我需要一个可变的 Mutex
集合,需要在多个线程之间共享。这个集合的目的是为了根据给定的键返回 MutexGuard
列表(以便能够根据键值同步线程)。请注意,当初始化时,此集合中没有 Mutex
,这些需要根据键值在运行时创建。
我的代码如下:
use std::collections::HashMap;
use std::sync::{Arc, Mutex, MutexGuard};
struct Bucket {
locks: HashMap<String, Mutex<()>>,
}
impl Bucket {
// This method will create and add one or multiple Mutex to the
// collection (therefore it needs to take self as mutable), according
// to the give key (if the Mutex already exist it will just return
// its MutexGuard).
fn get_guards(
&mut self,
key: impl Into<String>,
) -> Vec<MutexGuard<'_, ()>> {
let lock = self.locks.entry(key.into()).or_default();
vec![lock.lock().unwrap()]
}
}
impl Default for Bucket {
fn default() -> Self {
Self {
locks: HashMap::new(),
}
}
}
fn main() {
// I need to wrap the bucket in a Arc<Mutex> since it's going to be shared
// between multiple threads
let mut bucket = Arc::new(Mutex::new(Bucket::default()));
// Here I need to get a list of guards, so that until they are dropped
// I can synchronize multiple threads with the same key (or to be more
// precise with the same MutexGuards, as different keys may yields the
// same MutexGuards).
let guards = {
// IMPORTANT: I need to unlock the Mutex used for the `bucket` (with
// write access) asap, otherwise it will nullify the purpose of the
// bucket, since the write lock would be hold for the whole `guards`
// scope.
let mut b = bucket.lock().unwrap();
b.get_guards("key")
};
}
我遇到的错误信息如下:
error[E0597]: `b` does not live long enough
--> src/main.rs:29:9
|
27 | let _guards = {
| ------- borrow later stored here
28 | let mut b = bucket.lock().unwrap();
29 | b.get_guards("key")
| ^ borrowed value does not live long enough
30 | };
| - `b` dropped here while still borrowed
error: aborting due to previous error
有没有一种方法可以设计我的Bucket
集合的Mutex
,以便我能够实现我的目标?