Advanced TypeScript Exercises - Answer 2

#typescript #javascript

In the question I have asked why below snippet doesn't compile

type User = {
  id: number;
  kind: string;
};

function makeCustomer<T extends User>(u: T): T {
  // Below error, why?
  return {
    id: u.id,
    kind: 'customer'
  }
}

At the first look object which is returned by makeCustomer is valid User type as it has both needed fields defined in the User. The crucial thing to understand is that we work here with type variable T which extends from User but it doesn't mean it is User. T is assignable to User, so it needs to have all fields which User has, but, it can have more fields!

Yes and this is exactly the issue, returned object is a User and pass all constraints of it, but doesn't pass all constraints of T which can have additional fields. We don't know what are those fields though, so in order to fix the typing we should make an object which has all fields of T, and we know all fields of T are in argument u. We can then use spread operator in order to spread all unknown fields to the newly created object.

function makeCustomer<T extends User>(u: T): T {
  // no error
    return {
    ...u, // spread all properties of u being T
    id: u.id, // yes redundant line, leaving it for consistency
    kind: 'customer'
  }
}

Now we are sure that all fields from T will be included in our returned object. But there is also a case where TypeScript has an issue, design issue probably. We can create here situation which will be just a bug, when value will not match the type. Consider that we can make another type which will extend User, lets say Admin

type Admin = User & {
  kind: 'admin';
}

We can ask if Admin extends truly the User:

type IsAdminAUser = Admin extends User ? true : false // evaluates to true

Ok so Admin is a User, then we can use makeCustomer with Admin? Looks that we should as T extends User and Admin extends User. Lets check:

const admin = makeCustomer({ id: 1, kind: 'admin' } as Admin)

And there is a bug, TS is saying admin has a type Admin but when we console.log it, the structure is {id: 1, kind: 'customer'}, so unfortunately we got to the situation where TS has wrong assumptions 😩.

The whole code can be found in the playground

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Top comments (8)

Nikita Monastyrskiy • Aug 9 '22

Good explanation, thanks!

Tugbay Atilla • Nov 7 '23 • Edited

Thanks for the good examples.

for the bug, i would say, we are overriding "kind" member with 'customer' no matter what. So the solution could be like checking if the kind member given by the argument. I would not call this as bug, i was expecting this behavior actually :)

function makeCustomer<T extends User>(u: T): T {
    return {
    ...u,
    id: u.id,
    kind: u.kind ?? 'customer' // <-- here is the changed part
  }
}

const admin = makeCustomer({ id: 1, kind: 'admin' } as Admin)
console.log(admin); // {id: 1, kind: "admin"}

but also the important thing is what is the expectation from the function. function says, makeCustomer, does not matter we are assigning to kind 'admin', 'hooman', or 'animal', eventually it sounds like we are making a customer :)
yet, it was good example :)

Tiago Miguel Costa Rodrigues • Jan 30 '23

Thank you for putting together these exercises, they are really great for learning purposes!

I have a question regarding the bug you mention please. Isn't TS assuming the wrong kind because in makeCustomer you are overwriting the kind to be customer ?

Wouldn't the following solve the bug?

function makeCustomer<T extends User>(u: T): T {
    return {
    ...u, // spread all properties of u being T
  }
}

Pragmatic Maciej • Feb 7 '23

That would indeed fixed the issue, but the idea was to have constructor of the customer. Besides sense of this function (which is discussable) we can fix it by saying we will return specific kind. Pay attention that I needed to omit kind property from Admin type as having a join between {kind:'a'} & {kind: 'b'} creates never type as there is no intersection between them. Here is the code - typescriptlang.org/play?#code/C4Tw...