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<h2 id="使用-use-关键字将路径引入作用域"><a class="header" href="#使用-use-关键字将路径引入作用域">使用 <code>use</code> 关键字将路径引入作用域</a></h2>
<!-- https://github.com/rust-lang/book/blob/main/src/ch07-04-bringing-paths-into-scope-with-the-use-keyword.md -->
<!-- commit 72ad14e4acb12438aa467c4cf256e0bc55df585a -->
<p>不得不编写路径来调用函数显得繁琐且重复。在示例 7-7 中,无论我们选择 <code>add_to_waitlist</code> 函数的绝对路径还是相对路径,每次我们想要调用 <code>add_to_waitlist</code> 时,都必须指定<code>front_of_house</code><code>hosting</code>。幸运的是,有一种方法可以简化这个过程。我们可以使用 <code>use</code> 关键字创建一个捷径,然后就可以在作用域中的任何地方使用这个更短的名字。</p>
<p>在示例 7-11 中,我们将 <code>crate::front_of_house::hosting</code> 模块引入了 <code>eat_at_restaurant</code> 函数的作用域,而我们只需要指定 <code>hosting::add_to_waitlist</code> 即可在 <code>eat_at_restaurant</code> 中调用 <code>add_to_waitlist</code> 函数。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground test_harness">mod front_of_house {
pub mod hosting {
pub fn add_to_waitlist() {}
}
}
use crate::front_of_house::hosting;
pub fn eat_at_restaurant() {
hosting::add_to_waitlist();
}</code></pre>
<p><span class="caption">示例 7-11: 使用 <code>use</code> 将模块引入作用域</span></p>
<p>在作用域中增加 <code>use</code> 和路径类似于在文件系统中创建软连接符号连接symbolic link。通过在 crate 根增加 <code>use crate::front_of_house::hosting</code>,现在 <code>hosting</code> 在作用域中就是有效的名称了,如同 <code>hosting</code> 模块被定义于 crate 根一样。通过 <code>use</code> 引入作用域的路径也会检查私有性,同其它路径一样。</p>
<p>注意 <code>use</code> 只能创建 <code>use</code> 所在的特定作用域内的捷径。示例 7-12 将 <code>eat_at_restaurant</code> 函数移动到了一个叫 <code>customer</code> 的子模块,这又是一个不同于 <code>use</code> 语句的作用域,所以函数体不能编译。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground test_harness does_not_compile ignore">mod front_of_house {
pub mod hosting {
pub fn add_to_waitlist() {}
}
}
use crate::front_of_house::hosting;
mod customer {
pub fn eat_at_restaurant() {
hosting::add_to_waitlist();
}
}</code></pre>
<p><span class="caption">示例 7-12: <code>use</code> 语句只适用于其所在的作用域</span></p>
<p>编译器错误显示捷径不再适用于 <code>customer</code> 模块中:</p>
<pre><code class="language-console">$ cargo build
Compiling restaurant v0.1.0 (file:///projects/restaurant)
error[E0433]: failed to resolve: use of undeclared crate or module `hosting`
--&gt; src/lib.rs:11:9
|
11 | hosting::add_to_waitlist();
| ^^^^^^^ use of undeclared crate or module `hosting`
|
help: consider importing this module through its public re-export
|
10 + use crate::hosting;
|
warning: unused import: `crate::front_of_house::hosting`
--&gt; src/lib.rs:7:5
|
7 | use crate::front_of_house::hosting;
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|
= note: `#[warn(unused_imports)]` on by default
For more information about this error, try `rustc --explain E0433`.
warning: `restaurant` (lib) generated 1 warning
error: could not compile `restaurant` (lib) due to 1 previous error; 1 warning emitted
</code></pre>
<p>注意这里还有一个警告说 <code>use</code> 在其作用域内不再被使用!为了修复这个问题,可以将 <code>use</code> 移动到 <code>customer</code> 模块内,或者在子模块 <code>customer</code> 内通过 <code>super::hosting</code> 引用父模块中的这个捷径。</p>
<h3 id="创建惯用的-use-路径"><a class="header" href="#创建惯用的-use-路径">创建惯用的 <code>use</code> 路径</a></h3>
<p>在示例 7-11 中,你可能会比较疑惑,为什么我们是指定 <code>use crate::front_of_house::hosting</code>,然后在 <code>eat_at_restaurant</code> 中调用 <code>hosting::add_to_waitlist</code> ,而不是通过指定一直到 <code>add_to_waitlist</code> 函数的 <code>use</code> 路径来得到相同的结果,如示例 7-13 所示。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground test_harness">mod front_of_house {
pub mod hosting {
pub fn add_to_waitlist() {}
}
}
use crate::front_of_house::hosting::add_to_waitlist;
pub fn eat_at_restaurant() {
add_to_waitlist();
}</code></pre>
<p><span class="caption">示例 7-13: 使用 <code>use</code><code>add_to_waitlist</code> 函数引入作用域,这并不符合习惯</span></p>
<p>虽然示例 7-11 和 7-13 都完成了相同的任务,但示例 7-11 是使用 <code>use</code> 将函数引入作用域的习惯用法。要想使用 <code>use</code> 将函数的父模块引入作用域,我们必须在调用函数时指定父模块,这样可以清晰地表明函数不是在本地定义的,同时使完整路径的重复度最小化。示例 7-13 中的代码不清楚 <code>add_to_waitlist</code> 是在哪里被定义的。</p>
<p>另一方面,使用 <code>use</code> 引入结构体、枚举和其他项时,习惯是指定它们的完整路径。示例 7-14 展示了将 <code>HashMap</code> 结构体引入二进制 crate 作用域的习惯用法。</p>
<p><span class="filename">文件名src/main.rs</span></p>
<pre class="playground"><code class="language-rust edition2024">use std::collections::HashMap;
fn main() {
let mut map = HashMap::new();
map.insert(1, 2);
}</code></pre>
<p><span class="caption">示例 7-14: 将 <code>HashMap</code> 引入作用域的习惯用法</span></p>
<p>这种习惯用法背后没有什么硬性要求:它只是一种惯例,人们已经习惯了以这种方式阅读和编写 Rust 代码。</p>
<p>这个习惯用法有一个例外,那就是我们想使用 <code>use</code> 语句将两个具有相同名称的项带入作用域,因为 Rust 不允许这样做。示例 7-15 展示了如何将两个具有相同名称但不同父模块的 <code>Result</code> 类型引入作用域,以及如何引用它们。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground">use std::fmt;
use std::io;
fn function1() -&gt; fmt::Result {
// --snip--
<span class="boring"> Ok(())
</span>}
fn function2() -&gt; io::Result&lt;()&gt; {
// --snip--
<span class="boring"> Ok(())
</span>}</code></pre>
<p><span class="caption">示例 7-15: 使用父模块将两个具有相同名称的类型引入同一作用域</span></p>
<p>如你所见,使用父模块可以区分这两个 <code>Result</code> 类型。如果我们是指定 <code>use std::fmt::Result</code><code>use std::io::Result</code>,我们将在同一作用域拥有了两个 <code>Result</code> 类型,当我们使用 <code>Result</code>Rust 则不知道我们要用的是哪个。</p>
<h3 id="使用-as-关键字提供新的名称"><a class="header" href="#使用-as-关键字提供新的名称">使用 <code>as</code> 关键字提供新的名称</a></h3>
<p>使用 <code>use</code> 将两个同名类型引入同一作用域这个问题还有另一个解决办法:在这个类型的路径后面,我们使用 <code>as</code> 指定一个新的本地名称或者<strong>别名</strong>。示例 7-16 展示了另一个编写示例 7-15 中代码的方法,通过 <code>as</code> 重命名其中一个 <code>Result</code> 类型。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground">use std::fmt::Result;
use std::io::Result as IoResult;
fn function1() -&gt; Result {
// --snip--
<span class="boring"> Ok(())
</span>}
fn function2() -&gt; IoResult&lt;()&gt; {
// --snip--
<span class="boring"> Ok(())
</span>}</code></pre>
<p><span class="caption">示例 7-16: 使用 <code>as</code> 关键字重命名引入作用域的类型</span></p>
<p>在第二个 <code>use</code> 语句中,我们选择 <code>IoResult</code> 作为 <code>std::io::Result</code> 的新名称,它与从 <code>std::fmt</code> 引入作用域的 <code>Result</code> 并不冲突。示例 7-15 和示例 7-16 都是惯用写法,如何选择都取决于你!</p>
<h3 id="使用-pub-use-重导出名称"><a class="header" href="#使用-pub-use-重导出名称">使用 <code>pub use</code> 重导出名称</a></h3>
<p>使用 <code>use</code> 关键字,将某个名称导入当前作用域后,该名称对此作用域之外还是私有的。若要让作用域之外的代码能够像在当前作用域中一样使用该名称,可以将 <code>pub</code><code>use</code> 组合使用。这种技术被称为<strong>重导出</strong><em>re-exporting</em>),因为在把某个项目导入当前作用域的同时,也将其暴露给其他作用域。</p>
<p>示例 7-17 将示例 7-11 根模块中的 <code>use</code> 改为 <code>pub use</code> 的代码。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground test_harness">mod front_of_house {
pub mod hosting {
pub fn add_to_waitlist() {}
}
}
pub use crate::front_of_house::hosting;
pub fn eat_at_restaurant() {
hosting::add_to_waitlist();
}</code></pre>
<p><span class="caption">示例 7-17: 通过 <code>pub use</code> 使名称可从新作用域中被导入至任何代码</span></p>
<p>在这个修改之前,外部代码需要使用路径 <code>restaurant::front_of_house::hosting::add_to_waitlist()</code> 来调用 <code>add_to_waitlist</code> 函数,并且还需要将 <code>front_of_house</code> 模块标记为 <code>pub</code>。现在这个 <code>pub use</code> 从根模块重导出了 <code>hosting</code> 模块,外部代码现在可以使用路径 <code>restaurant::hosting::add_to_waitlist</code></p>
<p>当你代码的内部结构与调用你代码的程序员所想象的结构不同时,重导出会很有用。例如,在这个餐馆的比喻中,经营餐馆的人会想到“前台”和“后台”。但顾客在光顾一家餐馆时,可能不会以这些术语来考虑餐馆的各个部分。使用 <code>pub use</code>,我们可以使用一种结构编写代码,却将不同的结构形式暴露出来。这样做使我们的库井井有条,也使开发这个库的程序员和调用这个库的程序员都更加方便。在<a href="ch14-02-publishing-to-crates-io.html#使用-pub-use-导出便捷的公有-api">“使用 <code>pub use</code> 导出便捷的公有 API”</a>部分让我们再看另一个 <code>pub use</code> 的例子来了解这如何影响 crate 的文档。</p>
<h3 id="使用外部包"><a class="header" href="#使用外部包">使用外部包</a></h3>
<p>在第二章中我们编写了一个猜猜看游戏。那个项目使用了一个外部包 <code>rand</code> 来生成随机数。为了在项目中使用 <code>rand</code>,在 <em>Cargo.toml</em> 中加入了如下行:</p>
<p><span class="filename">文件名Cargo.toml</span></p>
<pre><code class="language-toml">rand = "0.8.5"
</code></pre>
<p><em>Cargo.toml</em> 中加入 <code>rand</code> 依赖告诉了 Cargo 要从 <a href="https://crates.io">crates.io</a> 下载 <code>rand</code> 和其依赖,并使其可在项目代码中使用。</p>
<p>接着,为了将 <code>rand</code> 定义引入项目包的作用域,我们加入一行 <code>use</code> 起始的包名,它以 <code>rand</code> 包名开头并列出了需要引入作用域的项。回忆一下第二章的“生成一个随机数”部分,我们曾将 <code>Rng</code> trait 引入作用域并调用了 <code>rand::thread_rng</code> 函数:</p>
<pre><code class="language-rust ignore"><span class="boring">use std::io;
</span><span class="boring">
</span>use rand::Rng;
fn main() {
<span class="boring"> println!("Guess the number!");
</span><span class="boring">
</span> let secret_number = rand::thread_rng().gen_range(1..=100);
<span class="boring">
</span><span class="boring"> println!("The secret number is: {secret_number}");
</span><span class="boring">
</span><span class="boring"> println!("Please input your guess.");
</span><span class="boring">
</span><span class="boring"> let mut guess = String::new();
</span><span class="boring">
</span><span class="boring"> io::stdin()
</span><span class="boring"> .read_line(&amp;mut guess)
</span><span class="boring"> .expect("Failed to read line");
</span><span class="boring">
</span><span class="boring"> println!("You guessed: {guess}");
</span>}</code></pre>
<p><a href="https://crates.io">crates.io</a> 上有很多 Rust 社区成员发布的包,将其引入你自己的项目都需要一道相同的步骤:在 <em>Cargo.toml</em> 列出它们并通过 <code>use</code> 将其中定义的项引入项目包的作用域中。</p>
<p>注意 <code>std</code> 标准库对于你的包来说也是外部 crate。因为标准库随 Rust 语言一同分发,无需修改 <em>Cargo.toml</em> 来引入 <code>std</code>,不过需要通过 <code>use</code> 将标准库中定义的项引入项目包的作用域中来引用它们。例如,对于 <code>HashMap</code>,我们会使用以下语句:</p>
<pre class="playground"><code class="language-rust edition2024"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>use std::collections::HashMap;
<span class="boring">}</span></code></pre>
<p>这是一个以标准库 crate 名 <code>std</code> 开头的绝对路径。</p>
<h3 id="使用嵌套路径来清理大量的-use-列表"><a class="header" href="#使用嵌套路径来清理大量的-use-列表">使用嵌套路径来清理大量的 <code>use</code> 列表</a></h3>
<p>当需要引入很多定义于相同包或相同模块的项时,为每一项单独列出一行会占用源码大量的垂直空间。例如猜猜看章节示例 2-4 中有两行 <code>use</code> 语句都从 <code>std</code> 引入项到作用域:</p>
<p><span class="filename">文件名src/main.rs</span></p>
<pre><code class="language-rust ignore"><span class="boring">use rand::Rng;
</span>// --snip--
use std::cmp::Ordering;
use std::io;
// --snip--
<span class="boring">
</span><span class="boring">fn main() {
</span><span class="boring"> println!("Guess the number!");
</span><span class="boring">
</span><span class="boring"> let secret_number = rand::thread_rng().gen_range(1..=100);
</span><span class="boring">
</span><span class="boring"> println!("The secret number is: {secret_number}");
</span><span class="boring">
</span><span class="boring"> println!("Please input your guess.");
</span><span class="boring">
</span><span class="boring"> let mut guess = String::new();
</span><span class="boring">
</span><span class="boring"> io::stdin()
</span><span class="boring"> .read_line(&amp;mut guess)
</span><span class="boring"> .expect("Failed to read line");
</span><span class="boring">
</span><span class="boring"> println!("You guessed: {guess}");
</span><span class="boring">
</span><span class="boring"> match guess.cmp(&amp;secret_number) {
</span><span class="boring"> Ordering::Less =&gt; println!("Too small!"),
</span><span class="boring"> Ordering::Greater =&gt; println!("Too big!"),
</span><span class="boring"> Ordering::Equal =&gt; println!("You win!"),
</span><span class="boring"> }
</span><span class="boring">}</span></code></pre>
<p>相反,我们可以使用嵌套路径将相同的项在一行中引入作用域。这么做需要指定路径的相同部分,接着是两个冒号,接着是大括号中的各自不同的路径部分,如示例 7-18 所示。</p>
<p><span class="filename">文件名src/main.rs</span></p>
<pre><code class="language-rust ignore"><span class="boring">use rand::Rng;
</span>// --snip--
use std::{cmp::Ordering, io};
// --snip--
<span class="boring">
</span><span class="boring">fn main() {
</span><span class="boring"> println!("Guess the number!");
</span><span class="boring">
</span><span class="boring"> let secret_number = rand::thread_rng().gen_range(1..=100);
</span><span class="boring">
</span><span class="boring"> println!("The secret number is: {secret_number}");
</span><span class="boring">
</span><span class="boring"> println!("Please input your guess.");
</span><span class="boring">
</span><span class="boring"> let mut guess = String::new();
</span><span class="boring">
</span><span class="boring"> io::stdin()
</span><span class="boring"> .read_line(&amp;mut guess)
</span><span class="boring"> .expect("Failed to read line");
</span><span class="boring">
</span><span class="boring"> let guess: u32 = guess.trim().parse().expect("Please type a number!");
</span><span class="boring">
</span><span class="boring"> println!("You guessed: {guess}");
</span><span class="boring">
</span><span class="boring"> match guess.cmp(&amp;secret_number) {
</span><span class="boring"> Ordering::Less =&gt; println!("Too small!"),
</span><span class="boring"> Ordering::Greater =&gt; println!("Too big!"),
</span><span class="boring"> Ordering::Equal =&gt; println!("You win!"),
</span><span class="boring"> }
</span><span class="boring">}</span></code></pre>
<p><span class="caption">示例 7-18: 指定嵌套的路径在一行中将多个带有相同前缀的项引入作用域</span></p>
<p>在较大的程序中,使用嵌套路径从相同包或模块中引入很多项,可以显著减少所需的独立 <code>use</code> 语句的数量!</p>
<p>我们可以在路径的任何层级使用嵌套路径,这在组合两个共享子路径的 <code>use</code> 语句时非常有用。例如,示例 7-19 中展示了两个 <code>use</code> 语句:一个将 <code>std::io</code> 引入作用域,另一个将 <code>std::io::Write</code> 引入作用域:</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground">use std::io;
use std::io::Write;</code></pre>
<p><span class="caption">示例 7-19: 通过两行 <code>use</code> 语句引入两个路径,其中一个是另一个的子路径</span></p>
<p>两个路径的相同部分是 <code>std::io</code>,这正是第一个路径。为了在一行 <code>use</code> 语句中引入这两个路径,可以在嵌套路径中使用 <code>self</code>,如示例 7-20 所示。</p>
<p><span class="filename">文件名src/lib.rs</span></p>
<pre><code class="language-rust noplayground">use std::io::{self, Write};</code></pre>
<p><span class="caption">示例 7-20: 将示例 7-19 中部分重复的路径合并为一个 <code>use</code> 语句</span></p>
<p>这一行便将 <code>std::io</code><code>std::io::Write</code> 同时引入作用域。</p>
<h3 id="glob-运算符"><a class="header" href="#glob-运算符">glob 运算符</a></h3>
<p>如果希望将一个路径下<strong>所有</strong>公有项引入作用域,可以指定路径后跟 <code>*</code> glob 运算符:</p>
<pre class="playground"><code class="language-rust edition2024"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>use std::collections::*;
<span class="boring">}</span></code></pre>
<p>这个 <code>use</code> 语句将 <code>std::collections</code> 中定义的所有公有项引入当前作用域。使用 glob 运算符时请多加小心Glob 会使得我们难以推导作用域中有什么名称和它们是在何处定义的。</p>
<p>glob 运算符经常用于测试模块 <code>tests</code>这时会将所有内容引入作用域我们将在第十一章“如何编写测试”部分讲解。glob 运算符有时也用于 prelude 模式;查看<a href="https://doc.rust-lang.org/std/prelude/index.html#other-preludes">标准库中的文档</a>了解这个模式的更多细节。</p>
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