CODEWITHSUNDEEP
typescripttypesinterface
thetarnav
thetarnav

Reputation: 898

Is there a way to "extend an interface while creating an object" in typescript?

Let's say there is an interface that restricts what kind of keys or values can be used in an object. For example, an interface like that could be used for storing some calculations.

interface CheckNumberMethods {
    [key: string]: (n: number) => boolean
}

So while creating an actual object with some methods, I would apply that interface to have the IDE check for type mismatches.

const myMethods:CheckNumberMethods = {
    greaterThanFive: (n: number) => n > 5,
    isEven: (n: number) => n % 2 === 0,
}

But now the object is stuck with the type MyMethodsType. While I want the type to represent the actual object I've created. In this example, the object's type should look like this:

{
   greaterThanFive: (n: number) => boolean;
   isEven: (n: number) => boolean;
}

Of course, that this is natural behavior, but is there a way to have type checking while creating the object, but to keep the original object's type for "outside" usage?

In this example simply extending the interface and applying the extended one to the object would work just fine, but on a bigger scale or with more complex types it is troublesome.

Upvotes: 1

Views: 1000

Answers (1)

captain-yossarian from Ukraine
captain-yossarian from Ukraine

Reputation: 33091

In order to infer exact properties and make object assignable to CheckNumberMethods interface, you should create a function with appropriate constraints. Example:

interface CheckNumberMethods {
  [key: string]: (n: number) => boolean
}

const myMethods = {
  greaterThanFive: (n: number) => n > 5,
  isEven: (n: number) => n % 2 === 0,
}

const handle = <Obj extends CheckNumberMethods>(obj: Obj) => obj

// {
//     greaterThanFive: (n: number) => boolean;
//     isEven: (n: number) => boolean;
// }
const result = handle(myMethods)

Playground

I have removed explicit type CheckNumberMethods from myMethods because it affects return type of handle function.

Hence, you are able to infer exact properties only with extra function.

Upvotes: 2

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