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< h1 id = "构建多线程-web-服务器" > < a class = "header" href = "#构建多线程-web-服务器" > 构建多线程 Web 服务器< / a > < / h1 >
< p > 目前的单线程版本只能依次处理用户的请求：一时间只能处理一个请求连接。随着用户的请求数增多，可以预料的是排在后面的用户可能要等待数十秒甚至超时！< / p >
< p > 本章我们将解决这个问题，但是首先来模拟一个慢请求场景，看看单线程是否真的如此糟糕。< / p >
< h2 id = "基于单线程模拟慢请求" > < a class = "header" href = "#基于单线程模拟慢请求" > 基于单线程模拟慢请求< / a > < / h2 >
< p > 下面的代码中，使用 sleep 的方式让每次请求持续 5 秒，模拟真实的慢请求:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > // in main.rs
use std::{
fs,
io::{prelude::*, BufReader},
net::{TcpListener, TcpStream},
thread,
time::Duration,
};
// --snip--
fn handle_connection(mut stream: TcpStream) {
// --snip--
let (status_line, filename) = match & request_line[..] {
"GET / HTTP/1.1" => ("HTTP/1.1 200 OK", "hello.html"),
"GET /sleep HTTP/1.1" => {
thread::sleep(Duration::from_secs(5));
("HTTP/1.1 200 OK", "hello.html")
}
_ => ("HTTP/1.1 404 NOT FOUND", "404.html"),
};
// --snip--
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 由于增加了新的请求路径 < code > /sleep< / code > ，之前的 < code > if else< / code > 被修改为 < code > match< / code > ，需要注意的是，由于 < code > match< / code > 不会像方法那样自动做引用或者解引用，因此我们需要显式调用: < code > match & request_line[..]< / code > ，来获取所需的 < code > & str< / code > 类型。< / p >
< p > 可以看出，当用户访问 < code > /sleep< / code > 时，请求会持续 5 秒后才返回，下面来试试，启动服务器后，打开你的浏览器，这次要分别打开两个页面(tab页): < code > http://127.0.0.1:7878/< / code > 和 < code > http://127.0.0.1:7878/sleep< / code > 。< / p >
< p > 此时，如果我们连续访问 < code > /< / code > 路径，那效果跟之前一样：立刻看到请求的页面。但假如先访问 < code > /sleep< / code > ，接着在另一个页面访问 < code > /< / code > ，就会看到 < code > /< / code > 的页面直到 5 秒后才会刷出来，验证了请求排队这个糟糕的事实。< / p >
< p > 至于如何解决，其实办法不少，本章我们来看看一个经典解决方案：线程池。< / p >
< h2 id = "使用线程池改善吞吐" > < a class = "header" href = "#使用线程池改善吞吐" > 使用线程池改善吞吐< / a > < / h2 >
< p > 线程池包含一组已生成的线程，它们时刻等待着接收并处理新的任务。当程序接收到新任务时，它会将线程池中的一个线程指派给该任务，在该线程忙着处理时，新来的任务会交给池中剩余的线程进行处理。最终，当执行任务的线程处理完后，它会被重新放入到线程池中，准备处理新任务。< / p >
< p > 假设线程池中包含 N 个线程，那么可以推断出，服务器将拥有并发处理 N 个请求连接的能力，从而增加服务器的吞吐量。< / p >
< p > 同时， ， （ ） ： ， ， ， ， < / p >
< p > 因此，还需对线程池进行一定的架构设计，首先是设定最大线程数的上限，其次维护一个请求队列。池中的线程去队列中依次弹出请求并处理。这样就可以同时并发处理 N 个请求，其中 N 是线程数。< / p >
< p > 但聪明的读者可能会想到，假如每个请求依然耗时很长，那请求队列依然会堆积，后续的用户请求还是需要等待较长的时间，毕竟你也就 N 个线程，但总归比单线程要强 N 倍吧 :D< / p >
< p > 当然，线程池依然是较为传统的提升吞吐方法，比较新的有：单线程异步 IO， ； ， < / p >
< h3 id = "为每个请求生成一个线程" > < a class = "header" href = "#为每个请求生成一个线程" > 为每个请求生成一个线程< / a > < / h3 >
< p > 这显然不是我们的最终方案，原因在于它会生成无上限的线程数，最终导致资源耗尽。但它确实是一个好的起点:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > fn main() {
let listener = TcpListener::bind("127.0.0.1:7878").unwrap();
for stream in listener.incoming() {
let stream = stream.unwrap();
thread::spawn(|| {
handle_connection(stream);
});
}
}< / code > < / pre >
< p > 这种实现下，依次访问 < code > /sleep< / code > 和 < code > /< / code > 就无需再等待，不错的开始。< / p >
< h3 id = "限制创建线程的数量" > < a class = "header" href = "#限制创建线程的数量" > 限制创建线程的数量< / a > < / h3 >
< p > 原则上，我们希望在上面代码的基础上，尽量少的去修改，下面是一个假想的线程池 API 实现:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > fn main() {
let listener = TcpListener::bind("127.0.0.1:7878").unwrap();
let pool = ThreadPool::new(4);
for stream in listener.incoming() {
let stream = stream.unwrap();
pool.execute(|| {
handle_connection(stream);
});
}
}< / code > < / pre >
< p > 代码跟之前的类似，也非常简洁明了， < code > ThreadPool::new(4)< / code > 创建一个包含 4 个线程的线程池，接着通过 < code > pool.execute< / code > 去分发执行请求。< / p >
< p > 显然，上面的代码无法编译，下面来逐步实现。< / p >
< h3 id = "使用编译器驱动的方式开发-threadpool" > < a class = "header" href = "#使用编译器驱动的方式开发-threadpool" > 使用编译器驱动的方式开发 ThreadPool< / a > < / h3 >
< p > 你可能听说过测试驱动开发，但听过编译器驱动开发吗？来见识下 Rust 中的绝招吧。< / p >
< p > 检查之前的代码，看看报什么错:< / p >
< pre > < code class = "language-shell" > $ cargo check
Checking hello v0.1.0 (file:///projects/hello)
error[E0433]: failed to resolve: use of undeclared type `ThreadPool`
--> src/main.rs:11:16
|
11 | let pool = ThreadPool::new(4);
| ^^^^^^^^^^ use of undeclared type `ThreadPool`
For more information about this error, try `rustc --explain E0433`.
error: could not compile `hello` due to previous error
< / code > < / pre >
< p > 俗话说，不怕敌人很强，就怕他们不犯错，很好，编译器漏出了破绽。看起来我们需要实现 < code > ThreadPool< / code > 类型。看起来，还需要添加一个库包，未来线程池的代码都将在这个独立的包中完成，甚至于未来你要实现其它的服务，也可以复用这个多线程库包。< / p >
< p > 创建 < code > src/lib.rs< / code > 文件并写入如下代码:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub struct ThreadPool;
< span class = "boring" > }< / span > < / code > < / pre >
< p > 接着在 < code > main.rs< / code > 中引入:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > // main.rs
use hello::ThreadPool;
< span class = "boring" > }< / span > < / code > < / pre >
< p > 编译后依然报错:< / p >
< pre > < code class = "language-shell" > $ cargo check
Checking hello v0.1.0 (file:///projects/hello)
error[E0599]: no function or associated item named `new` found for struct `ThreadPool` in the current scope
--> src/main.rs:12:28
|
12 | let pool = ThreadPool::new(4);
| ^^^ function or associated item not found in `ThreadPool`
For more information about this error, try `rustc --explain E0599`.
error: could not compile `hello` due to previous error
< / code > < / pre >
< p > 好，继续实现 < code > new< / code > 函数 :< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub struct ThreadPool;
impl ThreadPool {
pub fn new(size: usize) -> ThreadPool {
ThreadPool
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 继续检查：< / p >
< pre > < code class = "language-shell" > $ cargo check
Checking hello v0.1.0 (file:///projects/hello)
error[E0599]: no method named `execute` found for struct `ThreadPool` in the current scope
--> src/main.rs:17:14
|
17 | pool.execute(|| {
| ^^^^^^^ method not found in `ThreadPool`
For more information about this error, try `rustc --explain E0599`.
error: could not compile `hello` due to previous error
< / code > < / pre >
< p > 这个方法类似于 < code > thread::spawn< / code > ，用于将闭包中的任务交给某个空闲的线程去执行。< / p >
< p > 其实这里有一个小难点：< code > execute< / code > 的参数是一个闭包，回忆下之前学过的内容，闭包作为参数时可以由三个特征进行约束: < code > Fn< / code > 、< code > FnMut< / code > 和 < code > FnOnce< / code > ，选哪个就成为一个问题。由于 < code > execute< / code > 在实现上类似 < code > thread::spawn< / code > ，我们可以参考下后者的签名如何声明。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub fn spawn< F, T> (f: F) -> JoinHandle< T>
where
F: FnOnce() -> T,
F: Send + 'static,
T: Send + 'static,
< span class = "boring" > }< / span > < / code > < / pre >
< p > 可以看出，< code > spawn< / code > 选择 < code > FnOnce< / code > 作为 < code > F< / code > 闭包的特征约束，原因是闭包作为任务只需被线程执行一次即可。< / p >
< p > < code > F< / code > 还有一个特征约束 < code > Send< / code > ，也可以照抄过来，毕竟闭包需要从一个线程传递到另一个线程，至于生命周期约束 < code > 'static< / code > ，是因为我们并不知道线程需要多久时间来执行该任务。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > impl ThreadPool {
// --snip--
pub fn execute< F> (& self, f: F)
where
F: FnOnce() + Send + 'static,
{
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 在理解 < code > spawn< / code > 后，就可以轻松写出如上的 < code > execute< / code > 实现，注意这里的 < code > FnOnce()< / code > 跟 < code > spawn< / code > 有所不同，原因是要 < code > execute< / code > 传入的闭包没有参数也没有返回值。< / p >
< pre > < code class = "language-shell" > $ cargo check
Checking hello v0.1.0 (file:///projects/hello)
Finished dev [unoptimized + debuginfo] target(s) in 0.24s
< / code > < / pre >
< p > 成功编译，但在浏览器访问依然会报之前类似的错误，下面来实现 < code > execute< / code > 。< / p >
< h3 id = "new-还是-build" > < a class = "header" href = "#new-还是-build" > < code > new< / code > 还是 < code > build< / code > < / a > < / h3 >
< p > 关于 < code > ThreadPool< / code > 的构造函数，存在两个选择 < code > new< / code > 和 < code > build< / code > 。< / p >
< p > < code > new< / code > 往往用于简单初始化一个实例，而 < code > build< / code > 往往会完成更加复杂的构建工作，例如入门实战中的 < code > Config::build< / code > 。< / p >
< p > 在这个项目中，我们并不需要在初始化线程池的同时创建相应的线程，因此 < code > new< / code > 是更适合的选择:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > impl ThreadPool {
/// Create a new ThreadPool.
///
/// The size is the number of threads in the pool.
///
/// # Panics
///
/// The `new` function will panic if the size is zero.
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
ThreadPool
}
// --snip--
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 这里有两点值得注意:< / p >
< ul >
< li > < code > usize< / code > 类型包含 < code > 0< / code > ，但是创建没有任何线程的线程池显然是无意义的，因此做一下 < code > assert!< / code > 验证< / li >
< li > < code > ThreadPool< / code > 拥有不错的< a href = "https://course.rs/basic/comment.html#文档注释" > 文档注释< / a > ，甚至包含了可能 < code > panic< / code > 的情况，通过 < code > cargo doc --open< / code > 可以访问文档注释< / li >
< / ul >
< h3 id = "存储线程" > < a class = "header" href = "#存储线程" > 存储线程< / a > < / h3 >
< p > 创建 < code > ThreadPool< / code > 后，下一步就是存储具体的线程，既然要存放线程，一个绕不过去的问题就是：用什么类型来存放，例如假如使用 < code > Vec< T> < / code > 来存储，那这个 < code > T< / code > 应该是什么？< / p >
< p > 估计还得探索下 < code > thread::spawn< / code > 的签名，毕竟它生成并返回一个线程:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub fn spawn< F, T> (f: F) -> JoinHandle< T>
where
F: FnOnce() -> T,
F: Send + 'static,
T: Send + 'static,
< span class = "boring" > }< / span > < / code > < / pre >
< p > 看起来 < code > JoinHandle< T> < / code > 是我们需要的，这里的 < code > T< / code > 是传入的闭包任务所返回的，我们的任务无需任何返回，因此 < code > T< / code > 直接使用 < code > ()< / code > 即可。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > use std::thread;
pub struct ThreadPool {
threads: Vec< thread::JoinHandle< ()> > ,
}
impl ThreadPool {
// --snip--
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
let mut threads = Vec::with_capacity(size);
for _ in 0..size {
// create some threads and store them in the vector
}
ThreadPool { threads }
}
// --snip--
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 如上所示，最终我们使用 < code > Vec< thread::JoinHandle< ()> > < / code > 来存储线程，同时设定了容量上限 < code > with_capacity(size)< / code > ，该方法还可以提前分配好内存空间，比 < code > Vec::new< / code > 的性能要更好一点。< / p >
< h3 id = "将代码从-threadpool-发送到线程中" > < a class = "header" href = "#将代码从-threadpool-发送到线程中" > 将代码从 ThreadPool 发送到线程中< / a > < / h3 >
< p > 上面的代码留下一个未实现的 < code > for< / code > 循环，用于创建和存储线程。< / p >
< p > 学过多线程一章后，大家应该知道 < code > thread::spawn< / code > 虽然是生成线程最好的方式，但是它会立即执行传入的任务，然而，在我们的使用场景中，创建线程和执行任务明显是要分离的，因此标准库看起来不再适合。< / p >
< p > 可以考虑创建一个 < code > Worker< / code > 结构体，作为 < code > ThreadPool< / code > 和任务线程联系的桥梁，它的任务是获得将要执行的代码，然后在具体的线程中去执行。想象一个场景：一个餐馆，< code > Worker< / code > 等待顾客的点餐，然后将具体的点餐信息传递给厨房，感觉类似服务员？< / p >
< p > 引入 < code > Worker< / code > 后，就无需再存储 < code > JoinHandle< ()> < / code > 实例，直接存储 < code > Worker< / code > 实例：该实例内部会存储 < code > JoinHandle< ()> < / code > 。下面是新的线程池创建流程:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > use std::thread;
pub struct ThreadPool {
workers: Vec< Worker> ,
}
impl ThreadPool {
// --snip--
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
let mut workers = Vec::with_capacity(size);
for id in 0..size {
workers.push(Worker::new(id));
}
ThreadPool { workers }
}
// --snip--
}
struct Worker {
id: usize,
thread: thread::JoinHandle< ()> ,
}
impl Worker {
fn new(id: usize) -> Worker {
// 尚未实现..
let thread = thread::spawn(|| {});
// 每个 `Worker` 都拥有自己的唯一 id
Worker { id, thread }
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 由于外部调用者无需知道 < code > Worker< / code > 的存在，因此这里使用了私有的声明。< / p >
< p > 大家可以编译下代码，如果出错了，请仔细检查下，是否遗漏了什么，截止目前，代码是完全可以通过编译的，但是任务该怎么执行依然还没有实现。< / p >
< h3 id = "将请求发送给线程" > < a class = "header" href = "#将请求发送给线程" > 将请求发送给线程< / a > < / h3 >
< p > 在上面的代码中， < code > thread::spawn(|| {})< / code > 还没有给予实质性的内容，现在一起来完善下。< / p >
< p > 首先 < code > Worker< / code > 结构体需要从线程池 < code > ThreadPool< / code > 的队列中获取待执行的代码，对于这类场景，消息传递非常适合：我们将使用消息通道( channel )作为任务队列。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > use std::{sync::mpsc, thread};
pub struct ThreadPool {
workers: Vec< Worker> ,
sender: mpsc::Sender< Job> ,
}
struct Job;
impl ThreadPool {
// --snip--
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
let (sender, receiver) = mpsc::channel();
let mut workers = Vec::with_capacity(size);
for id in 0..size {
workers.push(Worker::new(id));
}
ThreadPool { workers, sender }
}
// --snip--
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 阅读过之前内容的同学应该知道，消息通道有发送端和接收端，其中线程池 < code > ThreadPool< / code > 持有发送端，通过 < code > execute< / code > 方法来发送任务。那么问题来了，谁持有接收端呢？答案是 < code > Worker< / code > ，它的内部线程将接收任务，然后进行处理。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > impl ThreadPool {
// --snip--
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
let (sender, receiver) = mpsc::channel();
let mut workers = Vec::with_capacity(size);
for id in 0..size {
workers.push(Worker::new(id, receiver));
}
ThreadPool { workers, sender }
}
// --snip--
}
// --snip--
impl Worker {
fn new(id: usize, receiver: mpsc::Receiver< Job> ) -> Worker {
let thread = thread::spawn(|| {
receiver;
});
Worker { id, thread }
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 看起来很美好，但是很不幸，它会报错:< / p >
< pre > < code class = "language-shell" > $ cargo check
Checking hello v0.1.0 (file:///projects/hello)
error[E0382]: use of moved value: `receiver`
--> src/lib.rs:26:42
|
21 | let (sender, receiver) = mpsc::channel();
| -------- move occurs because `receiver` has type `std::sync::mpsc::Receiver< Job> `, which does not implement the `Copy` trait
...
26 | workers.push(Worker::new(id, receiver));
| ^^^^^^^^ value moved here, in previous iteration of loop
For more information about this error, try `rustc --explain E0382`.
error: could not compile `hello` due to previous error
< / code > < / pre >
< p > 原因也很简单，< code > receiver< / code > 并没有实现 < code > Copy< / code > ，因此它的所有权在第一次循环中，就被传入到第一个 < code > Worker< / code > 实例中，后续自然无法再使用。< / p >
< p > 报错就解决呗，但 Rust 中的 channel 实现是 mpsc， ， ， ， ？ < / p >
< p > 雪上加霜的是，就算 < code > receiver< / code > 可以克隆，但是你得保证同一个时间只有一个< code > receiver< / code > 能接收消息，否则一个任务可能同时被多个 < code > Worker< / code > 执行，因此多个线程需要安全的共享和使用 < code > receiver< / code > ，等等，安全的共享？听上去 < code > Arc< / code > 这个多所有权结构非常适合，互斥使用？貌似 < code > Mutex< / code > 很适合，结合一下，< code > Arc< Mutex< T> > < / code > ，这不就是我们之前见过多次的线程安全类型吗？< / p >
< p > 总之，< code > Arc< / code > 允许多个 < code > Worker< / code > 同时持有 < code > receiver< / code > ，而 < code > Mutex< / code > 可以确保一次只有一个 < code > Worker< / code > 能从 < code > receiver< / code > 接收消息。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > use std::{
sync::{mpsc, Arc, Mutex},
thread,
};
// --snip--
impl ThreadPool {
// --snip--
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
let (sender, receiver) = mpsc::channel();
let receiver = Arc::new(Mutex::new(receiver));
let mut workers = Vec::with_capacity(size);
for id in 0..size {
workers.push(Worker::new(id, Arc::clone(& receiver)));
}
ThreadPool { workers, sender }
}
// --snip--
}
// --snip--
impl Worker {
fn new(id: usize, receiver: Arc< Mutex< mpsc::Receiver< Job> > > ) -> Worker {
// --snip--
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 修改后，每一个 Worker 都可以安全的持有 < code > receiver< / code > ，同时不必担心一个任务会被重复执行多次，完美！< / p >
< h3 id = "实现-execute-方法" > < a class = "header" href = "#实现-execute-方法" > 实现 execute 方法< / a > < / h3 >
< p > 首先，需要为一个很长的类型创建一个别名, 有多长呢？< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > // --snip--
type Job = Box< dyn FnOnce() + Send + 'static> ;
impl ThreadPool {
// --snip--
pub fn execute< F> (& self, f: F)
where
F: FnOnce() + Send + 'static,
{
let job = Box::new(f);
self.sender.send(job).unwrap();
}
}
// --snip--
< span class = "boring" > }< / span > < / code > < / pre >
< p > 创建别名的威力暂时还看不到，敬请期待。总之，这里的工作很简单，将传入的任务包装成 < code > Job< / code > 类型后，发送出去。< / p >
< p > 但是还没完，接收的代码也要完善下:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > // --snip--
impl Worker {
fn new(id: usize, receiver: Arc< Mutex< mpsc::Receiver< Job> > > ) -> Worker {
let thread = thread::spawn(move || loop {
let job = receiver.lock().unwrap().recv().unwrap();
println!("Worker {id} got a job; executing.");
job();
});
Worker { id, thread }
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 修改后，就可以不停地循环去接收任务，最后进行执行。还可以看到因为之前 < code > Job< / code > 别名的引入， < code > new< / code > 函数的签名才没有过度复杂，否则你将看到的是 < code > fn new(id: usize, receiver: Arc< Mutex< mpsc::Receiver< Box< dyn FnOnce() + Send + 'static> > > > ) -> Worker< / code > ，感受下类型别名的威力吧 :D< / p >
< p > < code > lock()< / code > 方法可以获得一个 < code > Mutex< / code > 锁，至于为何使用 < code > unwrap< / code > ，难道获取锁还能失败？没错，假如当前持有锁的线程 < code > panic< / code > 了，那么这些等待锁的线程就会获取一个错误，因此 通过 < code > unwrap< / code > 来让当前等待的线程 < code > panic< / code > 是一个不错的解决方案，当然你还可以换成 < code > expect< / code > 。< / p >
< p > 一旦获取到锁里的内容 < code > mpsc::Receiver< Job> > < / code > 后，就可以调用其上的 < code > recv< / code > 方法来接收消息，依然是一个 < code > unwrap< / code > ，原因在于持有发送端的线程可能会被关闭，这种情况下直接 < code > panic< / code > 也是不错的。< / p >
< p > < code > recv< / code > 的调用过程是阻塞的，意味着若没有任何任务，那当前的调用线程将一直等待，直到接收到新的任务。< code > Mutex< T> < / code > 可以保证同一个任务只会被一个 Worker 获取，不会被重复执行。< / p >
< pre > < code class = "language-shell" > $ cargo run
Compiling hello v0.1.0 (file:///projects/hello)
warning: field is never read: `workers`
--> src/lib.rs:7:5
|
7 | workers: Vec< Worker> ,
| ^^^^^^^^^^^^^^^^^^^^
|
= note: `#[warn(dead_code)]` on by default
warning: field is never read: `id`
--> src/lib.rs:48:5
|
48 | id: usize,
| ^^^^^^^^^
warning: field is never read: `thread`
--> src/lib.rs:49:5
|
49 | thread: thread::JoinHandle< ()> ,
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
warning: `hello` (lib) generated 3 warnings
Finished dev [unoptimized + debuginfo] target(s) in 1.40s
Running `target/debug/hello`
Worker 0 got a job; executing.
Worker 2 got a job; executing.
Worker 1 got a job; executing.
Worker 3 got a job; executing.
Worker 0 got a job; executing.
Worker 2 got a job; executing.
Worker 1 got a job; executing.
Worker 3 got a job; executing.
Worker 0 got a job; executing.
Worker 2 got a job; executing.
< / code > < / pre >
< p > 终于，程序如愿运行起来，我们的线程池可以并发处理任务了！从打印的数字可以看到，只有 4 个线程去执行任务，符合我们对线程池的要求，这样再也不用担心系统的线程资源会被消耗殆尽了！< / p >
< blockquote >
< p > 注意： 出于缓存的考虑，有些浏览器会对多次同样的请求进行顺序的执行，因此你可能还是会遇到访问 < code > /sleep< / code > 后，就无法访问另一个 < code > /sleep< / code > 的问题 :(< / p >
< / blockquote >
< h2 id = "while-let-的巨大陷阱" > < a class = "header" href = "#while-let-的巨大陷阱" > while let 的巨大陷阱< / a > < / h2 >
< p > 还有一个问题，为啥之前我们不用 < code > while let< / code > 来循环？例如：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > // --snip--
impl Worker {
fn new(id: usize, receiver: Arc< Mutex< mpsc::Receiver< Job> > > ) -> Worker {
let thread = thread::spawn(move || {
while let Ok(job) = receiver.lock().unwrap().recv() {
println!("Worker {id} got a job; executing.");
job();
}
});
Worker { id, thread }
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 这段代码编译起来没问题，但是并不会产生我们预期的结果：后续请求依然需要等待慢请求的处理完成后，才能被处理。奇怪吧，仅仅是从 < code > let< / code > 改成 < code > while let< / code > 就会变成这样？大家可以思考下为什么会这样，具体答案会在下一章节末尾给出，这里先出给一个小提示：< code > Mutex< / code > 获取的锁在作用域结束后才会被释放。< / p >
< / main >
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