How to create a sub type with values specified for some fields and eliminate optional member in typescript

Question

interface A{ 
prop1:string,
prop2:boolean,
prop3?:boolean
}
interface B extends A{
prop1='some real value',
prop2:boolean
}

I want to do something like above, I would also like to know if something similar can be achieved through type as well. Any help appreciated.

Answer 1

Assuming that your goal is to remove the optional properties using types instead of interfaces you could do something like this:

type A = { 
  prop1: string,
  prop2: boolean,
  prop3?: boolean
}

type B = {
    [K in keyof A as A[K] extends Required<A>[K] ? K : never]: A[K]
}