(2019-06-02)
When the producer sends messages too frequently, and the producer takes too much time to process each message. I wrote a code for
I want to discard aging messages and process only the latest one. So I wrote a code for repeating receive messages until the channel is empty for a short duration. However, I found my code was too complicated. So I use shared memory with mutex-lock for inter-thread communication instead of a channel.
use std::sync::{Mutex, Arc};
use std::{thread, time};
use std::collections::HashMap;
fn main() {
let r_shared_data = Arc::new(Mutex::new(None));
let w_shared_data = r_shared_data.clone();
thread::spawn(move || {
let tx_delay_time = time::Duration::from_millis(100);
loop {
for i in 1..1000 {
let mut h = HashMap::new();
h.insert("X", i);
let mut w = w_shared_data.lock().unwrap();
*w = Some(h);
if i % 10 == 0 {
thread::sleep(tx_delay_time);
}
}
}
});
let task_proc_time = time::Duration::from_millis(500);
loop {
let r = r_shared_data.lock().unwrap();
dbg!(r);
// long task
thread::sleep(task_proc_time);
}
}
`
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