TypeScript: Standalone function type reusable for many functions

Question

I want to create a re-usable type for functions, so that I can say:

type MyFunc = ...

const func1: MyFunc = ...
const func2: MyFunc = ...

The following doesn't work, but basically I want:

type Book = {
  id: string;
  title: string;
}

type createBook = (book: Partial<Book>) => Book;

const someBookFactory: createBook = ({ title }) => ({ id: /* autogenerated */, title });
const someOtherBookFactory: createBook = ({ id, title }) => ({ id, title });

Is this possible in TypeScript?

Ali Habibzadeh · Accepted Answer · 2020-12-06 15:19:48Z

1

type createBook = (book: Partial<Book>) => Book;

function getId(): string {
  return Math.random().toString();
}

// Partial means they can be undefined so you need to provide default values
const someBookFactory: createBook = ({ title = "untitled" }) => ({ id: getId(), title });
const someOtherBookFactory: createBook = ({ id = getId(), title = "untitled" }) => ({ id, title });

Bottom line is Book is a contract with 2 required fields and if you are not passing them all in, at some point before returning you are in breach of contract. That is the case in all languages with static types.

// You can also have both fallbacks handled inside
const anotherBookFactory: createBook = ({ id, title }) => {
  return {
    id: id ?? getId(),
    title: title ?? "untitled"
  };
};

edited Dec 6, 2020 at 15:19

answered Dec 6, 2020 at 13:57

Ali Habibzadeh

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