rust - 为什么没有为简单的异步 fn 返回状态实现处理程序特征？

Question

我认为应该为返回 StatusCode 的 async 函数实现 Handler:

use axum::{
    headers,
    http::StatusCode,
    routing::{delete, get, post, put},
    Router,
};

pub fn create_routes() -> Router {
    Router::new().route("/webhook", post(webhook_handler()))
}

#[debug_handler]
async fn webhook_handler() -> StatusCode {
    StatusCode::OK
}

error[E0277]: the trait bound `impl futures::Future<Output = StatusCode>: Handler<_, _>` is not satisfied
   --> src/webhook.rs:16:39
    |
16  |     Router::new().route("/webhook", post(webhook_handler()))
    |                                     ---- ^^^^^^^^^^^^^^^^^ the trait `Handler<_, _>` is not implemented for `impl futures::Future<Output = StatusCode>`
    |                                     |
    |                                     required by a bound introduced by this call
    |
    = help: the trait `Handler<T, ReqBody>` is implemented for `Layered<S, T>`
note: required by a bound in `axum::routing::post`
   --> /home/andrew/.cargo/registry/src/github.com-1ecc6299db9ec823/axum-0.5.11/src/routing/method_routing.rs:400:1
    |
400 | top_level_handler_fn!(post, POST);
    | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ required by this bound in `axum::routing::post`
    = note: this error originates in the macro `top_level_handler_fn` (in Nightly builds, run with -Z macro-backtrace for more info)

这看起来与 the documentation 中提供的示例几乎相同:

// `StatusCode` gives an empty response with that status code
async fn status() -> StatusCode {
    StatusCode::NOT_FOUND
}

Answer 1

函数本身需要作为参数传递给 post，而不是函数调用。删除了括号，如下所示: .route("webhook", post(webhook_handler))