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<h1 class="menu-title">Rust 程序设计语言 简体中文版</h1>
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<a class="header" href="#导入命名" name="导入命名"><h2>导入命名</h2></a>
<blockquote>
<p><a href="https://github.com/rust-lang/book/blob/master/second-edition/src/ch07-03-importing-names-with-use.md">ch07-03-importing-names-with-use.md</a>
<br>
commit 3f2a1bd8dbb19cc48b210fc4fb35c305c8d81b56</p>
</blockquote>
<p>我们已经讲到了如何使用模块名称作为调用的一部分,来调用模块中的函数,如列表 7-6 中所示的<code>nested_modules</code>函数调用。</p>
<p><span class="filename">Filename: src/main.rs</span></p>
<pre><code class="language-rust">pub mod a {
pub mod series {
pub mod of {
pub fn nested_modules() {}
}
}
}
fn main() {
a::series::of::nested_modules();
}
</code></pre>
<p><span class="caption">Listing 7-6: Calling a function by fully specifying its
enclosing modules namespaces</span></p>
<p>如你所见,指定函数的完全限定名称可能会非常冗长。所幸 Rust 有一个关键字使得这些调用显得更简洁。</p>
<a class="header" href="#使用use的简单导入" name="使用use的简单导入"><h3>使用<code>use</code>的简单导入</h3></a>
<p>Rust 的<code>use</code>关键字的工作是缩短冗长的函数调用,通过将想要调用的函数所在的模块引入到作用域中。这是一个将<code>a::series::of</code>模块导入一个二进制 crate 的根作用域的例子:</p>
<p><span class="filename">Filename: src/main.rs</span></p>
<pre><code class="language-rust">pub mod a {
pub mod series {
pub mod of {
pub fn nested_modules() {}
}
}
}
use a::series::of;
fn main() {
of::nested_modules();
}
</code></pre>
<p><code>use a::series::of;</code>这一行的意思是每当想要引用<code>of</code>模块时,不用使用完整的<code>a::series::of</code>路径,可以直接使用<code>of</code></p>
<p><code>use</code>关键字只将指定的模块引入作用域;它并不会将其子模块也引入。这就是为什么想要调用<code>nested_modules</code>函数时仍然必须写成<code>of::nested_modules</code></p>
<p>也可以将函数本身引入到作用域中,通过如下在<code>use</code>中指定函数的方式:</p>
<pre><code class="language-rust">pub mod a {
pub mod series {
pub mod of {
pub fn nested_modules() {}
}
}
}
use a::series::of::nested_modules;
fn main() {
nested_modules();
}
</code></pre>
<p>这使得我们可以忽略所有的模块并直接引用函数。</p>
<p>因为枚举也像模块一样组成了某种命名空间,也可以使用<code>use</code>来导入枚举的成员。对于任何类型的<code>use</code>语句,如果从一个命名空间导入多个项,可以使用大括号和逗号来列举他们,像这样:</p>
<pre><code class="language-rust">enum TrafficLight {
Red,
Yellow,
Green,
}
use TrafficLight::{Red, Yellow};
fn main() {
let red = Red;
let yellow = Yellow;
let green = TrafficLight::Green; // because we didnt `use` TrafficLight::Green
}
</code></pre>
<a class="header" href="#使用的全局引用导入" name="使用的全局引用导入"><h3>使用<code>*</code>的全局引用导入</h3></a>
<p>为了一次导入某个命名空间的所有项,可以使用<code>*</code>语法。例如:</p>
<pre><code class="language-rust">enum TrafficLight {
Red,
Yellow,
Green,
}
use TrafficLight::*;
fn main() {
let red = Red;
let yellow = Yellow;
let green = Green;
}
</code></pre>
<p><code>*</code>被称为<strong>全局导入</strong><em>glob</em>),它会导入命名空间中所有可见的项。全局导入应该保守的使用:他们是方便的,但是也可能会引入多于你预期的内容从而导致命名冲突。</p>
<a class="header" href="#使用super访问父模块" name="使用super访问父模块"><h3>使用<code>super</code>访问父模块</h3></a>
<p>正如我们已经知道的,当创建一个库 crate 时Cargo 会生成一个<code>tests</code>模块。现在让我们来深入了解一下。在<code>communicator</code>项目中,打开 <em>src/lib.rs</em></p>
<p><span class="filename">Filename: src/lib.rs</span></p>
<pre><code class="language-rust,ignore">pub mod client;
pub mod network;
#[cfg(test)]
mod tests {
#[test]
fn it_works() {
}
}
</code></pre>
<p>第十一章会更详细的解释测试,不过其部分内容现在应该可以理解了:有一个叫做<code>tests</code>的模块紧邻其他模块,同时包含一个叫做<code>it_works</code>的函数。即便存在一些特殊注解,<code>tests</code>也不过是另外一个模块!所以我们的模块层次结构看起来像这样:</p>
<pre><code>communicator
├── client
├── network
| └── client
└── tests
</code></pre>
<p>测试是为了检验库中的代码而存在的,所以让我们尝试在<code>it_works</code>函数中调用<code>client::connect</code>函数,即便现在不准备测试任何功能:</p>
<p><span class="filename">Filename: src/lib.rs</span></p>
<pre><code class="language-rust">#[cfg(test)]
mod tests {
#[test]
fn it_works() {
client::connect();
}
}
</code></pre>
<p>使用<code>cargo test</code>命令运行测试:</p>
<pre><code>$ cargo test
Compiling communicator v0.1.0 (file:///projects/communicator)
error[E0433]: failed to resolve. Use of undeclared type or module `client`
--&gt; src/lib.rs:9:9
|
9 | client::connect();
| ^^^^^^^^^^^^^^^ Use of undeclared type or module `client`
warning: function is never used: `connect`, #[warn(dead_code)] on by default
--&gt; src/network/server.rs:1:1
|
1 | fn connect() {
| ^
</code></pre>
<p>编译失败了,不过为什么呢?并不需要像 <em>src/main.rs</em> 那样将<code>communicator::</code>置于函数前,因为这里肯定是在<code>communicator</code>库 crate 之内的。之所以失败的原因是路径是相对于当前模块的,在这里就是<code>tests</code>。唯一的例外就是<code>use</code>语句,它默认是相对于 crate 根模块的。我们的<code>tests</code>模块需要<code>client</code>模块位于其作用域中!</p>
<p>那么如何在模块层次结构中回退一级模块,以便在<code>tests</code>模块中能够调用<code>client::connect</code>函数呢?在<code>tests</code>模块中,要么可以在开头使用双冒号来让 Rust 知道我们想要从根模块开始并列出整个路径:</p>
<pre><code class="language-rust,ignore">::client::connect();
</code></pre>
<p>要么可以使用<code>super</code>在层级中获取当前模块的上一级模块:</p>
<pre><code class="language-rust,ignore">super::client::connect();
</code></pre>
<p>在这个例子中这两个选择看不出有多么大的区别,不过随着模块层次的更加深入,每次都从根模块开始就会显得很长了。在这些情况下,使用<code>super</code>来获取当前模块的同级模块是一个好的捷径。再加上,如果在代码中的很多地方指定了从根开始的路径,那么当通过移动子树或到其他位置来重新排列模块时,最终就需要更新很多地方的路径,这就非常乏味无趣了。</p>
<p>在每一个测试中总是不得不编写<code>super::</code>也会显得很恼人,不过你已经见过解决这个问题的利器了:<code>use</code><code>super::</code>的功能改变了提供给<code>use</code>的路径,使其不再相对于根模块而是相对于父模块。</p>
<p>为此,特别是在<code>tests</code>模块,<code>use super::something</code>是常用的手段。所以现在的测试看起来像这样:</p>
<p><span class="filename">Filename: src/lib.rs</span></p>
<pre><code class="language-rust">#[cfg(test)]
mod tests {
use super::client;
#[test]
fn it_works() {
client::connect();
}
}
</code></pre>
<p>如果再次运行<code>cargo test</code>,测试将会通过而且测试结果输出的第一部分将会是:</p>
<pre><code>$ cargo test
Compiling communicator v0.1.0 (file:///projects/communicator)
Running target/debug/communicator-92007ddb5330fa5a
running 1 test
test tests::it_works ... ok
test result: ok. 1 passed; 0 failed; 0 ignored; 0 measured
</code></pre>
<a class="header" href="#总结" name="总结"><h2>总结</h2></a>
<p>现在你掌握了组织代码的核心科技!利用他们将相关的代码组合在一起、防止代码文件过长并将一个整洁的公有 API 展现给库的用户。</p>
<p>接下来,让我们看看一些标准库提供的集合数据类型,你可以利用他们编写出漂亮整洁的代码。</p>
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