TypeScript discriminating union - parameter of discrimating property type

Question

I have the following definitions:

interface Dog {
  bark(): void;
  type: 'dog';
}

interface Cat {
  meow(): void;
  type: 'cat';
}

type Animal = Cat | Dog;

Now I would like to make a function that uses Animal.type as a parameter:

function useAnimalType(type: string) {}

In this example, I used string, but I really mean 'cat' | 'dog'. I realize I could write out all the possible type of Animal, but imagine a scenario when there are hundreds of animal types defined in Animal. How do I reuse the Animal.type as a parameter here?

Answer 1

You can retrieve the type of a field in another object type by using bracket notation. For your example, simply:

function useAnimalType(type: Animal["type"]) {}

would suffice. In such case, Animal["type"] is inferred to be "cat" | "dog".