Typescript: Ensure a type has a property A if it also has property B

Question

I have a type that looks like something as follows:

type MyType = {
  a: string;
  b: string;
  c?: number;
  d?: string;
}

There are objects of this type which can look like:

const myObj1: MyType = { a, b };
const myObj2: MyType = { a, b, c, d };

So if an object of MyType has the property c, it will definitely have property d. Is there a way to define this type as such that I do not have to non-null assert or check for both types other than extending the type into a different type?

Answer 1

If you're allowed to change MyType, one approach is to separate the c and d properties into a different object where they're required, and alternate with an intersection with that object.

type AB = {
    a: string;
    b: string;
};
type MyType = AB | AB & {
    c: number;
    d: string;
}

This way, if something of MyType is seen to have a c property, TypeScript will see that it definitely also has a d property.

Answer 2

If you want property 'd' to exist only when 'c' is defined and vice-versa, it would be worth using extended interfaces and overwriting the optional properties as never as follows:

type Base = {
  a: string;
  b: string;
  c?: number;
  d?: string;
} 

interface Concrete1 extends Base {
  c?: never;
  d?: never;
}
    
interface Concrete2 extends Base {
  c: number;
  d: string;
}

type MyType = Concrete1 | Concrete2;

const myObj1: MyType = { a: 'a', b: 'b' }; ?
const myObj2: MyType = { a: 'a', b: 'b', c: 2 }; ?
const myObj3: MyType = { a: 'a', b: 'b', c: 2, d: 'd' }; ?

This will ensure that an object of MyType always contains either a, b or a,b,c,d properties.

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