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<main>
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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>
|
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<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><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--
|
|
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|
fn handle_connection(mut stream: TcpStream) {
|
|
|
// --snip--
|
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|
|
|
|
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></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，我们将限制线程池中的线程数量，以保护服务器免受拒绝服务攻击（DoS）的影响：如果针对每个请求创建一个新线程，那么一个人向我们的服务器发出1000万个请求，会直接耗尽资源，导致后续用户的请求无法被处理，这也是拒绝服务名称的来源。</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，例如 redis；多线程异步 IO，例如 Rust 的主流 web 框架。事实上，大家在下一个实战项目中，会看到相关技术的应用。</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><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></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><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></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><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></pre>
|
|
|
<p>接着在 <code>main.rs</code> 中引入:</p>
|
|
|
<pre><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></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><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></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><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></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><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></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><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></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#%E6%96%87%E6%A1%A3%E6%B3%A8%E9%87%8A">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><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></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><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></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><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></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><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></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><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></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，即多生产者单消费者，因此我们无法通过克隆消费者的方式来修复这个错误。当然，发送多条消息给多个接收者也不在考虑范畴，该怎么办？似乎陷入了绝境。</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><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></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><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></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>
|
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<p>但是还没完，接收的代码也要完善下:</p>
|
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|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
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|
</span><span class="boring">fn main() {
|
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|
</span>// --snip--
|
|
|
|
|
|
impl Worker {
|
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|
fn new(id: usize, receiver: Arc<Mutex<mpsc::Receiver<Job>>>) -> Worker {
|
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|
let thread = thread::spawn(move || loop {
|
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|
let job = receiver.lock().unwrap().recv().unwrap();
|
|
|
|
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println!("Worker {id} got a job; executing.");
|
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|
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|
job();
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|
});
|
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|
|
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Worker { id, thread }
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|
}
|
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|
}
|
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|
<span class="boring">}</span></code></pre></pre>
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<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>
|
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|
<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>
|
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<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>
|
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<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>
|
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|
<pre><code class="language-shell">$ cargo run
|
|
|
Compiling hello v0.1.0 (file:///projects/hello)
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|
warning: field is never read: `workers`
|
|
|
--> src/lib.rs:7:5
|
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|
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|
7 | workers: Vec<Worker>,
|
|
|
| ^^^^^^^^^^^^^^^^^^^^
|
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|
|
|
= note: `#[warn(dead_code)]` on by default
|
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|
|
|
warning: field is never read: `id`
|
|
|
--> src/lib.rs:48:5
|
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|
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|
48 | id: usize,
|
|
|
| ^^^^^^^^^
|
|
|
|
|
|
warning: field is never read: `thread`
|
|
|
--> src/lib.rs:49:5
|
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|
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|
49 | thread: thread::JoinHandle<()>,
|
|
|
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|
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|
|
|
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><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></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>
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