Question

我正在尝试利用TypeScript类型检查，但我坚持使用以下代码：

abstract class Mammal {
  abstract breed(other: Mammal);
}

class Dog extends Mammal {
  breed(other: Dog) {}
}

class Cat extends Mammal {
  breed(other: Cat) {}
}

const toby = new Dog();
const lucy = new Dog();
const luna = new Cat();

toby.breed(lucy); // OK
toby.breed(luna); // Works, but it shouldn't since luna is a Cat!

似乎TypeScript正在执行某种类型的鸭子打字并正在考虑Dog == Cat。如何让typechecker拒绝此代码？

Sidenote ：这就是我在Rust中的表现：

trait Mammal {
    fn breed(&self, other: &Self);
}

struct Cat {}

impl Mammal for Cat {
    fn breed(&self, _other: &Self) {}
}

struct Dog {}

impl Mammal for Dog {
    fn breed(&self, _other: &Self) {}
}

fn main() {
    let toby = Dog{};
    let lucy = Dog{};
    let luna = Cat{};

    toby.breed(&lucy);
    toby.breed(&luna); // expected struct `Dog`, found struct `Cat`
}

Answer 1

TypeScript是structurally typed（感谢@jcalz）。

结构类型背后的想法是，如果它们的成员兼容，则两种类型是兼容的。

不太优雅但有效：您可以添加虚拟属性以在类之间产生一些差异（Playground）：

abstract class Mammal {
  abstract breed(other: this): void; // Bonus: Here you can make use of Polymorphic this types
}

class Dog extends Mammal {
    private dummy1 = undefined;
    breed(other: Dog) {}
}

class Cat extends Mammal {
  private dummy2 = undefined;
  breed(other: Cat) {}
}

const toby = new Dog();
const lucy = new Dog();
const luna = new Cat();

toby.breed(lucy); // OK
toby.breed(luna); // Error

奖励：您可以在基类中使用Polymorphic this types（感谢@jcalz）。

abstract breed(other: this): void;

打字稿没有执行类型检查

1 个答案: