我想用hyper编写一个简单的微服务：
use hyper::service::{make_service_fn, service_fn};
use hyper::{Body, Request, Response, Server};
use hyper::{Method, StatusCode};
use std::{convert::Infallible, net::SocketAddr};
struct ApiServer {}
impl ApiServer {
pub async fn route(&mut self, req: Request<Body>) -> Result<Response<Body>, Infallible> {
let mut response = Response::new(Body::empty());
*response.body_mut() = req.into_body();
Ok(response)
}
}
#[tokio::main]
async fn main() {
let addr = SocketAddr::from(([127, 0, 0, 1], 3000));
let mut api = ApiServer {};
let make_svc = make_service_fn(|_conn| async {
Ok::<_, Infallible>(service_fn(|req| async { api.route(req) }))
});
let server = Server::bind(&addr).serve(make_svc);
if let Err(e) = server.await {
eprintln!("server error: {}", e);
}
}
我收到编译器错误：
error[E0271]: type mismatch resolving `<impl std::future::Future as std::future::Future>::Output == std::result::Result<http::response::Response<_>, _>`
--> src/master/bin/main.rs:46:38
|
46 | let server = Server::bind(&addr).serve(make_svc);
| ^^^^^ expected opaque type, found enum `std::result::Result`
|
如何重写代码以使route函数保留为ApiServer的成员函数？
Rust错误消息与您期望的相反：serve()期望为Result，但找到了opaque类型。您需要等待api.route(req)的返回值。修复其他编译错误，最终代码为：
use hyper::service::{make_service_fn, service_fn};
use hyper::{Body, Request, Response, Server};
use std::{convert::Infallible, net::SocketAddr};
#[derive(Clone, Copy)]
struct ApiServer {}
impl ApiServer {
pub async fn route(&mut self, req: Request<Body>) -> Result<Response<Body>, Infallible> {
let mut response = Response::new(Body::empty());
*response.body_mut() = req.into_body();
Ok(response)
}
}
#[tokio::main]
async fn main() {
let mut api = ApiServer {};
let make_svc = make_service_fn(move |_conn| async move {
Ok::<_, Infallible>(service_fn(move |req| async move { api.route(req).await }))
});
let addr = SocketAddr::from(([127, 0, 0, 1], 3000));
let server = Server::bind(&addr).serve(make_svc);
if let Err(e) = server.await {
eprintln!("server error: {}", e);
}
}
参见：