Multithreading 如何";解锁";一个锁?
我正在努力解决这个问题。在每个线程中,我读取令牌值Multithreading 如何";解锁";一个锁?,multithreading,rust,mutex,Multithreading,Rust,Mutex,我正在努力解决这个问题。在每个线程中,我读取令牌值 如果它不是我的,检查它是否是节目的结尾 如果是,则完成线程 否则,请再次阅读并重复 如果它是我的(即有我的id),那么获取写锁,增加令牌的值,检查它是否结束,然后告诉主线程我完成了它并完成了当前线程循环 如果没有结束,则释放写锁,然后重新开始读取 没有解锁。这里有我需要的那种解锁吗 似乎我也应该释放读锁,因为如果有人读取数据,写锁就不会发生。有必要吗 fn main() { use std::sync::{Arc, RwLock
- 如果它不是我的,检查它是否是节目的结尾
- 如果是,则完成线程
- 否则,请再次阅读并重复
- 如果它是我的(即有我的id),那么获取写锁,增加令牌的值,检查它是否结束,然后告诉主线程我完成了它并完成了当前线程循环
- 如果没有结束,则释放写锁,然后重新开始读取
fn main() {
use std::sync::{Arc, RwLock};
use std::thread;
use std::sync::mpsc::channel;
const N: usize = 5; //503;
const STOP_POINT: usize = 100;
let n = Arc::new(RwLock::new(1));
let (sender, reciever) = channel();
for i in 1..N {
let (n_c, channel) = (n.clone(), sender.clone());
// println!("Thread n.{} beeing created!", i);
let a = thread::Builder::new()
.name(i.to_string())
.spawn(move || -> () {
loop {
let mut read_only = n_c.read().unwrap();
let say_my_name = (*thread::current().name().unwrap()).to_string();
// println!("Thread {} says: gonna try!", say_my_name);
while (*read_only % N) != i {
if *read_only == 0 {
break;
}
// println!("Thread {} says: aint mine!", say_my_name);
read_only = n_c.read().unwrap();
} // WAIT
println!("Thread {} says: my turn!", say_my_name);
let mut ref_to_num = n_c.write().unwrap();
*ref_to_num += 1;
if *ref_to_num == STOP_POINT {
channel.send(say_my_name).unwrap();
break;
}
}
()
});
assert_eq!(a.is_ok(), true);
// thread::spawn();
// println!("Thread n.{} created!", i);
}
println!("{}", reciever.recv().unwrap());
}
要释放锁,可以让它脱离作用域,或者通过调用显式调用其析构函数 下面介绍如何在两个位置使用
drop
编写程序:
fn main() {
use std::sync::{Arc, RwLock};
use std::sync::mpsc::channel;
use std::thread;
use std::time::Duration;
const N: usize = 503;
const STOP_POINT: usize = 100;
let n = Arc::new(RwLock::new(1));
let (sender, receiver) = channel();
for i in 1..N {
let (n_c, channel) = (n.clone(), sender.clone());
// println!("Thread n.{} beeing created!", i);
thread::Builder::new()
.name(i.to_string())
.spawn(move || {
loop {
let mut read_only = n_c.read().unwrap();
let say_my_name = (*thread::current().name().unwrap()).to_string();
// println!("Thread {} says: gonna try!", say_my_name);
while (*read_only % N) != i {
if *read_only == 0 {
break;
}
drop(read_only); // release the lock before sleeping
// println!("Thread {} says: aint mine!", say_my_name);
thread::sleep(Duration::from_millis(1));
read_only = n_c.read().unwrap();
}
println!("Thread {} says: my turn!", say_my_name);
drop(read_only); // release the read lock before taking a write lock
let mut ref_to_num = n_c.write().unwrap();
*ref_to_num += 1;
if *ref_to_num == STOP_POINT {
channel.send(say_my_name).unwrap();
break;
}
}
})
.expect("failed to spawn a thread");
// println!("Thread n.{} created!", i);
}
println!("{}", receiver.recv().unwrap());
}
请注意,如果我们在while
循环中不重新分配read\u lock
,编译器将给出一个错误,因为调用drop(read\u lock)
后read\u lock
没有保存有效值。对于暂时未初始化的局部变量,Rust是可以接受的,但在再次使用它们之前,我们当然需要重新初始化它们
下面是如何编写线程的主循环,以使用作用域替换一个drop
s:
loop {
let say_my_name = (*thread::current().name().unwrap()).to_string();
{
let mut read_only = n_c.read().unwrap();
// println!("Thread {} says: gonna try!", say_my_name);
while (*read_only % N) != i {
if *read_only == 0 {
break;
}
drop(read_only);
thread::sleep(Duration::from_millis(1));
// println!("Thread {} says: aint mine!", say_my_name);
read_only = n_c.read().unwrap();
}
println!("Thread {} says: my turn!", say_my_name);
} // read_only is dropped here
let mut ref_to_num = n_c.write().unwrap();
*ref_to_num += 1;
if *ref_to_num == STOP_POINT {
channel.send(say_my_name).unwrap();
break;
}
}
谢谢我有点理解,如果一个锁不在作用域内,那么它就会被释放,但我可以想出如何使它发生在写锁上。这滴水正是我需要的。