RUST——互斥锁的使用

互斥锁是Rust语言中所提供的，基于共享内存的方式来实现线程之间通信的机制。

Mutex的简单使用

Mutex是rust标准库中提供的互斥锁的实现。使用的示例代码如下所示：

use std::sync::Mutex;

fn main() {
    let m = Mutex::new(5);

    {
        let mut num = m.lock().unwrap();
        *num = 6;
    }

    println!("m = {:?}", m);
}

共享变量是存储于Mutex对象的内部的，如果要访问共享的数据，需要先获取锁，然后才能访问内部的数据。

Mutex在多线程之间的共享

下面我们尝试在多个线程之间共享Mutex对象，以实现对数据的共享。示例代码如下：

use std::sync::Mutex;
use std::thread;

fn main() {
    let counter = Mutex::new(0);
    let mut handles = vec![];

    for _ in 0..10 {
        let handle = thread::spawn(move || {
            let mut num = counter.lock().unwrap();

            *num += 1;
        });
        handles.push(handle);
    }

    for handle in handles {
        handle.join().unwrap();
    }

    println!("Result: {}", *counter.lock().unwrap());
}

如果我们编译上述代码，会得到如下的结果：

   Compiling playground v0.0.1 (/playground)
error[E0382]: use of moved value: `counter`
  --> src/main.rs:9:36
   |
5  |     let counter = Mutex::new(0);
   |         ------- move occurs because `counter` has type `Mutex<i32>`, which does not implement the `Copy` trait
...
9  |         let handle = thread::spawn(move || {
   |                                    ^^^^^^^ value moved into closure here, in previous iteration of loop
10 |             let mut num = counter.lock().unwrap();
   |                           ------- use occurs due to use in closure

For more information about this error, try `rustc --explain E0382`.
error: could not compile `playground` due to previous error

显然，在创建线程的时候，使用move语义，Mutex对象的所有权发生了转移，因而无法实现在多个线程之间共享Mutex的对象。如果想要实现Mutex对象在多个线程之间共享，必须使用具有多所有权的对象，我们很自然的就想到了rust中基于引用计数实现的智能指针Rc。

基于Rc实现Mutex在多线程之间的共享

基于Rc的机制实现Mutex在多线程之间的共享的示例代码如下：

use std::rc::Rc;
use std::sync::Mutex;
use std::thread;

fn main() {
    let counter = Rc::new(Mutex::new(0));
    let mut handles = vec![];

    for _ in 0..10 {
        let counter = Rc::clone(&counter);
        let handle = thread::spawn(move || {
            let mut num = counter.lock().unwrap();

            *num += 1;
        });
        handles.push(handle);
    }

    for handle in handles {
        handle.join().unwrap();
    }

    println!("Result: {}", *counter.lock().unwrap());
}

如果我们编译上述代码，会得到如下的编译结果：

   Compiling playground v0.0.1 (/playground)
error[E0277]: `Rc<Mutex<i32>>` cannot be sent between threads safely
   --> src/main.rs:11:22
    |
11  |           let handle = thread::spawn(move || {
    |  ______________________^^^^^^^^^^^^^_-
    | |                      |
    | |                      `Rc<Mutex<i32>>` cannot be sent between threads safely
12  | |             let mut num = counter.lock().unwrap();
13  | |
14  | |             *num += 1;
15  | |         });
    | |_________- within this `[closure@src/main.rs:11:36: 15:10]`
    |
    = help: within `[closure@src/main.rs:11:36: 15:10]`, the trait `Send` is not implemented for `Rc<Mutex<i32>>`
    = note: required because it appears within the type `[closure@src/main.rs:11:36: 15:10]`
note: required by a bound in `spawn`

For more information about this error, try `rustc --explain E0277`.
error: could not compile `playground` due to previous error

从上述结果中，我们可以看到，Rc的实现的引用计数，并不是线程安全的，无法在多线程之间传递。很自然地，我们可以使用rust中的线程安全的引用计数实现机制Arc，A代表的是Atomic，是原子性的操作，因而可以保证线程的安全。

基于Arc实现Mutex在多线程之间的共享

基于Arc实现的引用计数，来实现在Mutex在多线程之间的共享的示例代码如下：

use std::sync::{Arc, Mutex};
use std::thread;

fn main() {
    let counter = Arc::new(Mutex::new(0));
    let mut handles = vec![];

    for _ in 0..10 {
        let counter = Arc::clone(&counter);
        let handle = thread::spawn(move || {
            let mut num = counter.lock().unwrap();

            *num += 1;
        });
        handles.push(handle);
    }

    for handle in handles {
        handle.join().unwrap();
    }

    println!("Result: {}", *counter.lock().unwrap());
}

运行上述代码，结果如下：

Result: 10

标签：handle,unwrap,counter,互斥,num,let,使用,Mutex,RUST
来源： https://blog.csdn.net/z974656361/article/details/122508749