Typescript: A type that allows only objects which implement the same interface

Question

How can I force environment.service to only accept values which implement ServiceInterface?

interface ServiceInterface {
  id: string
}

class ServiceA implements ServiceInterface {
  id = 'a'
}

class ServiceB implements ServiceInterface {
  id = 'b'
}

interface Environment {
  service: ServiceInterface
}

const environment: Environment = {
  service: ServiceA // TS2741: Property 'id' is missing in type 'typeof ServiceA'...
}

I want people to be able to switch out the service in the environment for their own implementation of the service, but only as long as their new implementation adheres to ServiceInterface.

The reason I cannot use an instance service: new ServiceA() is because I am using Angular, where the dependency injection takes care of instantiating the class. I have to supply the class to the dependency injection system. To do that I import my environment file, where I need to make sure that the class added there implements the interface.

const environment: Environment = { service: { id: 'a' } }

Roberto Zvjerković
– Roberto Zvjerković

2020-03-19 11:41:31 +00:00
Commented Mar 19, 2020 at 11:41 — Roberto Zvjerković
– Roberto Zvjerković, Commented Mar 19, 2020 at 11:41

bugs · Accepted Answer · 2020-03-19 11:41:48Z

2

When you write

const environment: Environment = {
  service: ServiceA
}

the value of the service key is not an instance of the ServiceA class, but the class itself.

You need to pass an actual instance of the class to satisfy the constraint:

const environment: Environment = {
  service: new ServiceA()
}

answered Mar 19, 2020 at 11:41

bugs

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2 Comments

Martin Jaskulla Over a year ago

I understand, thanks. Is there a way around this though? Maybe a custom type to achieve what I want instead?

Maciej Sikora Over a year ago

Custom type is only compile time. You need to have value. When you do new X you get instance value.

Martin Jaskulla · Accepted Answer · 2020-03-19 14:21:19Z

1

Found the solutions thanks to this

interface ServiceInterface {
  id: string
}

type Implements<T> = new (...args: any[]) => T;

class ServiceA implements ServiceInterface {
  id = 'a'
}

class ServiceB implements ServiceInterface {
  id = 'b'
}

interface Environment {
  service: Implements<ServiceInterface>
}

const environment: Environment = {
  service: ServiceB
}

answered Mar 19, 2020 at 14:21

Martin Jaskulla

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