Simple future example - rust

Question

I'm trying to do a simple future example. I followed https://blog.logrocket.com/a-practical-guide-to-async-in-rust/
So I did:
use futures::Future;
type Result<T> = std::result::Result<T, Box<dyn std::error::Error + Send + Sync>>;
pub async fn receive_fut(
f: &mut dyn FnMut(&[u8])
) -> impl Future<Output = Result<Vec<u8>>> {
futures::future::ok(Ok(Vec::new())).await
}
but I get
error[E0277]: `std::result::Result<std::result::Result<Vec<_>, _>, _>` is not a future
--> src/lib.rs:6:10
|
6 | ) -> impl Future<Output = Result<Vec<u8>>> {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `std::result::Result<std::result::Result<Vec<_>, _>, _>` is not a future
|
= help: the trait `futures::Future` is not implemented for `std::result::Result<std::result::Result<Vec<_>, _>, _>`
error: aborting due to previous error
Why std::result::Result should implement future, if I used futures::future::ok to turn it into a future result?

Answer 1

This is answered in the article:
Async functions differ in one important way: all your return types are “wrapped” into a Future.
You might read the documentation about Futures in Rust and think your async function needs to look like this:
async fn our_async_program() -> impl Future<Output = Result<String>> {
future::ok("Hello world".to_string()).await
}
This is wrong! If you’re doing this, you’re overthinking it. An async function already wraps the return type, so you can write functions the way you’re used to.
This is what you actually want:
async fn our_async_program() -> Result<String> {
future::ok("Hello world".to_string()).await
}
So you'd want to drop the impl Future<... part of the return type:
pub async fn receive_fut(
f: &mut dyn FnMut(&[u8])
) -> Result<Vec<u8>> {
futures::future::ok(Ok(Vec::new())).await
}
And then realize that future::ok already yields a Result::Ok, so you don't need to wrap it in Ok() again.
pub async fn receive_fut(
f: &mut dyn FnMut(&[u8])
) -> Result<Vec<u8>> {
futures::future::ok(Vec::new()).await
}