rust - Rust 中的 "Subclassing"特征

Question

我有这样一种情况，我的几个结构应该实现多个特征，但它们都至少实现了一个共同特征。当我掌握这些结构的混合包时，我想将它们全部视为具有共同特征:将它们作为类型化为该特征的方法参数传递，将它们存储在为该特征类型化的集合中，等等。

我还没想好怎么办。这是我尝试按照此处建议的方式执行的一些代码，但无法编译:

trait ThingWithKeys {
    fn use_keys (&self) -> String;
}

//////

trait CorrectionsOfficer {
    fn hitch_up_pants (&self) -> String;
}

trait CorrectionsOfficerWithKeys: ThingWithKeys + CorrectionsOfficer {}

struct CorrectionsOfficerReal {}

impl ThingWithKeys for CorrectionsOfficerReal {
    fn use_keys (&self) -> String {
        String::from ("Clank, clank")
    }
}

impl CorrectionsOfficer for CorrectionsOfficerReal {
    fn hitch_up_pants (&self) -> String {
        String::from ("Grunt")
    }
}

impl <T: ThingWithKeys + CorrectionsOfficer> CorrectionsOfficerWithKeys for T {}

//////

trait Piano {
    fn close_lid (&self) -> String;
}

trait PianoWithKeys: Piano + ThingWithKeys {}

struct PianoReal {}

impl ThingWithKeys for PianoReal {
    fn use_keys (&self) -> String {
        String::from ("Tinkle, tinkle")
    }
}

impl Piano for PianoReal {
    fn close_lid (&self) -> String {
        String::from ("Bang!")
    }
}

impl <T: ThingWithKeys + Piano> PianoWithKeys for T {}

//////

trait Florida {
    fn hurricane (&self) -> String;
}

trait FloridaWithKeys: ThingWithKeys + Florida {}

struct FloridaReal {}

impl ThingWithKeys for FloridaReal {
    fn use_keys (&self) -> String {
        String::from ("Another margarita, please")
    }
}

impl Florida for FloridaReal {
    fn hurricane (&self) -> String {
        String::from ("Ho-hum...")
    }
}

impl <T: ThingWithKeys + Florida> FloridaWithKeys for T {}

//////

fn main() {
    let corrections_officer_ref: &CorrectionsOfficerWithKeys = &CorrectionsOfficerReal {};
    let piano_ref: &PianoWithKeys = &PianoReal {};
    let florida_ref: &FloridaWithKeys = &FloridaReal {};

    use_keys (corrections_officer_ref);
    use_keys (piano_ref);
    use_keys (florida_ref);
}

fn use_keys (thing_with_keys: &ThingWithKeys) {
    println! ("{}", thing_with_keys.use_keys ());
}

这里是编译错误:

Compiling playground v0.0.1 (file:///playground)
error[E0308]: mismatched types
  --> src/main.rs:80:19
   |
80 |         use_keys (corrections_officer_ref);
   |                   ^^^^^^^^^^^^^^^^^^^^^^^ expected trait `ThingWithKeys`, found trait `CorrectionsOfficerWithKeys`
   |
   = note: expected type `&ThingWithKeys`
              found type `&CorrectionsOfficerWithKeys`

error[E0308]: mismatched types
  --> src/main.rs:81:19
   |
81 |         use_keys (piano_ref);
   |                   ^^^^^^^^^ expected trait `ThingWithKeys`, found trait `PianoWithKeys`
   |
   = note: expected type `&ThingWithKeys`
              found type `&PianoWithKeys`

error[E0308]: mismatched types
  --> src/main.rs:82:19
   |
82 |         use_keys (florida_ref);
   |                   ^^^^^^^^^^^ expected trait `ThingWithKeys`, found trait `FloridaWithKeys`
   |
   = note: expected type `&ThingWithKeys`
              found type `&FloridaWithKeys`

error: aborting due to 3 previous errors

本质上，它仍然无法在 XxxWithKeys 实现中找到 ThingWithKeys 实现。

Answer 1

Rust 中的 Trait 继承不同于 OOP 继承。特征继承只是一种指定要求的方式。 trait B: A 并不意味着如果一个类型实现了B它会自动执行 A ;这意味着如果一个类型实现了 B它必须实现 A .这也意味着您将必须实现 A 分别如果B已实现。

举个例子，

trait A {}
trait B: A {}

struct S;

impl B for S {}

// Commenting this line will result in a "trait bound unsatisfied" error
impl A for S {}

fn main() {
    let _x: &B = &S;
}

但是，如果想要一个类型自动实现C如果它实现 A和 B (从而避免为该类型手动实现 C)，那么您可以使用通用的 impl :

impl<T: A + B> C for T {}

在你的例子中，这转化为

impl<T: Florida + ThingWithKeys> FloridaWithKeys for T {}

看看this forum thread获取更多信息。

顺便说一句，您不需要 ThingWithKeys开往PianoWithKeys作为Piano已经需要 ThingWithKeys .

编辑(根据您的评论和问题编辑):

如前所述，Rust 中的特征继承不同于 OOP 继承。 即使 trait B: A , 你不能强制 B 的特征对象到 A 的特征对象.如果您别无选择，只能按原样将特征对象传递给方法，那么使用泛型是可行的:

fn use_keys<T: ThingWithKeys + ?Sized>(thing_with_keys: &T) {
    println! ("{}", thing_with_keys.use_keys ());
}

泛型方法也适用于类型引用(非特征对象)。

同时检查:Why doesn't Rust support trait object upcasting?