |     fn connect(&self) -> Self::ConnectResult {
   |                          ------------------- expected `<Self as RtspClient>::ConnectResult` because of return type
13 |         ready(Ok(()))
   |         ^^^^^^^^^^^^^ expected associated type, found struct `std::future::Ready`
   |
   = note: expected associated type `<Self as RtspClient>::ConnectResult`
                       found struct `std::future::Ready<Result<(), _>>`

error[E0308]: mismatched types
  --> src/lib.rs:17:9
   |
16 |     fn disconnect(&self) ->  Self::DisconnectResult {
   |                              ---------------------- expected `<Self as RtspClient>::DisconnectResult` because of return type
17 |         ready(Ok(()))
   |         ^^^^^^^^^^^^^ expected associated type, found struct `std::future::Ready`
   |
   = note: expected associated type `<Self as RtspClient>::DisconnectResult`
                       found struct `std::future::Ready<Result<(), _>>`

即使这确实有效，你也会很难把它作为一个特征对象使用。每个实现都可以有不同的具体关联类型，但是您需要指定具体的关联类型来使用它作为一个属性对象。它们无疑将无法与之相匹配：

struct MyDefaultClient;

impl RtspClient for MyDefaultClient {
    // just uses the default types
    
    fn connect(&self) -> Self::ConnectResult {
        ready(Ok(()))
    }
    
    fn disconnect(&self) -> Self::DisconnectResult {
        ready(Ok(()))
    }
}

struct MyCustomClient;

impl RtspClient for MyCustomClient {
    type ConnectResult = Pending<Result<(), RtspConnectionError>>; // different associated types
    type DisconnectResult = Pending<Result<(), RtspDisconnectionError>>;
    
    fn connect(&self) -> Self::ConnectResult {
        pending()
    }
    
    fn disconnect(&self) -> Self::DisconnectResult {
        pending()
    }
}

fn main() {
    let data = [
        Box::new(MyDefaultClient) as Box<dyn RtspClient<ConnectResult = _, DisconnectResult = _>>,
        Box::new(MyCustomClient),
    ];
}

_部件允许编译器推断正确的类型，但会产生错误，因为dyn RtspClient<ConnectResult = A, ...>与dyn RtspClient<ConnectResult = B>不同。

error[E0271]: type mismatch resolving `<MyDefaultClient as RtspClient>::DisconnectResult == std::future::Pending<Result<(), RtspDisconnectionError>>`
  --> src/main.rs:48:9
   |
48 |         Box::new(MyDefaultClient) as Box<dyn RtspClient<ConnectResult = _, DisconnectResult = _>>,
   |         ^^^^^^^^^^^^^^^^^^^^^^^^^ expected struct `std::future::Pending`, found struct `std::future::Ready`
   |
   = note: expected struct `std::future::Pending<Result<(), RtspDisconnectionError>>`
              found struct `std::future::Ready<Result<(), RtspDisconnectionError>>`
   = note: required for the cast to the object type `dyn RtspClient<DisconnectResult = std::future::Pending<Result<(), RtspDisconnectionError>>, ConnectResult = std::future::Pending<Result<(), RtspConnectionError>>>`

error[E0271]: type mismatch resolving `<MyDefaultClient as RtspClient>::ConnectResult == std::future::Pending<Result<(), RtspConnectionError>>`
  --> src/main.rs:48:9
   |
48 |         Box::new(MyDefaultClient) as Box<dyn RtspClient<ConnectResult = _, DisconnectResult = _>>,
   |         ^^^^^^^^^^^^^^^^^^^^^^^^^ expected struct `std::future::Pending`, found struct `std::future::Ready`
   |
   = note: expected struct `std::future::Pending<Result<(), RtspConnectionError>>`
              found struct `std::future::Ready<Result<(), RtspConnectionError>>`
   = note: required for the cast to the object type `dyn RtspClient<DisconnectResult = std::future::Pending<Result<(), RtspDisconnectionError>>, ConnectResult = std::future::Pending<Result<(), RtspConnectionError>>>`

关联类型和特征对象不会以这种方式混在一起。您可以将它们视为常规泛型类型：Struct<A>是与Struct<B>完全不同的类型。

如果您想要交替地使用不同的实现，您应该遵循链接答案中的其他建议，并使用装箱期货、Pin<Box<Future<...>>>或助手crate async-trait，而不是使用关联类型。