相关代码，部分已省略

• main.rs

#[derive(Clone)]
pub struct State {
    db_pool: sqlx::PgPool,
}

let state = State{……} //省略连接池

let mut app = tide::with_state(state);
    app.add_route(); // add_route为自己定义的方法
    app.listen(config::api_address()).await?;

• route.rs

pub trait Route
{
    fn add_route(&mut self) -> &mut Self;
}

impl<T> Route for tide::Server<T>
    where
        T: Clone + Send + Sync + 'static,
// impl Route for tide::Server<State>  换成此行，一切正常
{
    fn add_route(&mut self) -> &mut Self {
        use controller::*;

        self.at("/auth").get(auth);

• auth.rs

pub async fn auth(req: tide::Request<State>) -> tide::Result
{
    let db_pool = req.state().db_pool.clone();
    ......
}

• 报错

error[E0631]: type mismatch in function arguments
  --> xxx/src/route.rs:17:30
   |
17 |         self.at("/auth").get(auth::auth);
   |                              ^^^^^^^^^^ expected signature of `fn(tide::Request<T>) -> _`
   | 
  ::: http-api/src/controller/auth.rs:6:1
   |
6  | pub async fn auth(req: tide::Request<State>) -> tide::Result
   | ------------------------------------------------------------ found signature of `fn(tide::Request<State>) -> _`
   |
   = note: required because of the requirements on the impl of `Endpoint<T>` for `fn(tide::Request<State>) -> impl Future {controller::auth::auth}`

各位前辈，请问为啥route.rs里不用范型就能正常，使用范型后，明明这里的auth里的State满足T所需的Trait，但还是报错？