What methods can we employ to ensure that Typescript adheres to the default generic argument type in this situation?

const EventKeys = {
  openItem: 'openItem',
  changeActiveItem: 'changeActiveItem',
  selectionToggled: 'selectionToggled',
} as const

type EventKeys = keyof typeof EventKeys


class Test<Evmt>{

    subscribe<CurrentEeventKey = Evmt>(arg:CurrentEeventKey){
      // In this section, the correct type is being detected
      console.log(arg)
      return arg
            //     ^?

  }
}


const t1 = new Test<EventKeys>
//     ^?const t1: Test<"openItem" | "changeActiveItem" | "selectionToggled">

const abc = t1.subscribe('11223')
// The type parameter should cause an error here since it should take on the constructor value
//     ^? const abc: "11223"

t1.subscribe<'11223'>('11223')
// However, the type parameter should not cause an error here as it should use the supplied value instead


I attempted to pass the argument and set a default value. My intention was for the function to automatically pick up the generic type from the default value passed during the constructor call, eliminating the need to define it again. But when provided during the function call, it should accept that specific type. The inference works correctly inside the function where the type is accurate.

Answer №1

Utilize two different versions,
One standard version, with inherited argument Another without generic inference, defaulting to void Interactive Example

import { F } from 'ts-toolbelt'
class Test<Evmt>{
    subscribe<CurrentEeventKey extends Evmt>(arg: CurrentEeventKey): CurrentEeventKey;
    subscribe<CurrentEeventKey = void>(arg: F.NoInfer<CurrentEeventKey>): CurrentEeventKey;
    subscribe<CurrentEeventKey = Evmt>(arg: CurrentEeventKey) {
        //correctly identifies the type here
        console.log(arg)
        return arg
        //     ^?
    }
}

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