Async v5: Printing the payload

[iot-resister]

Is this how the simplest/idiomatic way to get the payload?

#[tokio::main(worker_threads = 1)]
async fn main() -> Result<(), Box<dyn Error>> {
    let mut options = MqttOptions::new("test-id", "localhost", 1883);
    options.set_keep_alive(Duration::from_secs(5));
    let (client, mut eventloop) = AsyncClient::new(options, 10);
    client.subscribe("foo", QoS::AtMostOnce).await.unwrap();
    client
        .publish("foo", QoS::AtMostOnce, false, "bar")
        .await
        .unwrap();
    while let Ok(event) = eventloop.poll().await {
        if let Event::Incoming(b) = event {
            if let Packet::Publish(p, None) = *b {
                println!("Received: {:?}", p.payload);
            }
        }
    }
    Ok(())
}

[de-sh]

Looks good, but I'd usually do it like this to make the branches more explicit:

loop {
        let packet = match eventloop.poll().await {
            Ok(Event::Incoming(b)) => *b,
            Err(_) => break,
            _ => continue,
        };

        if let Packet::Publish(p, None) = packet {
            println!("Received: {:?}", p.payload);
        }
    }