typescript - Defining an interface property as the evaluated version of a generic conditional type

Question

I'm trying to define an interface that has a method with a conditional function definition.

I.e.,

interface Test<T> {
    func: T extends string ? () => string : () => number;
}

class TestClass<T extends string> implements Test<T> {
    func = () => "Cats";
}

I get the error:

Property 'func' in type 'TestClass<T>' is not assignable to the same property in base type 'Test<T>'.
  Type '() => string' is not assignable to type 'T extends string ? () => string : () => number'`

My intention is that TS should know I want () => string because T is declared as extending string. However, it seems to want me to declare a type that covers both sides of the conditional even though (I think) I've already narrowed it.

question from:https://stackoverflow.com/questions/65947618/defining-an-interface-property-as-the-evaluated-version-of-a-generic-conditional

Answer 1

I've come across this quite a few times, and it looks like the typescript compiler just doesn't support conditional types very well. There's this issue on the typescript repo: https://github.com/microsoft/TypeScript/issues/24929, but it has been closed without a fix.

If you want to keep the TestClass generic, there is no way around an any cast: Playground Link

however, if you can specify the type of TestClass, then this is solvable: Playground Link