我想知道如何创建死锁。
我试图在Rust中创建一个有死锁的程序。
如何创建一个?
答案 0 :(得分:1)
一个非常简单的变体:
use std::sync::{Arc, Mutex};
fn main() {
let data = Arc::new(Mutex::new(0));
let d1 = data.lock();
let d2 = data.lock(); // cannot lock, since d1 is still active
}
答案 1 :(得分:0)
与其他编程语言一样,Rust 可能会出现无声杀手死锁。
如果某个线程锁定一个互斥体正在等待另一个,这不是一个好兆头:如果另一个不能来,第一个互斥体将永远不会被释放。
use std::{sync::{Mutex, MutexGuard}, thread};
use std::thread::sleep;
use std::time::Duration;
use lazy_static::lazy_static;
lazy_static! {
static ref MUTEX1: Mutex<i64> = Mutex::new(0);
static ref MUTEX2: Mutex<i64> = Mutex::new(0);
}
fn main() {
// Spawn thread and store handles
let mut children = vec![];
for i_thread in 0..2 {
children.push(thread::spawn(move || {
for _ in 0..1 {
// Thread 1
if i_thread % 2 == 0 {
// Lock mutex1
// No need to specify type but yes create a dummy variable to prevent rust
// compiler from being lazy
let _guard: MutexGuard<i64> = MUTEX1.lock().unwrap();
// Just log
println!("Thread {} locked mutex1 and will try to lock the mutex2, after a nap !", i_thread);
// Here I sleep to let Thread 2 lock mutex2
sleep(Duration::from_millis(10));
// Lock mutex 2
let _guard = MUTEX2.lock().unwrap();
// Thread 2
} else {
// Lock mutex 1
let _guard = MUTEX2.lock().unwrap();
println!("Thread {} locked mutex2 and will try to lock the mutex1", i_thread);
// Here I freeze !
let _guard = MUTEX1.lock().unwrap();
}
}
}));
}
// Wait
for child in children {
let _ = child.join();
}
println!("This is not printed");
}
哪些输出
Thread 0 locked mutex1 and will try to lock the mutex2, after a nap !
Thread 1 locked mutex2 and will try to lock the mutex1
然后永远等待