一个模块如何具有多个文件?

问题描述：

我不明白为什么每个模块只有一个文件.

I don't understand why we have only one file per module.

// main.rs

mod dog; // Find dog in dog.rs or dog/mod.rs
fn main() {
    dog::sonic_bark();
}

当dog变大并具有很多功能时，将所有功能都放在一个文件中是不好的.

When dog grows larger and has lots of functions, it's not good to have all of them in one file.

我如何不使用它们就将它们分开

How I can separate them without using

dog::leg::walk();
dog::head::nose::smell();
dog::tail::iron_tail();
dog::mouth::sonic_bark();

我只想使用dog::sonic_bark();

答

您不能.

您可以拥有比文件更多的模块(典型示例是mod tests嵌套在文件中)，但不能相反.

You can have more modules than files (the typical examples being mod tests nested in the file), but not the reverse.

但是，这并不重要，因为您可以使用封装+重新导出.

However, this does not matter because you can use encapsulation + re-export.

使用mod xxx;声明子模块时的默认值是xxx是 private :当前模块的任何用户都不会知道它依赖于xxx.

The default when declaring a submodule with mod xxx; is that xxx is private: no user of the current module will know that it depends on xxx.

通过选择重新导出符号将其合并:

Combine this with selecting re-exporting symbols:

pub use self::leg::walk;
pub use self::head::nose::smell;
pub use self::tail::iron_tail;
pub use self::mouth::sonic_bark;

您可以直接调用它们:dog::walk()，dog::smell()，...

And you can call those directly: dog::walk(), dog::smell(), ...

因此，私人进口和公共再出口可帮助您隐藏内部层次结构，同时公开一个统一的公共界面.

Therefore, private imports and public re-exports help you have a hidden internal hierarchy while exposing a flat public interface.

完整示例:

mod dog {
    pub use self::head::nose::smell;
    pub use self::leg::walk;
    pub use self::mouth::sonic_bark;
    pub use self::tail::iron_tail;

    mod leg {
        pub fn walk() {}
    }

    mod head {
        pub mod nose {
            pub fn smell() {}
        }
    }

    mod tail {
        pub fn iron_tail() {}
    }

    mod mouth {
        pub fn sonic_bark() {}
    }
}

fn main() {
    dog::sonic_bark();
}

一个模块如何具有多个文件?

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