我正在使用Rust编写一个小程序，连接到RabbitMQ以发送消息（使用lapin-futures），使用以下代码会有些困难：
fn interpret_action(
channel: &channel::Channel<TcpStream>,
action: Action,
) -> impl Future<Item = (), Error = std::io::Error> + Send + 'static {
match action {
Action::DoNothing => {
debug!("Doing nothing");
futures::future::ok(())
}
Action::SendMsg(MessageDispatch { message, delay: _ }) => {
info!("Publishing a message: {:?}", message);
let mut rng = thread_rng();
let message_id: String = iter::repeat(())
.map(|()| rng.sample(Alphanumeric))
.take(16)
.collect();
channel
.basic_publish(
&*message.route.exchange,
&*message.route.routing_key,
message.body.0.into_bytes(),
channel::BasicPublishOptions::default(),
channel::BasicProperties::default()
.with_message_id(message_id)
.with_user_id("guest".to_string()),
)
.map(|_| ())
}
_ => {
debug!("Don't know what to do");
futures::future::ok(())
}
}
}
编译器在抱怨：
error[E0308]: match arms have incompatible types
--> lib/src/server/mod.rs:127:3
|
127 | / match action {
128 | | Action::DoNothing => {
129 | | debug!("Doing nothing");
130 | | futures::future::ok(())
... |
154 | | }
155 | | }
| |___^ expected struct `futures::FutureResult`, found struct `futures::Map`
|
= note: expected type `futures::FutureResult<(), _>`
found type `futures::Map<impl std::marker::Send+futures::Future, [closure@lib/src/server/mod.rs:149:14:149:20]>`
note: match arm with an incompatible type
--> lib/src/server/mod.rs:132:63
|
132 | Action::SendMsg(MessageDispatch { message, delay: _ }) => {
| _______________________________________________________________^
133 | | info!("Publishing a message: {:?}", message);
134 | | let mut rng = thread_rng();
135 | | let message_id: String = iter::repeat(())
... |
149 | | .map(|_| ())
150 | | }
| |_____^
据我了解，我需要统一分支之间的类型，但是我希望情况已经如此，因为Map和FutureResult都应与impl特性匹配。一种选择是使用有效的Box。但是我真的很想知道我是否可以避免使用Box，如果不可能，为什么不可能呢。我怀疑使用闭包类型会使事情变得不可能，但不确定。
谢谢您的帮助！