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<main>
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<h1 id="动态数组-vector"><a class="header" href="#动态数组-vector">动态数组 Vector</a></h1>
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<p>动态数组类型用 <code>Vec<T></code> 表示，事实上，在之前的章节，它的身影多次出现，我们一直没有细讲，只是简单的把它当作数组处理。</p>
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<p>动态数组允许你存储多个值，这些值在内存中一个紧挨着另一个排列，因此访问其中某个元素的成本非常低。动态数组只能存储相同类型的元素，如果你想存储不同类型的元素，可以使用之前讲过的枚举类型或者特征对象。</p>
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<p>总之，当我们想拥有一个列表，里面都是相同类型的数据时，动态数组将会非常有用。</p>
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<h2 id="创建动态数组"><a class="header" href="#创建动态数组">创建动态数组</a></h2>
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<p>在 Rust 中，有多种方式可以创建动态数组。</p>
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<h3 id="vecnew"><a class="header" href="#vecnew">Vec::new</a></h3>
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<p>使用 <code>Vec::new</code> 创建动态数组是最 rusty 的方式，它调用了 <code>Vec</code> 中的 <code>new</code> 关联函数：</p>
|
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<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>let v: Vec<i32> = Vec::new();
|
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<span class="boring">}</span></code></pre>
|
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|
<p>这里，<code>v</code> 被显式地声明了类型 <code>Vec<i32></code>，这是因为 Rust 编译器无法从 <code>Vec::new()</code> 中得到任何关于类型的暗示信息，因此也无法推导出 <code>v</code> 的具体类型，但是当你向里面增加一个元素后，一切又不同了：</p>
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<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
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|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut v = Vec::new();
|
|
|
v.push(1);
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>此时，<code>v</code> 就无需手动声明类型，因为编译器通过 <code>v.push(1)</code>，推测出 <code>v</code> 中的元素类型是 <code>i32</code>，因此推导出 <code>v</code> 的类型是 <code>Vec<i32></code>。</p>
|
|
|
<blockquote>
|
|
|
<p>如果预先知道要存储的元素个数，可以使用 <code>Vec::with_capacity(capacity)</code> 创建动态数组，这样可以避免因为插入大量新数据导致频繁的内存分配和拷贝，提升性能</p>
|
|
|
</blockquote>
|
|
|
<h3 id="vec"><a class="header" href="#vec">vec![]</a></h3>
|
|
|
<p>还可以使用宏 <code>vec!</code> 来创建数组，与 <code>Vec::new</code> 有所不同，前者能在创建同时给予初始化值：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let v = vec![1, 2, 3];
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>同样，此处的 <code>v</code> 也无需标注类型，编译器只需检查它内部的元素即可自动推导出 <code>v</code> 的类型是 <code>Vec<i32></code> （Rust 中，整数默认类型是 <code>i32</code>，在<a href="https://course.rs/basic/base-type/numbers.html#整数类型">数值类型</a>中有详细介绍）。</p>
|
|
|
<h2 id="更新-vector"><a class="header" href="#更新-vector">更新 Vector</a></h2>
|
|
|
<p>向数组尾部添加元素，可以使用 <code>push</code> 方法：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut v = Vec::new();
|
|
|
v.push(1);
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>与其它类型一样，必须将 <code>v</code> 声明为 <code>mut</code> 后，才能进行修改。</p>
|
|
|
<h2 id="vector-与其元素共存亡"><a class="header" href="#vector-与其元素共存亡">Vector 与其元素共存亡</a></h2>
|
|
|
<p>跟结构体一样，<code>Vector</code> 类型在超出作用域范围后，会被自动删除：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>{
|
|
|
let v = vec![1, 2, 3];
|
|
|
|
|
|
// ...
|
|
|
} // <- v超出作用域并在此处被删除
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>当 <code>Vector</code> 被删除后，它内部存储的所有内容也会随之被删除。目前来看，这种解决方案简单直白，但是当 <code>Vector</code> 中的元素被引用后，事情可能会没那么简单。</p>
|
|
|
<h2 id="从-vector-中读取元素"><a class="header" href="#从-vector-中读取元素">从 Vector 中读取元素</a></h2>
|
|
|
<p>读取指定位置的元素有两种方式可选：</p>
|
|
|
<ul>
|
|
|
<li>通过下标索引访问。</li>
|
|
|
<li>使用 <code>get</code> 方法。</li>
|
|
|
</ul>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let v = vec![1, 2, 3, 4, 5];
|
|
|
|
|
|
let third: &i32 = &v[2];
|
|
|
println!("第三个元素是 {}", third);
|
|
|
|
|
|
match v.get(2) {
|
|
|
Some(third) => println!("第三个元素是 {third}"),
|
|
|
None => println!("去你的第三个元素，根本没有！"),
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>和其它语言一样，集合类型的索引下标都是从 <code>0</code> 开始，<code>&v[2]</code> 表示借用 <code>v</code> 中的第三个元素，最终会获得该元素的引用。而 <code>v.get(2)</code> 也是访问第三个元素，但是有所不同的是，它返回了 <code>Option<&T></code>，因此还需要额外的 <code>match</code> 来匹配解构出具体的值。</p>
|
|
|
<blockquote>
|
|
|
<p>细心的同学会注意到这里使用了两种格式化输出的方式，其中第一种我们在之前已经见过，而第二种是后续新版本中引入的写法，也是更推荐的用法，具体介绍请参见<a href="https://course.rs/basic/formatted-output.html">格式化输出章节</a>。</p>
|
|
|
</blockquote>
|
|
|
<h3 id="下标索引与-get-的区别"><a class="header" href="#下标索引与-get-的区别">下标索引与 <code>.get</code> 的区别</a></h3>
|
|
|
<p>这两种方式都能成功的读取到指定的数组元素，既然如此为什么会存在两种方法？何况 <code>.get</code> 还会增加使用复杂度，这就涉及到数组越界的问题了，让我们通过示例说明：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let v = vec![1, 2, 3, 4, 5];
|
|
|
|
|
|
let does_not_exist = &v[100];
|
|
|
let does_not_exist = v.get(100);
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>运行以上代码，<code>&v[100]</code> 的访问方式会导致程序无情报错退出，因为发生了数组越界访问。 但是 <code>v.get</code> 就不会，它在内部做了处理，有值的时候返回 <code>Some(T)</code>，无值的时候返回 <code>None</code>，因此 <code>v.get</code> 的使用方式非常安全。</p>
|
|
|
<p>既然如此，为何不统一使用 <code>v.get</code> 的形式？因为实在是有些啰嗦，Rust 语言的设计者和使用者在审美这方面还是相当统一的：简洁即正义，何况性能上也会有轻微的损耗。</p>
|
|
|
<p>既然有两个选择，肯定就有如何选择的问题，答案很简单，当你确保索引不会越界的时候，就用索引访问，否则用 <code>.get</code>。例如，访问第几个数组元素并不取决于我们，而是取决于用户的输入时，用 <code>.get</code> 会非常适合，天知道那些可爱的用户会输入一个什么样的数字进来！</p>
|
|
|
<h2 id="同时借用多个数组元素"><a class="header" href="#同时借用多个数组元素">同时借用多个数组元素</a></h2>
|
|
|
<p>既然涉及到借用数组元素，那么很可能会遇到同时借用多个数组元素的情况，还记得在<a href="https://course.rs/basic/ownership/borrowing.html#借用规则总结">所有权和借用</a>章节咱们讲过的借用规则嘛？如果记得，就来看看下面的代码 :)</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut v = vec![1, 2, 3, 4, 5];
|
|
|
|
|
|
let first = &v[0];
|
|
|
|
|
|
v.push(6);
|
|
|
|
|
|
println!("The first element is: {first}");
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>先不运行，来推断下结果，首先 <code>first = &v[0]</code> 进行了不可变借用，<code>v.push</code> 进行了可变借用，如果 <code>first</code> 在 <code>v.push</code> 之后不再使用，那么该段代码可以成功编译（原因见<a href="https://course.rs/basic/ownership/borrowing.html#可变引用与不可变引用不能同时存在">引用的作用域</a>）。</p>
|
|
|
<p>可是上面的代码中，<code>first</code> 这个不可变借用在可变借用 <code>v.push</code> 后被使用了，那么妥妥的，编译器就会报错：</p>
|
|
|
<pre><code class="language-console">$ cargo run
|
|
|
Compiling collections v0.1.0 (file:///projects/collections)
|
|
|
error[E0502]: cannot borrow `v` as mutable because it is also borrowed as immutable 无法对v进行可变借用，因此之前已经进行了不可变借用
|
|
|
--> src/main.rs:6:5
|
|
|
|
|
|
|
4 | let first = &v[0];
|
|
|
| - immutable borrow occurs here // 不可变借用发生在此处
|
|
|
5 |
|
|
|
6 | v.push(6);
|
|
|
| ^^^^^^^^^ mutable borrow occurs here // 可变借用发生在此处
|
|
|
7 |
|
|
|
8 | println!("The first element is: {}", first);
|
|
|
| ----- immutable borrow later used here // 不可变借用在这里被使用
|
|
|
|
|
|
For more information about this error, try `rustc --explain E0502`.
|
|
|
error: could not compile `collections` due to previous error
|
|
|
</code></pre>
|
|
|
<p>其实，按理来说，这两个引用不应该互相影响的：一个是查询元素，一个是在数组尾部插入元素，完全不相干的操作，为何编译器要这么严格呢？</p>
|
|
|
<p>原因在于：数组的大小是可变的，当旧数组的大小不够用时，Rust 会重新分配一块更大的内存空间，然后把旧数组拷贝过来。这种情况下，之前的引用显然会指向一块无效的内存，这非常 rusty —— 对用户进行严格的教育。</p>
|
|
|
<p>其实想想，<strong>在长大之后，我们感激人生路上遇到过的严师益友，正是因为他们，我们才在正确的道路上不断前行，虽然在那个时候，并不能理解他们</strong>，而 Rust 就如那个良师益友，它不断的在纠正我们不好的编程习惯，直到某一天，你发现自己能写出一次性通过的漂亮代码时，就能明白它的良苦用心。</p>
|
|
|
<blockquote>
|
|
|
<p>若读者想要更深入的了解 <code>Vec<T></code>，可以看看<a href="https://nomicon.purewhite.io/vec/vec.html">Rustonomicon</a>，其中从零手撸一个动态数组，非常适合深入学习。</p>
|
|
|
</blockquote>
|
|
|
<h2 id="迭代遍历-vector-中的元素"><a class="header" href="#迭代遍历-vector-中的元素">迭代遍历 Vector 中的元素</a></h2>
|
|
|
<p>如果想要依次访问数组中的元素，可以使用迭代的方式去遍历数组，这种方式比用下标的方式去遍历数组更安全也更高效（每次下标访问都会触发数组边界检查）：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let v = vec![1, 2, 3];
|
|
|
for i in &v {
|
|
|
println!("{i}");
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>也可以在迭代过程中，修改 <code>Vector</code> 中的元素：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut v = vec![1, 2, 3];
|
|
|
for i in &mut v {
|
|
|
*i += 10
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<h2 id="存储不同类型的元素"><a class="header" href="#存储不同类型的元素">存储不同类型的元素</a></h2>
|
|
|
<p>在本节开头，有讲到数组的元素必须类型相同，但是也提到了解决方案：那就是通过使用枚举类型和特征对象来实现不同类型元素的存储。先来看看通过枚举如何实现：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">#[derive(Debug)]
|
|
|
enum IpAddr {
|
|
|
V4(String),
|
|
|
V6(String)
|
|
|
}
|
|
|
fn main() {
|
|
|
let v = vec![
|
|
|
IpAddr::V4("127.0.0.1".to_string()),
|
|
|
IpAddr::V6("::1".to_string())
|
|
|
];
|
|
|
|
|
|
for ip in v {
|
|
|
show_addr(ip)
|
|
|
}
|
|
|
}
|
|
|
|
|
|
fn show_addr(ip: IpAddr) {
|
|
|
println!("{:?}",ip);
|
|
|
}</code></pre>
|
|
|
<p>数组 <code>v</code> 中存储了两种不同的 <code>ip</code> 地址，但是这两种都属于 <code>IpAddr</code> 枚举类型的成员，因此可以存储在数组中。</p>
|
|
|
<p>再来看看特征对象的实现：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">trait IpAddr {
|
|
|
fn display(&self);
|
|
|
}
|
|
|
|
|
|
struct V4(String);
|
|
|
impl IpAddr for V4 {
|
|
|
fn display(&self) {
|
|
|
println!("ipv4: {:?}",self.0)
|
|
|
}
|
|
|
}
|
|
|
struct V6(String);
|
|
|
impl IpAddr for V6 {
|
|
|
fn display(&self) {
|
|
|
println!("ipv6: {:?}",self.0)
|
|
|
}
|
|
|
}
|
|
|
|
|
|
fn main() {
|
|
|
let v: Vec<Box<dyn IpAddr>> = vec![
|
|
|
Box::new(V4("127.0.0.1".to_string())),
|
|
|
Box::new(V6("::1".to_string())),
|
|
|
];
|
|
|
|
|
|
for ip in v {
|
|
|
ip.display();
|
|
|
}
|
|
|
}</code></pre>
|
|
|
<p>比枚举实现要稍微复杂一些，我们为 <code>V4</code> 和 <code>V6</code> 都实现了特征 <code>IpAddr</code>，然后将它俩的实例用 <code>Box::new</code> 包裹后，存在了数组 <code>v</code> 中，需要注意的是，这里必须手动地指定类型：<code>Vec<Box<dyn IpAddr>></code>，表示数组 <code>v</code> 存储的是特征 <code>IpAddr</code> 的对象，这样就实现了在数组中存储不同的类型。</p>
|
|
|
<p>在实际使用场景中，<strong>特征对象数组要比枚举数组常见很多</strong>，主要原因在于<a href="https://course.rs/basic/trait/trait-object.html">特征对象</a>非常灵活，而编译器对枚举的限制较多，且无法动态增加类型。</p>
|
|
|
<h2 id="vector-常用方法"><a class="header" href="#vector-常用方法">Vector 常用方法</a></h2>
|
|
|
<p>初始化 vec 的更多方式：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let v = vec![0; 3]; // 默认值为 0，初始长度为 3
|
|
|
let v_from = Vec::from([0, 0, 0]);
|
|
|
assert_eq!(v, v_from);
|
|
|
}</code></pre>
|
|
|
<p>动态数组意味着我们增加元素时，如果<strong>容量不足就会导致 vector 扩容</strong>（目前的策略是重新申请一块 2 倍大小的内存，再将所有元素拷贝到新的内存位置，同时更新指针数据），显然，当频繁扩容或者当元素数量较多且需要扩容时，大量的内存拷贝会降低程序的性能。</p>
|
|
|
<p>可以考虑在初始化时就指定一个实际的预估容量，尽量减少可能的内存拷贝：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let mut v = Vec::with_capacity(10);
|
|
|
v.extend([1, 2, 3]); // 附加数据到 v
|
|
|
println!("Vector 长度是: {}, 容量是: {}", v.len(), v.capacity());
|
|
|
|
|
|
v.reserve(100); // 调整 v 的容量，至少要有 100 的容量
|
|
|
println!("Vector（reserve） 长度是: {}, 容量是: {}", v.len(), v.capacity());
|
|
|
|
|
|
v.shrink_to_fit(); // 释放剩余的容量，一般情况下，不会主动去释放容量
|
|
|
println!("Vector（shrink_to_fit） 长度是: {}, 容量是: {}", v.len(), v.capacity());
|
|
|
}</code></pre>
|
|
|
<p>Vector 常见的一些方法示例：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut v = vec![1, 2];
|
|
|
assert!(!v.is_empty()); // 检查 v 是否为空
|
|
|
|
|
|
v.insert(2, 3); // 在指定索引插入数据，索引值不能大于 v 的长度， v: [1, 2, 3]
|
|
|
assert_eq!(v.remove(1), 2); // 移除指定位置的元素并返回, v: [1, 3]
|
|
|
assert_eq!(v.pop(), Some(3)); // 删除并返回 v 尾部的元素，v: [1]
|
|
|
assert_eq!(v.pop(), Some(1)); // v: []
|
|
|
assert_eq!(v.pop(), None); // 记得 pop 方法返回的是 Option 枚举值
|
|
|
v.clear(); // 清空 v, v: []
|
|
|
|
|
|
let mut v1 = [11, 22].to_vec(); // append 操作会导致 v1 清空数据，增加可变声明
|
|
|
v.append(&mut v1); // 将 v1 中的所有元素附加到 v 中, v1: []
|
|
|
v.truncate(1); // 截断到指定长度，多余的元素被删除, v: [11]
|
|
|
v.retain(|x| *x > 10); // 保留满足条件的元素，即删除不满足条件的元素
|
|
|
|
|
|
let mut v = vec![11, 22, 33, 44, 55];
|
|
|
// 删除指定范围的元素，同时获取被删除元素的迭代器, v: [11, 55], m: [22, 33, 44]
|
|
|
let mut m: Vec<_> = v.drain(1..=3).collect();
|
|
|
|
|
|
let v2 = m.split_off(1); // 指定索引处切分成两个 vec, m: [22], v2: [33, 44]
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>当然也可以像<a href="/basic/compound-type/array.html#数组切片">数组切片</a>的方式获取 vec 的部分元素：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let v = vec![11, 22, 33, 44, 55];
|
|
|
let slice = &v[1..=3];
|
|
|
assert_eq!(slice, &[22, 33, 44]);
|
|
|
}</code></pre>
|
|
|
<p>更多细节，阅读 Vector 的<a href="https://doc.rust-lang.org/std/vec/struct.Vec.html#">标准库文档</a>。</p>
|
|
|
<h2 id="vector-的排序"><a class="header" href="#vector-的排序">Vector 的排序</a></h2>
|
|
|
<p>在 rust 里，实现了两种排序算法，分别为稳定的排序 <code>sort</code> 和 <code>sort_by</code>，以及非稳定排序 <code>sort_unstable</code> 和 <code>sort_unstable_by</code>。</p>
|
|
|
<p>当然，这个所谓的 <code>非稳定</code> 并不是指排序算法本身不稳定，而是指在排序过程中对相等元素的处理方式。在 <code>稳定</code> 排序算法里，对相等的元素，不会对其进行重新排序。而在 <code>不稳定</code> 的算法里则不保证这点。</p>
|
|
|
<p>总体而言，<code>非稳定</code> 排序的算法的速度会优于 <code>稳定</code> 排序算法，同时，<code>稳定</code> 排序还会额外分配原数组一半的空间。</p>
|
|
|
<h3 id="整数数组的排序"><a class="header" href="#整数数组的排序">整数数组的排序</a></h3>
|
|
|
<p>以下是对整数列进行排序的例子。</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let mut vec = vec![1, 5, 10, 2, 15];
|
|
|
vec.sort_unstable();
|
|
|
assert_eq!(vec, vec![1, 2, 5, 10, 15]);
|
|
|
}</code></pre>
|
|
|
<h3 id="浮点数数组的排序"><a class="header" href="#浮点数数组的排序">浮点数数组的排序</a></h3>
|
|
|
<p>我们尝试使用上面的方法来对浮点数进行排序：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let mut vec = vec![1.0, 5.6, 10.3, 2.0, 15f32];
|
|
|
vec.sort_unstable();
|
|
|
assert_eq!(vec, vec![1.0, 2.0, 5.6, 10.3, 15f32]);
|
|
|
}</code></pre>
|
|
|
<p>结果，居然报错了，</p>
|
|
|
<pre><code>error[E0277]: the trait bound `f32: Ord` is not satisfied
|
|
|
--> src/main.rs:29:13
|
|
|
|
|
|
|
29 | vec.sort_unstable();
|
|
|
| ^^^^^^^^^^^^^ the trait `Ord` is not implemented for `f32`
|
|
|
|
|
|
|
= help: the following other types implement trait `Ord`:
|
|
|
i128
|
|
|
i16
|
|
|
i32
|
|
|
i64
|
|
|
i8
|
|
|
isize
|
|
|
u128
|
|
|
u16
|
|
|
and 4 others
|
|
|
note: required by a bound in `core::slice::<impl [T]>::sort_unstable`
|
|
|
--> /home/keijack/.rustup/toolchains/stable-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/core/src/slice/mod.rs:2635:12
|
|
|
|
|
|
|
2635 | T: Ord,
|
|
|
| ^^^ required by this bound in `core::slice::<impl [T]>::sort_unstable`
|
|
|
|
|
|
For more information about this error, try `rustc --explain E0277`.
|
|
|
</code></pre>
|
|
|
<p>原来，在浮点数当中，存在一个 <code>NAN</code> 的值，这个值无法与其他的浮点数进行对比，因此，浮点数类型并没有实现全数值可比较 <code>Ord</code> 的特性，而是实现了部分可比较的特性 <code>PartialOrd</code>。</p>
|
|
|
<p>如此，如果我们确定在我们的浮点数数组当中，不包含 <code>NAN</code> 值，那么我们可以使用 <code>partial_cmp</code> 来作为大小判断的依据。</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let mut vec = vec![1.0, 5.6, 10.3, 2.0, 15f32];
|
|
|
vec.sort_unstable_by(|a, b| a.partial_cmp(b).unwrap());
|
|
|
assert_eq!(vec, vec![1.0, 2.0, 5.6, 10.3, 15f32]);
|
|
|
}</code></pre>
|
|
|
<p>OK，现在可以正确执行了。</p>
|
|
|
<h3 id="对结构体数组进行排序"><a class="header" href="#对结构体数组进行排序">对结构体数组进行排序</a></h3>
|
|
|
<p>有了上述浮点数排序的经验，我们推而广之，那么对结构体是否也可以使用这种自定义对比函数的方式来进行呢？马上来试一下：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">#[derive(Debug)]
|
|
|
struct Person {
|
|
|
name: String,
|
|
|
age: u32,
|
|
|
}
|
|
|
|
|
|
impl Person {
|
|
|
fn new(name: String, age: u32) -> Person {
|
|
|
Person { name, age }
|
|
|
}
|
|
|
}
|
|
|
|
|
|
fn main() {
|
|
|
let mut people = vec![
|
|
|
Person::new("Zoe".to_string(), 25),
|
|
|
Person::new("Al".to_string(), 60),
|
|
|
Person::new("John".to_string(), 1),
|
|
|
];
|
|
|
// 定义一个按照年龄倒序排序的对比函数
|
|
|
people.sort_unstable_by(|a, b| b.age.cmp(&a.age));
|
|
|
|
|
|
println!("{:?}", people);
|
|
|
}</code></pre>
|
|
|
<p>执行后输出：</p>
|
|
|
<pre><code>[Person { name: "Al", age: 60 }, Person { name: "Zoe", age: 25 }, Person { name: "John", age: 1 }]
|
|
|
</code></pre>
|
|
|
<p>结果正确。</p>
|
|
|
<p>从上面我们学习过程当中，排序需要我们实现 <code>Ord</code> 特性，那么如果我们把我们的结构体实现了该特性，是否就不需要我们自定义对比函数了呢？</p>
|
|
|
<p>是，但不完全是，实现 <code>Ord</code> 需要我们实现 <code>Ord</code>、<code>Eq</code>、<code>PartialEq</code>、<code>PartialOrd</code> 这些属性。好消息是，你可以 <code>derive</code> 这些属性：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">#[derive(Debug, Ord, Eq, PartialEq, PartialOrd)]
|
|
|
struct Person {
|
|
|
name: String,
|
|
|
age: u32,
|
|
|
}
|
|
|
|
|
|
impl Person {
|
|
|
fn new(name: String, age: u32) -> Person {
|
|
|
Person { name, age }
|
|
|
}
|
|
|
}
|
|
|
|
|
|
fn main() {
|
|
|
let mut people = vec![
|
|
|
Person::new("Zoe".to_string(), 25),
|
|
|
Person::new("Al".to_string(), 60),
|
|
|
Person::new("Al".to_string(), 30),
|
|
|
Person::new("John".to_string(), 1),
|
|
|
Person::new("John".to_string(), 25),
|
|
|
];
|
|
|
|
|
|
people.sort_unstable();
|
|
|
|
|
|
println!("{:?}", people);
|
|
|
}</code></pre>
|
|
|
<p>执行输出</p>
|
|
|
<pre><code>[Person { name: "Al", age: 30 }, Person { name: "Al", age: 60 }, Person { name: "John", age: 1 }, Person { name: "John", age: 25 }, Person { name: "Zoe", age: 25 }]
|
|
|
</code></pre>
|
|
|
<p>需要 <code>derive</code> <code>Ord</code> 相关特性，需要确保你的结构体中所有的属性均实现了 <code>Ord</code> 相关特性，否则会发生编译错误。<code>derive</code> 的默认实现会依据属性的顺序依次进行比较，如上述例子中，当 <code>Person</code> 的 <code>name</code> 值相同，则会使用 <code>age</code> 进行比较。</p>
|
|
|
<h2 id="课后练习"><a class="header" href="#课后练习">课后练习</a></h2>
|
|
|
<blockquote>
|
|
|
<p><a href="https://practice-zh.course.rs/collections/vector.html">Rust By Practice</a>，支持代码在线编辑和运行，并提供详细的<a href="https://github.com/sunface/rust-by-practice/blob/master/solutions/collections/Vector.md">习题解答</a>。</p>
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