TypeScript 如何为子构造函数使用更严格的类型？

Question

Suppose I have two classes, a Dog and an Animal class. In this example, Dog extends Animal . Now suppose I have two additional classes (one called Parent , and another called Child ). In TypeScript, I am able to type the arguments of my child class's constructor function to be that of my parent class's constructor function's arguments using ConstructorParameters like so:

class Parent {
  constructor(a: Animal, b: number, c: number) {}
}

class Child extends Parent {
  constructor(...args: ConstructorParameters<typeof Parent>) { // same as parent's arguments: a: Animal, b: number, c: number
    super(...args); // doesn't complain, as args matches the arguments for Parent ([a: Animal, b: number, c: number])
  }
}

In this example, new Child() and new Parent() would accept the same argument types. But how could I type the Child constructor function to accept a stricter type for a , such as Dog ?

class Parent {
  constructor(a: Animal, b: number, c: number) {}
}

class Child extends Parent {
  //          v----- want to access a in `Child`'s constructor
  constructor(a: Dog, ...args: ConstructorParameters<typeof Parent>) {
    super(a, ...args); // <-- complains, I want to pass `a: Dog` and ...args (containing [b: number, c: number])
    a.bark(); // use specific Dog methods (that don't exist on Animal)
  }
}
// Using: 
// - new Child(new Animal(), 0, 0); should complain as Animal is not of type Dog
// - new Child(new Dog(), 0, 0); should work as Dog is of type Dog

The above doesn't work, as Child now expects to be passed a Dog instance followed by the three arguments for the Parent constructor. I was thinking of using Omit<> or Partial<> on ConstructorParameters to remove the expected a: Animal type but couldn't seem to get those to work. I'm also looking for a solution that doesn't involve me needing to pass through the arguments individually to the super() call (ie: if I add arguments in Parent's constructor, I won't need to add them to the Child).

Answer 1

try conditional types

type RestWithoutOne<K> = K extends [infer WithThis, ...infer WithRest] ? WithRest : never;

class Child extends Parent {
  //          v----- want to access a in `Child`'s constructor
  constructor(a: Dog, ...args: RestWithoutOne<ConstructorParameters<typeof Parent>>) {
    const a = args[0];
    super(a, ...args); // <-- complains, I want to pass `a: Dog` and ...args (containing [b: number, c: number])
    a.bark(); // use specific Dog methods (that don't exist on Animal)
  }
}

Playground link

Answer 2

But how could I type the Child constructor function to accept a stricter type for a , such as Dog ?

That sounds like a bad idea, violating the Liskov substitution principle : one should be able to use Child everywhere where Parent can be used.

I would therefore recommend generics:

class Parent<T extends Animal> {
  constructor(a: T, b: number, c: number) {}
}

class Child extends Parent<Dog> {
  constructor(...args: ConstructorParameters<typeof Parent<Dog>>) {
    super(...args);
    const a = args[0];
    a.bark(); // use specific Dog methods (that don't exist on Animal)
  }
}