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如何将async fn放入Rust中的映射?

hyper async-await rust

1

Ukonn Ra · 技术社区 · 6 年前

hyper .

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>>>`

似乎两者之间有冲突 impl Future dyn Future 但我不知道怎么处理。

0 回复 | 直到 6 年前

1

5

Ömer Erden Eugen Dück 6 年前

这是因为 impl Future 是一个具体的独特类型 dyn Future 是一种抽象类型。 HashMap 应为抽象类型,因为它只能包含单个类型的实例。

如果我们可以将异步函数的返回类型装箱,我们将能够将这些未来添加到 .

CalcFn

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

这样就可以做到:

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 complete example 简化 Future Item 类型,使用 i32 而不是 Result . 请同时检查 full code for your case .