如何在Rust中将异步函数放入地图中？

为编写异步路由器时，我无法处理异步功能hyper。

这段代码：

use std::collections::HashMap;
use std::future::Future;

type BoxedResult<T> = Result<T, Box<dyn std::error::Error + Send + Sync>>;
type CalcFn = Box<dyn Fn(i32, i32) -> dyn Future<Output = BoxedResult<i32>>>;

async fn add(a: i32, b: i32) -> BoxedResult<i32> {
    Ok(a + b)
}

async fn sub(a: i32, b: i32) -> BoxedResult<i32> {
    Ok(a - b)
}

fn main() {
    let mut map: HashMap<&str, CalcFn> = Default::default();
    map.insert("add", Box::new(add));
    map.insert("sub", Box::new(sub));

    println!("map size: {}", map.len());
}

生成以下编译器错误：

error[E0271]: type mismatch resolving `<fn(i32, i32) -> impl std::future::Future {add} as std::ops::FnOnce<(i32, i32)>>::Output == dyn std::future::Future<Output = std::result::Result<i32, std::boxed::Box<dyn std::error::Error + std::marker::Send + std::marker::Sync>>>`
  --> src/main.rs:17:23
   |
17 |     map.insert("add", Box::new(add));
   |                       ^^^^^^^^^^^^^ expected opaque type, found trait std::future::Future
   |
   = note: expected type `impl std::future::Future`
              found type `dyn std::future::Future<Output = std::result::Result<i32, std::boxed::Box<dyn std::error::Error + std::marker::Send + std::marker::Sync>>>`
   = note: required for the cast to the object type `dyn std::ops::Fn(i32, i32) -> dyn std::future::Future<Output = std::result::Result<i32, std::boxed::Box<dyn std::error::Error + std::marker::Send + std::marker::Sync>>>`

error[E0271]: type mismatch resolving `<fn(i32, i32) -> impl std::future::Future {sub} as std::ops::FnOnce<(i32, i32)>>::Output == dyn std::future::Future<Output = std::result::Result<i32, std::boxed::Box<dyn std::error::Error + std::marker::Send + std::marker::Sync>>>`
  --> src/main.rs:18:23
   |
18 |     map.insert("sub", Box::new(sub));
   |                       ^^^^^^^^^^^^^ expected opaque type, found trait std::future::Future
   |
   = note: expected type `impl std::future::Future`
              found type `dyn std::future::Future<Output = std::result::Result<i32, std::boxed::Box<dyn std::error::Error + std::marker::Send + std::marker::Sync>>>`
   = note: required for the cast to the object type `dyn std::ops::Fn(i32, i32) -> dyn std::future::Future<Output = std::result::Result<i32, std::boxed::Box<dyn std::error::Error + std::marker::Send + std::marker::Sync>>>`

impl Future和之间似乎存在冲突dyn Future，但是我不知道如何处理。

发生这种情况是因为impl Future它是具体的唯一类型，dyn Future而是抽象类型。HashMap期望抽象类型，因为它只能容纳单个类型的实例。

如果我们可以对异步函数的返回类型进行装箱，则可以将这些期货添加到中HashMap。

首先，我们需要更改以下类型CalcFn：

type CalcFn = Box<dyn Fn(i32, i32) -> Pin<Box<dyn Future<Output = i32>>>>;

然后可以做到这一点：

let mut map: HashMap<&str, CalcFn> = Default::default();
map.insert("add", Box::new(|a, b| Box::pin(add(a, b))));
map.insert("sub", Box::new(|a, b| Box::pin(sub(a, b))));

println!("map size: {}", map.len());

//map.get("add").unwrap()(2, 3).await

这个完整的示例 简化Future的Item类型，使用i32的而不是一个Result。请也检查您的情况的完整代码。