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<main>
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<h1 id="解构-option"><a class="header" href="#解构-option">解构 Option</a></h1>
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<p>在枚举那章，提到过 <code>Option</code> 枚举，它用来解决 Rust 中变量是否有值的问题，定义如下：</p>
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<pre><pre class="playground"><code class="language-rust edition2021">
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<span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
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</span>enum Option<T> {
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Some(T),
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None,
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}
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<span class="boring">}
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</span></code></pre></pre>
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<p>简单解释就是：<strong>一个变量要么有值：<code>Some(T)</code>, 要么为空：<code>None</code></strong>。</p>
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<p>那么现在的问题就是该如何去使用这个 <code>Option</code> 枚举类型，根据我们上一节的经验，可以通过 <code>match</code> 来实现。</p>
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<blockquote>
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<p>因为 <code>Option</code>，<code>Some</code>，<code>None</code> 都包含在 <code>prelude</code> 中，因此你可以直接通过名称来使用它们，而无需以 <code>Option::Some</code> 这种形式去使用，总之，千万不要因为调用路径变短了，就忘记 <code>Some</code> 和 <code>None</code> 也是 <code>Option</code> 底下的枚举成员！</p>
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</blockquote>
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<h2 id="匹配-optiont"><a class="header" href="#匹配-optiont">匹配 <code>Option<T></code></a></h2>
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<p>使用 <code>Option<T></code>，是为了从 <code>Some</code> 中取出其内部的 <code>T</code> 值以及处理没有值的情况，为了演示这一点，下面一起来编写一个函数，它获取一个 <code>Option<i32></code>，如果其中含有一个值，将其加一；如果其中没有值，则函数返回 <code>None</code> 值：</p>
|
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<pre><pre class="playground"><code class="language-rust edition2021">
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<span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
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</span>fn plus_one(x: Option<i32>) -> Option<i32> {
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match x {
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None => None,
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Some(i) => Some(i + 1),
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}
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}
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let five = Some(5);
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let six = plus_one(five);
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let none = plus_one(None);
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<span class="boring">}
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</span></code></pre></pre>
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<p><code>plus_one</code> 接受一个 <code>Option<i32></code> 类型的参数，同时返回一个 <code>Option<i32></code> 类型的值(这种形式的函数在标准库内随处所见)，在该函数的内部处理中，如果传入的是一个 <code>None</code> ，则返回一个 <code>None</code> 且不做任何处理；如果传入的是一个 <code>Some(i32)</code>，则通过模式绑定，把其中的值绑定到变量 <code>i</code> 上，然后返回 <code>i+1</code> 的值，同时用 <code>Some</code> 进行包裹。</p>
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<p>为了进一步说明，假设 <code>plus_one</code> 函数接受的参数值 x 是 <code>Some(5)</code>，来看看具体的分支匹配情况：</p>
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<h4 id="传入参数-some5"><a class="header" href="#传入参数-some5">传入参数 <code>Some(5)</code></a></h4>
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<pre><code class="language-rust ignore">None => None,
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</code></pre>
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<p>首先是匹配 <code>None</code> 分支，因为值 <code>Some(5)</code> 并不匹配模式 <code>None</code>，所以继续匹配下一个分支。</p>
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<pre><code class="language-rust ignore">Some(i) => Some(i + 1),
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</code></pre>
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<p><code>Some(5)</code> 与 <code>Some(i)</code> 匹配吗？当然匹配！它们是相同的成员。<code>i</code> 绑定了 <code>Some</code> 中包含的值，因此 <code>i</code> 的值是 <code>5</code>。接着匹配分支的代码被执行，最后将 <code>i</code> 的值加一并返回一个含有值 <code>6</code> 的新 <code>Some</code>。</p>
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<h4 id="传入参数-none"><a class="header" href="#传入参数-none">传入参数 None</a></h4>
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<p>接着考虑下 <code>plus_one</code> 的第二个调用，这次传入的 <code>x</code> 是 <code>None</code>， 我们进入 <code>match</code> 并与第一个分支相比较。</p>
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<pre><code class="language-rust ignore">None => None,
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</code></pre>
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<p>匹配上了！接着程序继续执行该分支后的代码：返回表达式 <code>None</code> 的值，也就是返回一个 <code>None</code>，因为第一个分支就匹配到了，其他的分支将不再比较。</p>
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