Typing rest operator object in TypeScript

user_9824703 0
Viewed 1190

I was trying to remove a property from an Object using the rest operator and could not figure out how to force a type on the new Object.

interface Ab {
  a: string;
  b: number;
}

interface Bc {
  b: number;
  c: boolean | undefined;
}

const ab: Ab = { a: 'a', b: 1};
const {a, ...bc} = {...ab, c: true};

I know that bc now has a type

{
  b: number;
  c: boolean;
}

is there a way to force bc to be explicitly of type Bc?

spread-syntax typescript interface
2 Answers

Actually it's pretty fun and tricky question :) So here is my solution:

interface Abc extends Ab, Bc {
  bc: Bc;
}

const ab: Ab = { a: 'a', b: 1};
const { a, ...bc }: Abc = { ...ab, c: true } as Abc;

Playground link

user_35510350

It feels like what you're trying to achieve is very particular to the case you're handling at the moment and not something that can be applied in general.

If I were to take your original example and rename some variables and types (without changing any code), I could get this:

interface Human {
  intelligenceQuotient: number;
  speed: number;
}

interface Dog {
  speed: number;
  name: boolean | undefined;
}

const david: Human = { intelligenceQuotient: 100, speed: 5};
const {intelligenceQuotient, ...bc} = {...david, name: "David"};

Your question then becomes:
If I remove the intelligenceQuotient from david and give him a name, then why doesn't he become a Dog?

A simplified version of the code in your question would be:

interface Human {
  numberOfLegs: number;
}

interface Dog {
  numberOfLegs: number;
}

interface DogWithName extends Dog {
  petName: string;
}

const dog: DogWithName = { numberOfLegs: 4, petName: "Snowy" };
const { petName, ...rest } = dog;

rest is neither Dog nor Human, it is of type { numberOfLegs: number; }.

user_21490930
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