Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

289
Views
Intersection types with interfaces fails to throw TypeScript error

By mixing TypeScript types with interfaces in an intersection, I seem to loose the stricter behaviour of interfaces in my code. I want to be able to compose types using intersections (such that I can narrow down a type in specific instances), but I want to be able to keep the exactness of interfaces.

Take this code:

type Vehicle = {
  name: string
  properties: Record<string, unknown>
}

interface CarProperties {
  electric?: boolean
}

type Car = Vehicle & {
  name: 'car'
  properties: CarProperties
}

const car: Car = {
  name: 'car',
  properties: {
    electric: false,
    someOther: 'bs' // <= I want this to throw a TS Error
  }
}

As the CarProperties is an interface, I would expect it to disallow the inclusion of the someOther key.

Why is this not the case? And how else would one go about achieving the same thing?

about 4 years ago · Juan Pablo Isaza
2 answers
Answer question

0

The problem is the intersection type Car. This leads to properties of car being a Record<string, unknown> & CarProperties and thus allows string keys.

To achieve what you want, you can add a generic to your Vehicle type:


type Vehicle<T extends object> = {
  name: string
  properties: T
}

type GenericVehicle = Vehicle<Record<string, unknown>>;

interface CarProperties {
  electric?: boolean
}

interface Car extends Vehicle<CarProperties> {
  name: 'car'
}

PS: I noticed you call this a "union type", which it is not. Intersection types share the properties of the intersected types, whereas union types can be either of the unionized types.

about 4 years ago · Juan Pablo Isaza Report

0

You can make electric property a required one and use object type instead of Record<string, unknown>.

type Vehicle = {
    name: string
    properties: object, // <---- change
}

type CarProperties = {
    electric: boolean // is required now
}

type Car = Vehicle & {
    name: 'car'
    properties: CarProperties
}


// ok
const car2: Car = {
    name: 'car',
    properties: {
        electric: false,
    }
}

// expected error
const car4: Car = {
    name: 'car',
    properties: []
}

// expected error
const car5: Car = {
    name: 'car',
}

// error
const car3: Car = {
    name: 'car',
    properties: {}
}

// error
const car: Car = {
    name: 'car',
    properties: {
        electric: false,
        someOther: 'bs' // error
    }
}

I know, using object type is controversial a bit, because of this rule, but you can find more information about pros and cons in this answer.

If you have more variants of Vehicle, not only one Car, you may want to remove properties from Vehicle type and create a discriminated union of allowed Vehicles. It is only my guess, since I'm not aware of any other requirements.

It is also worth knowing about excess-property-checks. It will disallow you to provide any other extra properties for literal argument.

If you put all your requirements and test cases into your question I believe you will get what you are looking for

about 4 years ago · Juan Pablo Isaza Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!