在TypeScript中，如何根据其参数之一的返回类型来确定函数的返回类型？

Question

Is there some way I can have a function whose return type varies based on one of its arguments? For example:

interface Person {
  name: string;
  job: string;
}

interface Animal {
  name: string;
  deadly: boolean;
}

function getPerson(): Person {
  return {
    name: 'Bob',
    job: 'Manager',
  };
}

function getAnimal(): Animal {
  return {
    name: 'Spot',
    deadly: false,
  };
}

function create(generator: () => Person | Animal): ReturnType<typeof generator>[] {
  return []
}

const people = create(getPerson);

This is kind of close, except the type of people is (Person | Animal)[] . I need it to be Person[] - in other words, it can look at the getPerson argument and know that this function will return a Person . I realize, of course, I could rewrite the function so I can do this:

const people = create<Person>(getPerson);

But I'd rather have it infer this from the arguments somehow.

Answer 1

Figured it out! I just modified this part of the above code:

function create<M>(generator: () => M): M[] {
  return []
}

const people = create(getPerson);
const animals = create(getAnimal);

And that worked.

Answer 2

您已经弄清楚了，如果您想查找/了解更多信息，我只是想补充一下，该概念称为“泛型”。