TypeScript is there some elegant way to reflect child class type in the parent class?-CodePudding

I has a base class

class Animal {
  public abstract clone(): Animal
}

and some derived classes

class Dog extends Aniaml {
  public clone(): Dog {
    return clone(this)
  }
}
class Cat extends Aniaml {
  public clone(): Cat {
    return clone(this)
  }
}

Since the clone() implementations in the derived classes is identical each other. DRY priciple tells me that the code is less elegent.

So the version 2:

class Animal {
  public clone(): Animal {
    return clone(this)
  }
}
class Dog extends Aniaml {}
class Cat extends Aniaml {}

In this way, it's more elegant. But there is one thing bother me which, see below:

const dog = new Dog()
const clonedDog = dog.clone() // typeof clonedDog is 'Animal'
const clonedDog = dog.clone() as Animal // typeof clonedDog is 'Dog' but I dont want type assertion here

So, the question is: Is there the elegant way to auto infer clone() 's concrete return type in the derived classes?

I already had one solution but I dont think it's elegant :(

// still less elegant
class Animal<T> {
  public clone(): <T> {
    return clone(this)
  }
}
class Dog extends Aniaml<Dog> {}
class Cat extends Aniaml<Cat> {}

THANKS for the any help!!!

CodePudding user response：

Use the this type. It is the type of the current implementation.

class Animal {
    clone(): this {
        return clone(this)
    }
}

class Dog extends Animal {}

const dog = new Dog()
const clonedDog = dog.clone() // Dog