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< h1 id = "模块-module" > < a class = "header" href = "#模块-module" > 模块 Module< / a > < / h1 >
< p > 在本章节，我们将深入讲讲 Rust 的代码构成单元：模块。使用模块可以将包中的代码按照功能性进行重组，最终实现更好的可读性及易用性。同时，我们还能非常灵活地去控制代码的可见性，进一步强化 Rust 的安全性。< / p >
< h2 id = "创建嵌套模块" > < a class = "header" href = "#创建嵌套模块" > 创建嵌套模块< / a > < / h2 >
< p > 小旅馆， ， ， ， ， ， < / p >
< p > 使用 < code > cargo new --lib restaurant< / code > 创建一个小餐馆，注意，这里创建的是一个库类型的 < code > Package< / code > ，然后将以下代码放入 < code > src/lib.rs< / code > 中：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > // 餐厅前厅，用于吃饭
mod front_of_house {
mod hosting {
fn add_to_waitlist() {}
fn seat_at_table() {}
}
mod serving {
fn take_order() {}
fn serve_order() {}
fn take_payment() {}
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 以上的代码创建了三个模块，有几点需要注意的：< / p >
< ul >
< li > 使用 < code > mod< / code > 关键字来创建新模块，后面紧跟着模块名称< / li >
< li > 模块可以嵌套，这里嵌套的原因是招待客人和服务都发生在前厅，因此我们的代码模拟了真实场景< / li >
< li > 模块中可以定义各种 Rust 类型，例如函数、结构体、枚举、特征等< / li >
< li > 所有模块均定义在同一个文件中< / li >
< / ul >
< p > 类似上述代码中所做的，使用模块，我们就能将功能相关的代码组织到一起，然后通过一个模块名称来说明这些代码为何被组织在一起。这样其它程序员在使用你的模块时，就可以更快地理解和上手。< / p >
< h2 id = "模块树" > < a class = "header" href = "#模块树" > 模块树< / a > < / h2 >
< p > 在< a href = "https://course.rs/basic/crate-module/crate.html" > 上一节< / a > 中，我们提到过 < code > src/main.rs< / code > 和 < code > src/lib.rs< / code > 被称为包根(crate root)，这个奇葩名称的来源（我不想承认是自己翻译水平太烂-,-）是由于这两个文件的内容形成了一个模块 < code > crate< / code > ，该模块位于包的树形结构（由模块组成的树形结构）的根部：< / p >
< pre > < code class = "language-console" > crate
└── front_of_house
├── hosting
│ ├── add_to_waitlist
│ └── seat_at_table
└── serving
├── take_order
├── serve_order
└── take_payment
< / code > < / pre >
< p > 这颗树展示了模块之间< strong > 彼此的嵌套< / strong > 关系，因此被称为< strong > 模块树< / strong > 。其中 < code > crate< / code > 包根是 < code > src/lib.rs< / code > 文件，包根文件中的三个模块分别形成了模块树的剩余部分。< / p >
< h4 id = "父子模块" > < a class = "header" href = "#父子模块" > 父子模块< / a > < / h4 >
< p > 如果模块 < code > A< / code > 包含模块 < code > B< / code > ，那么 < code > A< / code > 是 < code > B< / code > 的父模块，< code > B< / code > 是 < code > A< / code > 的子模块。在上例中，< code > front_of_house< / code > 是 < code > hosting< / code > 和 < code > serving< / code > 的父模块，反之，后两者是前者的子模块。< / p >
< p > 聪明的读者，应该能联想到，模块树跟计算机上文件系统目录树的相似之处。不仅仅是组织结构上的相似，就连使用方式都很相似：每个文件都有自己的路径，用户可以通过这些路径使用它们，在 Rust 中，我们也通过路径的方式来引用模块。< / p >
< h2 id = "用路径引用模块" > < a class = "header" href = "#用路径引用模块" > 用路径引用模块< / a > < / h2 >
< p > 想要调用一个函数，就需要知道它的路径，在 Rust 中，这种路径有两种形式：< / p >
< ul >
< li > < strong > 绝对路径< / strong > ，从包根开始，路径名以包名或者 < code > crate< / code > 作为开头< / li >
< li > < strong > 相对路径< / strong > ，从当前模块开始，以 < code > self< / code > ， < code > super< / code > 或当前模块的标识符作为开头< / li >
< / ul >
< p > 让我们继续经营那个惨淡的小餐馆，这次为它实现一个小功能：
< span class = "filename" > 文件名： < / span > < / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > mod front_of_house {
mod hosting {
fn add_to_waitlist() {}
}
}
pub fn eat_at_restaurant() {
// 绝对路径
crate::front_of_house::hosting::add_to_waitlist();
// 相对路径
front_of_house::hosting::add_to_waitlist();
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 上面的代码为了简化实现，省去了其余模块和函数，这样可以把关注点放在函数调用上。< code > eat_at_restaurant< / code > 是一个定义在包根中的函数，在该函数中使用了两种方式对 < code > add_to_waitlist< / code > 进行调用。< / p >
< h4 id = "绝对路径引用" > < a class = "header" href = "#绝对路径引用" > 绝对路径引用< / a > < / h4 >
< p > 因为 < code > eat_at_restaurant< / code > 和 < code > add_to_waitlist< / code > 都定义在一个包中，因此在绝对路径引用时，可以直接以 < code > crate< / code > 开头，然后逐层引用，每一层之间使用 < code > ::< / code > 分隔：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > crate::front_of_house::hosting::add_to_waitlist();
< span class = "boring" > }< / span > < / code > < / pre >
< p > 对比下之前的模块树：< / p >
< pre > < code class = "language-console" > crate
└── eat_at_restaurant
└── front_of_house
├── hosting
│ ├── add_to_waitlist
│ └── seat_at_table
└── serving
├── take_order
├── serve_order
└── take_payment
< / code > < / pre >
< p > 可以看出，绝对路径的调用，完全符合了模块树的层级递进，非常符合直觉，如果类比文件系统，就跟使用绝对路径调用可执行程序差不多：< code > /front_of_house/hosting/add_to_waitlist< / code > ，使用 < code > crate< / code > 作为开始就和使用 < code > /< / code > 作为开始一样。< / p >
< h4 id = "相对路径引用" > < a class = "header" href = "#相对路径引用" > 相对路径引用< / a > < / h4 >
< p > 再回到模块树中，因为 < code > eat_at_restaurant< / code > 和 < code > front_of_house< / code > 都处于包根 < code > crate< / code > 中，因此相对路径可以使用 < code > front_of_house< / code > 作为开头：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > front_of_house::hosting::add_to_waitlist();
< span class = "boring" > }< / span > < / code > < / pre >
< p > 如果类比文件系统，那么它类似于调用同一个目录下的程序，你可以这么做：< code > front_of_house/hosting/add_to_waitlist< / code > ，嗯也很符合直觉。< / p >
< h4 id = "绝对还是相对" > < a class = "header" href = "#绝对还是相对" > 绝对还是相对？< / a > < / h4 >
< p > 如果只是为了引用到指定模块中的对象，那么两种都可以，但是在实际使用时，需要遵循一个原则：< strong > 当代码被挪动位置时，尽量减少引用路径的修改< / strong > ，相信大家都遇到过，修改了某处代码，导致所有路径都要挨个替换，这显然不是好的路径选择。< / p >
< p > 回到之前的例子，假设要把 < code > front_of_house< / code > 模块和 < code > eat_at_restaurant< / code > 移动到一个模块中 < code > customer_experience< / code > ，那么绝对路径的引用方式就必须进行修改：< code > crate::customer_experience::front_of_house ...< / code > ，但如果使用的是相对路径，那该路径就无需修改，因为它们的相对位置其实没有变：< / p >
< pre > < code class = "language-console" > crate
└── customer_experience
└── eat_at_restaurant
└── front_of_house
├── hosting
│ ├── add_to_waitlist
│ └── seat_at_table
< / code > < / pre >
< p > 从新的模块树中可以很清晰的看出这一点。< / p >
< p > 再比如，其它的都不动，把 < code > eat_at_restaurant< / code > 移动到模块 < code > dining< / code > 中，如果使用相对路径，你需要修改该路径，但如果使用的是绝对路径，就无需修改：< / p >
< pre > < code class = "language-console" > crate
└── dining
└── eat_at_restaurant
└── front_of_house
├── hosting
│ ├── add_to_waitlist
< / code > < / pre >
< p > 不过，如果不确定哪个好，你可以考虑优先使用绝对路径，因为调用的地方和定义的地方往往是分离的，而定义的地方较少会变动。< / p >
< h2 id = "代码可见性" > < a class = "header" href = "#代码可见性" > 代码可见性< / a > < / h2 >
< p > 让我们运行下面（之前）的代码：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > mod front_of_house {
mod hosting {
fn add_to_waitlist() {}
}
}
pub fn eat_at_restaurant() {
// 绝对路径
crate::front_of_house::hosting::add_to_waitlist();
// 相对路径
front_of_house::hosting::add_to_waitlist();
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 运行 < code > cargo build< / code > 编译此库类型的 < code > Package< / code > ，意料之外的报错了，毕竟看上去确实很简单且没有任何问题：< / p >
< pre > < code class = "language-console" > error[E0603]: module `hosting` is private
--> src/lib.rs:9:28
|
9 | crate::front_of_house::hosting::add_to_waitlist();
| ^^^^^^^ private module
< / code > < / pre >
< p > 错误信息很清晰：< code > hosting< / code > 模块是私有的，无法在包根进行访问，那么为何 < code > front_of_house< / code > 模块就可以访问？因为它和 < code > eat_at_restaurant< / code > 同属于一个包根作用域内，同一个模块内的代码自然不存在私有化问题（所以我们之前章节的代码都没有报过这个错误！）。< / p >
< p > 模块不仅仅对于组织代码很有用，它还能定义代码的私有化边界：在这个边界内，什么内容能让外界看到，什么内容不能，都有很明确的定义。因此，如果希望让函数或者结构体等类型变成私有化的，可以使用模块。< / p >
< p > Rust 出于安全的考虑，默认情况下，所有的类型都是私有化的，包括函数、方法、结构体、枚举、常量，是的，就连模块本身也是私有化的。在中国，父亲往往不希望孩子拥有小秘密，但是在 Rust 中，< strong > 父模块完全无法访问子模块中的私有项，但是子模块却可以访问父模块、父父..模块的私有项< / strong > 。< / p >
< h4 id = "pub-关键字" > < a class = "header" href = "#pub-关键字" > pub 关键字< / a > < / h4 >
< p > 类似其它语言的 < code > public< / code > 或者 Go 语言中的首字母大写， < code > pub< / code > 关键字，通过它你可以控制模块和模块中指定项的可见性。< / p >
< p > 由于之前的解释，我们知道了只需要将 < code > hosting< / code > 模块标记为对外可见即可：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > mod front_of_house {
pub mod hosting {
fn add_to_waitlist() {}
}
}
/*--- snip ----*/
< span class = "boring" > }< / span > < / code > < / pre >
< p > 但是不幸的是，又报错了：< / p >
< pre > < code class = "language-console" > error[E0603]: function `add_to_waitlist` is private
--> src/lib.rs:12:30
|
12 | front_of_house::hosting::add_to_waitlist();
| ^^^^^^^^^^^^^^^ private function
< / code > < / pre >
< p > 哦？难道模块可见还不够，还需要将函数 < code > add_to_waitlist< / code > 标记为可见的吗？ 是的，没错，模块可见性不代表模块内部项的可见性，模块的可见性仅仅是允许其它模块去引用它，但是想要引用它内部的项，还得继续将对应的项标记为 < code > pub< / code > 。< / p >
< p > 在实际项目中，一个模块需要对外暴露的数据和 API 往往就寥寥数个，如果将模块标记为可见代表着内部项也全部对外可见，那你是不是还得把那些不可见的，一个一个标记为 < code > private< / code > ？反而是更麻烦的多。< / p >
< p > 既然知道了如何解决，那么我们为函数也标记上 < code > pub< / code > ： < / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > mod front_of_house {
pub mod hosting {
pub fn add_to_waitlist() {}
}
}
/*--- snip ----*/
< span class = "boring" > }< / span > < / code > < / pre >
< p > Bang， ， < / p >
< h2 id = "使用-super-引用模块" > < a class = "header" href = "#使用-super-引用模块" > 使用 < code > super< / code > 引用模块< / a > < / h2 >
< p > 在< a href = "#用路径引用模块" > 用路径引用模块< / a > 中，我们提到了相对路径有三种方式开始：< code > self< / code > 、< code > super< / code > 和 < code > crate< / code > 或者模块名，其中第三种在前面已经讲到过，现在来看看通过 < code > super< / code > 的方式引用模块项。< / p >
< p > < code > super< / code > 代表的是父模块为开始的引用方式，非常类似于文件系统中的 < code > ..< / code > 语法：< code > ../a/b< / code >
< span class = "filename" > 文件名： < / span > < / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > fn serve_order() {}
// 厨房模块
mod back_of_house {
fn fix_incorrect_order() {
cook_order();
super::serve_order();
}
fn cook_order() {}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 嗯，我们的小餐馆又完善了，终于有厨房了！看来第一个客人也快可以有了。。。在厨房模块中，使用 < code > super::serve_order< / code > 语法，调用了父模块（包根）中的 < code > serve_order< / code > 函数。< / p >
< p > 那么你可能会问，为何不使用 < code > crate::serve_order< / code > 的方式？额，其实也可以，不过如果你确定未来这种层级关系不会改变，那么 < code > super::serve_order< / code > 的方式会更稳定，未来就算它们都不在包根了，依然无需修改引用路径。所以路径的选用，往往还是取决于场景，以及未来代码的可能走向。< / p >
< h2 id = "使用-self-引用模块" > < a class = "header" href = "#使用-self-引用模块" > 使用 < code > self< / code > 引用模块< / a > < / h2 >
< p > < code > self< / code > 其实就是引用自身模块中的项，也就是说和我们之前章节的代码类似，都调用同一模块中的内容，区别在于之前章节中直接通过名称调用即可，而 < code > self< / code > ，你得多此一举：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > fn serve_order() {
self::back_of_house::cook_order()
}
mod back_of_house {
fn fix_incorrect_order() {
cook_order();
crate::serve_order();
}
pub fn cook_order() {}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 是的，多此一举，因为完全可以直接调用 < code > back_of_house< / code > ，但是 < code > self< / code > 还有一个大用处，在下一节中我们会讲。< / p >
< h2 id = "结构体和枚举的可见性" > < a class = "header" href = "#结构体和枚举的可见性" > 结构体和枚举的可见性< / a > < / h2 >
< p > 为何要把结构体和枚举的可见性单独拎出来讲呢？因为这两个家伙的成员字段拥有完全不同的可见性：< / p >
< ul >
< li > 将结构体设置为 < code > pub< / code > ，但它的所有字段依然是私有的< / li >
< li > 将枚举设置为 < code > pub< / code > ，它的所有字段也将对外可见< / li >
< / ul >
< p > 原因在于，枚举和结构体的使用方式不一样。如果枚举的成员对外不可见，那该枚举将一点用都没有，因此枚举成员的可见性自动跟枚举可见性保持一致，这样可以简化用户的使用。< / p >
< p > 而结构体的应用场景比较复杂，其中的字段也往往部分在 A 处被使用，部分在 B 处被使用，因此无法确定成员的可见性，那索性就设置为全部不可见，将选择权交给程序员。< / p >
< h2 id = "模块与文件分离" > < a class = "header" href = "#模块与文件分离" > 模块与文件分离< / a > < / h2 >
< p > 在之前的例子中，我们所有的模块都定义在 < code > src/lib.rs< / code > 中，但是当模块变多或者变大时，需要将模块放入一个单独的文件中，让代码更好维护。< / p >
< p > 现在，把 < code > front_of_house< / code > 前厅分离出来，放入一个单独的文件中 < code > src/front_of_house.rs< / code > ： < / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub mod hosting {
pub fn add_to_waitlist() {}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 然后，将以下代码留在 < code > src/lib.rs< / code > 中：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > mod front_of_house;
pub use crate::front_of_house::hosting;
pub fn eat_at_restaurant() {
hosting::add_to_waitlist();
hosting::add_to_waitlist();
hosting::add_to_waitlist();
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > so easy！ ， ： < / p >
< ul >
< li > < code > mod front_of_house;< / code > 告诉 Rust 从另一个和模块 < code > front_of_house< / code > 同名的文件中加载该模块的内容< / li >
< li > 使用绝对路径的方式来引用 < code > hosting< / code > 模块：< code > crate::front_of_house::hosting;< / code > < / li >
< / ul >
< p > 需要注意的是，和之前代码中 < code > mod front_of_house{..}< / code > 的完整模块不同，现在的代码中，模块的声明和实现是分离的，实现是在单独的 < code > front_of_house.rs< / code > 文件中，然后通过 < code > mod front_of_house;< / code > 这条声明语句从该文件中把模块内容加载进来。因此我们可以认为，模块 < code > front_of_house< / code > 的定义还是在 < code > src/lib.rs< / code > 中，只不过模块的具体内容被移动到了 < code > src/front_of_house.rs< / code > 文件中。< / p >
< p > 在这里出现了一个新的关键字 < code > use< / code > ，联想到其它章节我们见过的标准库引入 < code > use std::fmt;< / code > ，可以大致猜测，该关键字用来将外部模块中的项引入到当前作用域中来，这样无需冗长的父模块前缀即可调用：< code > hosting::add_to_waitlist();< / code > ，在下节中，我们将对 < code > use< / code > 进行详细的讲解。< / p >
< p > 当一个模块有许多子模块时，我们也可以通过文件夹的方式来组织这些子模块。< / p >
< p > 在上述例子中，我们可以创建一个目录 < code > front_of_house< / code > ，然后在文件夹里创建一个 < code > hosting.rs< / code > 文件，< code > hosting.rs< / code > 文件现在就剩下：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub fn add_to_waitlist() {}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 现在，我们尝试编译程序，很遗憾，编译器报错：< / p >
< pre > < code > error[E0583]: file not found for module `front_of_house`
--> src/lib.rs:3:1
|
1 | mod front_of_house;
| ^^^^^^^^^^^^^^^^^^
|
= help: to create the module `front_of_house`, create file "src/front_of_house.rs" or "src/front_of_house/mod.rs"
< / code > < / pre >
< p > 是的，如果需要将文件夹作为一个模块，我们需要进行显式指定暴露哪些子模块。按照上述的报错信息，我们有两种方法：< / p >
< ul >
< li > 在 < code > front_of_house< / code > 目录里创建一个 < code > mod.rs< / code > ，如果你使用的 < code > rustc< / code > 版本 < code > 1.30< / code > 之前，这是唯一的方法。< / li >
< li > 在 < code > front_of_house< / code > < strong > 同级< / strong > 目录里创建一个与模块（目录）< strong > 同名< / strong > 的 rs 文件 < code > front_of_house.rs< / code > ，在新版本里，更建议使用这样的命名方式来避免项目中存在大量同名的 < code > mod.rs< / code > 文件（ Python 点了个 < code > 踩< / code > ）。< / li >
< / ul >
< p > 如果使用第二种方式，文件结构将如下所示：< / p >
< pre > < code class = "language-shell" > src
├── front_of_house
│ └── hosting.rs
├── front_of_house.rs
└── lib.rs
< / code > < / pre >
< p > 而无论是上述哪个方式创建的文件，其内容都是一样的，你需要在定义你(< code > mod.rs< / code > 或 < code > front_of_house.rs< / code > )的子模块（子模块名与文件名相同）：< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub mod hosting;
// pub mod serving;
< span class = "boring" > }< / span > < / code > < / pre >
< 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/crate-module/module.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/crate-module/module.md" > 习题解答< / a > 。< / p >
< / blockquote >
< / main >
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