I'm trying to figure out how to send a function through a channel, and how to avoid extra cloning in order to execute the function at the other end. If I remove the extra cloning operation inside the closure, I get the following error:
error: cannot move out of captured outer variable in an 'Fn' closure
Ignoring the fact that this code does absolutely nothing, and makes use of a global mutable static Sender<T>
, it represents what I'm trying to achieve while giving the proper compiler errors. This code is not meant to be ran, just compiled.
use std::ops::DerefMut;
use std::sync::{Arc, Mutex};
use std::collections::LinkedList;
use std::sync::mpsc::{Sender, Receiver};
type SafeList = Arc<Mutex<LinkedList<u8>>>;
type SendableFn = Arc<Mutex<(Fn() + Send + Sync + 'static)>>;
static mut tx: *mut Sender<SendableFn> = 0 as *mut Sender<SendableFn>;
fn main() {
let list: SafeList = Arc::new(Mutex::new(LinkedList::new()));
loop {
let t_list = list.clone();
run(move || {
foo(t_list.clone());
});
}
}
fn run<T: Fn() + Send + Sync + 'static>(task: T) {
unsafe {
let _ = (*tx).send(Arc::new(Mutex::new(task)));
}
}
#[allow(dead_code)]
fn execute(rx: Receiver<SendableFn>) {
for t in rx.iter() {
let mut guard = t.lock().unwrap();
let task = guard.deref_mut();
task();
}
}
#[allow(unused_variables)]
fn foo(list: SafeList) { }
Is there a better method to getting around that error and/or another way I should be sending functions through channels?
See Question&Answers more detail:
os 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…