从功能返回特征

Question

我看一下生锈lib： https ： //github.com/cyderize/rust-websocket/源文件https://github.com/cyderize/rust-websocket/blob/master/src/client/response .rs我看到方法begin（）：

pub fn begin(self) -> Client<DataFrame, Sender<W>, Receiver<R>> {
    let (reader, writer) = self.into_inner();
    let sender = Sender::new(writer);
    let receiver = Receiver::new(reader);
    Client::new(sender, receiver)
}

所以......我决定创建一些简单的包装函数：

fn get_transport(url: &str)  ->  Client<DataFrame, Sender<Write>, Receiver<Read>> {
    let url = Url::parse(url).unwrap();
    let request = Client::connect(url).unwrap();
    let response = request.send().unwrap();
    let mut ws = response.begin();

    ws
}

但不幸的是我有错误的结果：

error: the trait `core::marker::Sized` is not implemented for the type `websocket::ws::sender::Sender<std::io::Write>` [E0277]
src/lib.rs:36 fn get_transport(url: &str)  ->  Client<DataFrame, Sender<Write>, Receiver<Read>> {

有人可以形容我为什么不能以这种方式回报价值。 与response.rs中的begin（）方法有什么区别？ 在我的情况下，我应该如何回报价值？

UPD1：在@ker推荐之后我有下一个代码：

fn get_transport<W: Write, R: Read, S: Sender<W>, RE: Receiver<R>>(url: &str) -> Client<DataFrame, S, RE> {

    let url = Url::parse(url).unwrap();
    let request = Client::connect(url).unwrap();
    let response = request.send().unwrap();
    let mut ws = response.begin();

    ws
}

但是在下一次编译中收到下一个错误：

src/lib.rs:45:5: 45:7 error: mismatched types:
 expected `websocket::client::Client<websocket::dataframe::DataFrame, S, RE>`,
    found `websocket::client::Client<websocket::dataframe::DataFrame, websocket::client::sender::Sender<websocket::stream::WebSocketStream>, websocket::client::receiver::Receiver<websocket::stream::WebSocketStream>>`
(expected type parameter,
    found struct `websocket::client::sender::Sender`) [E0308]
src/lib.rs:45     ws
                  ^~
src/lib.rs:45:5: 45:7 help: run `rustc --explain E0308` to see a detailed explanation
error: aborting due to previous error

似乎我使用错误的发件人和错误的接收器：\\经过更正后使用部分我已经得到了结果代码：

fn get_transport(url: &str) -> Client<DataFrame, Sender<WebSocketStream>, Receiver<WebSocketStream>> {
    let url = Url::parse(url).unwrap();
    let request = Client::connect(url).unwrap();
    let response = request.send().unwrap();
    let mut ws = response.begin();

    ws
}

不幸的是我无法找出为什么我不能使用泛型作为结果值。 像这样：

fn get_transport<R: Read, W: Write>(url: &str) -> Client<DataFrame, Sender<W>, Receiver<R>>

在这种情况下我收到错误消息：

expected `websocket::client::Client<websocket::dataframe::DataFrame, websocket::client::sender::Sender<W>, websocket::client::receiver::Receiver<R>>`,
    found `websocket::client::Client<websocket::dataframe::DataFrame, websocket::client::sender::Sender<websocket::stream::WebSocketStream>, websocket::client::receiver::Receiver<websocket::stream::WebSocketStream>>`
(expected type parameter,
    found enum `websocket::stream::WebSocketStream`) [E0308]
src/lib.rs:45     ws
                  ^~

Answer 1

由于任何类型都可能影响特征，因此您无法通过了解对象实现的特征来了解对象的大小。 你可以做的是，使你的函数分别通用于实现Write和Read的类型。 这样您就可以返回对象，而不知道对象的实际类型是什么。

fn get_transport<W: Write, R: Read, S: Sender<W>, RE: Receiver<R>>(url: &str) -> Client<DataFrame, S, RE> {
    let url = Url::parse(url).unwrap();
    let request = Client::connect(url).unwrap();
    let response = request.send().unwrap();
    let mut ws = response.begin();

    ws
}

添加泛型的原因之一是因为泛型标签指向实际类型，而不知道除了它们实现的特征之外的任何类型。

Answer 2

您希望使用具体类型替换返回类型中的Write和Send 。

尝试更改行let mut ws = response.begin(); let mut ws: () = response.begin() ，您将看到一条错误消息，告诉您从begin函数返回的具体类型。