Suppose I have the following code snippet:
class Arachnid {
numberOfLegs: number = 8;
}
export class Spider extends Arachnid {
numberOfEyes: number = 2;
}
Is there a method to create a type that only includes the numberOfEyes property from Spider?
I cannot use:
type WolfSpider = Omit<Spider, keyof Arachnid>since Arachnid is not exported. How can this be achieved without relying on modifications by the library author?
Unfortunately, it's not feasible. TypeScript employs structural typing, where a type retains only its structure without any data on class hierarchy. This implies that if the base class is not publicly accessible, there's no means to access it.
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