`?` 运算符只能在返回 `Result` 或 `Option 的异步块中使用

Question

我正在学习 Rust 并尝试通过发送一些 POST 数据来抓取随机站点，但我收到了一堆错误，例如：

error[E0277]: the `?` operator can only be used in an async block that returns `Result` or `Option` (or another type that implements `FromResidual`)

37  | |     let text = res.text().await?;
    | |                                ^ cannot use the `?` operator in an async block that returns `()`

代码如下。我已经多次浏览文档，但我返回了 Result<> 所以我不确定出了什么问题。

use std::sync::Arc;


#[tokio::main]
async fn main() -> Result<(), Box<dyn std::error::Error>> {
    println!("Hello, world!");

    use reqwest::{cookie::Jar, Url};
    use reqwest::header;

    let mut headers = header::HeaderMap::new();
    headers.insert("User-Agent", header::HeaderValue::from_static("Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.81 Safari/537.36"));
    headers.insert("Content-Type", header::HeaderValue::from_static("application/json"));
    
    let cookie = "cookies";

    let jar = Jar::default();
    let cookie_url = "https://url.com/".parse::<Url>()?;
    jar.add_cookie_str(cookie, &cookie_url);

    let body = String::from("post body"); 

    let client = reqwest::Client::builder()
    .default_headers(headers)
    .cookie_store(true)
    .cookie_provider(Arc::new(jar))
    .build()?;

    let path = "https://long_url.com".parse::<Url>()?;

    let res = client
        .post(path)
        .body(body)
        .send()
        .await?;

    let text = res.text().await?;
    println!("{}", text);

    Ok(());
}

谢谢你！

Answer 1

由于正在执行转换，该消息有点令人困惑，您可能想在错误跟踪器上报告这一点，但请注意措辞：

?^ 不能在返回的异步块中使用运算符()

它使用“块”这个词，而不是功能。因为这实际上是一个级联错误：在幕后，async函数被转换为状态机，其中每个状态机await都是代码块之间的“屈服点”，因此

let res = client
        .post(path)
        .body(body)
        .send()
        .await?;

    let text = res.text().await?;
    println!("{}", text);

    Ok(());

可以这样想（这不是实际的转换，如果你想要真正的解构，Jon Gjengset 有一个关于这个主题的大量视频）

let future = async {
    client.post(path).body(body).send()
};
yield future;
let future = async {
    let res = future.value()?;
    res.text()
};
yield future;
let future = async {
    let text = future.value()?;
    println!("{}", text);

    Ok(());
};
return future;

请注意最后一个块的最后一个表达式，这就是问题发生的地方：它是Ok(());.

尾随分号将“抑制”表达式的正常值并导致 a ()， not Result<...>，因此类型不兼容并且您会看到错误：块的返回值不连贯。

在一堆异步块错误的末尾，编译器实际上显示了“源”错误：

error[E0308]: mismatched types
 --> src/main.rs:6:20
  |
6 | async fn main() -> Result<(), Box<dyn std::error::Error>> {
  |          ----      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected enum `Result`, found `()`
  |          |
  |          implicitly returns `()` as its body has no tail or `return` expression
  |
  = note:   expected enum `Result<(), Box<(dyn std::error::Error + 'static)>>`
          found unit type `()`

Ok(()); 如果您查看整个“异步块”错误，它每次都会小心地显示错误：

error[E0277]: the `?` operator can only be used in an async block that returns `Result` or `Option` (or another type that implements `FromResidual`)
   --> src/lib.rs:19:55
    |
6   |   async fn main() -> Result<(), Box<dyn std::error::Error>> {
    |  ___________________________________________________________-
7   | |     println!("Hello, world!");
8   | |
9   | |     use reqwest::{cookie::Jar, Url};
...   |
19  | |     let cookie_url = "https://url.com/".parse::<Url>()?;
    | |                                                       ^ cannot use the `?` operator in an async block that returns `()`
...   |
41  | |     Ok(());
42  | | }
    | |_- this function should return `Result` or `Option` to accept `?`
    |
    = help: the trait `FromResidual<Result<Infallible, url::parser::ParseError>>` is not implemented for `()`
note: required by `from_residual`