Setting a variable based on TypeScript Generic

Question

Is it possible to set a variable based on a TypeScript Generic?

class Foo<T> {
  constructor() {
    // This doesn't actually work but it's essentially what I would like to do
    const myKey = (T extends number ? 'myNumber' : null);
  }
}

Here is a more complete example that will work in an external IDE. The objective here is to instantiate the Animal class with the appropriate generic (Dog | Cat) while also giving the constructor the ability to get a subset of the Object returned from getUpdate. In the below example, this is done with myKey. To reiterate on my original question, is there a way to perform all of this without having "Dog" appear 2 times when instantiating the class?

interface Dog {
  bark: string;
}

interface Cat {
  meow: string;
}

type AnimalType<T> =
  T extends Dog ? Dog :
    T extends Cat ? Cat : null;

const getUpdate = (): Record<string, Dog | Cat> => ({
  dog: { bark: 'woof' },
  cat: { meow: 'meow' },
});

class Animal<T> {
  public state!: AnimalType<T>;

  constructor(myKey: 'dog' | 'cat') {
    this.state = getUpdate()[myKey] as AnimalType<T>;
  }
}
// dogInstance is of type `Animal<Dog>` with a state of `{ bark: 'woof' }`
const dogInstance = new Animal<Dog>('dog');

I'm afraid it is not. TypeScript/JavaScript doesn't have reflection based on types. Possibly you can check here property which type T extends from number. — Anton
– Anton, Commented Feb 12, 2020 at 14:05
What's the use case? Maybe you want multiple classes? class FooNumber { ... myKey = 'myNumber' and class FooNull { ... myKey = null? Right now it doesn't look like T serves much of a purpose; can you expand out to a minimal reproducible example if you want more targeted advice? — jcalz
– jcalz, Commented Feb 12, 2020 at 14:09
@jcalz I'm trying to remove redundant code. My generic and constructor parameter are essentially the same in all cases so I want to drop one but keep the generic tagging. — J'e
– J'e, Commented Feb 12, 2020 at 14:12
Could you provide example code that demonstrates this? I don't know what you consider redundant without seeing it; it could be that what are trying to remove is necessary, or it could be that you could remove the redundancy in a different way. — jcalz
– jcalz, Commented Feb 12, 2020 at 14:16
You can probably change it so that new Foo('MyType') infers MyType for T, but without a true minimal reproducible example as described in How to Ask, I'm not going to guess about it. Looks like you've already accepted the answer below, which still requires new Foo<MyType>('MyType'), so I must be missing something about the use case. If you edit your question with an example that can be dropped into a standalone IDE and demonstrate the issue, I'll take a look. Otherwise, good luck! — jcalz
– jcalz, Commented Feb 12, 2020 at 15:03

Maciej Sikora · Accepted Answer · 2020-02-12 14:05:48Z

3

What I see is that you are mixing value and type level. We need to understand the difference, you need to have real value for given type to assign it, you cannot just assign a type.

I believe this is what you are looking for:

class Foo<T> {
  constructor(key: T extends number ? 'myNumber' : null) {
    const myKey = key;
  }
}

const foo = new Foo<number>('myNumber'); // only possible option is 'myNumber'
const fooNull = new Foo<null>(null); // only possible option is null

As you can see I am constraining the real value of the argument and pass it to myKey.

answered Feb 12, 2020 at 14:05

Maciej Sikora

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