Typescript force implementation of optional interface members [duplicate]

Question

Given an interface

interface IAmOptional {
   optional? : string,
   optional2?: string
   forced: string
}

Is there a way to covert, extend or similar IAmOptional in a way that this implementation fails?

class someClass {
    thisShouldHaveAllKeys : IAmOptional  = { // Forced<IAmOptional> ??
        forced: 'i am forced'
    }  // Here I want an error like 'thisShouldHaveAllKeys does not have optional and optional2' 
}

Answer 1

Yes, starting with TypeScript 2.8 there is a way to programmatically remove the optional modifier from properties using the -? syntax with mapped types:

type Required<T> = { [K in keyof T]-?: T[K] }

This gives you the behavior your want:

class someClass {
  thisShouldHaveAllKeys: Required<IAmOptional> = {  // error!
//~~~~~~~~~~~~~~~~~~~~~ <-- missing properties optional1, optional2
    forced: 'i am forced'
  }
}

In fact, this Required type alias is so useful that it's predefined for you in the standard library. So you can just use it without defining it.

Hope that helps. Good luck!