Describing a property of an interface as the determined form of a conditional type that is generic

I am currently working on defining an interface that includes a method with a conditional function definition.

For example:

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

class ExampleClass<T extends string> implements Example<T> {
    func = () => "Dogs";
}

However, I am encountering an error:

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

My goal was for TypeScript to understand that I specifically want () => string because T is defined as extending string. Even though I believe I have already narrowed it down, TypeScript seems to be asking me to declare a type that encompasses both conditions of the conditional statement.

Answer №1

I have encountered this issue multiple times, and it appears that the typescript compiler does not handle conditional types very effectively. There was a reported problem on the typescript repository: https://github.com/microsoft/TypeScript/issues/24929, but unfortunately, it has been closed without a resolution.

If you insist on keeping the generic TestClass, the only workaround seems to involve using an any cast: Playground Link

However, if you are able to specify the type of TestClass, then this issue can be resolved: Playground Link

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