Upon exploring the code in this playground link,
abstract class Base<F extends () => void> {
public abstract __call__: F;
}
type CallSignature<T> = {
(): T;
(value: T): void;
}
class Foo<T> extends Base<CallSignature<T>> {
public __call__(): T;
public __call__(value: T) : void;
public __call__(value?: T) {
return value;
}
}
Encountering the error message:
Class 'Base<CallSignature<T>>' defines instance member property '__call__', but extended class 'Foo<T>' defines it as instance member function.(2425)
The query arises on how to clarify that __call__ within Base is designated to be a method, while retaining its definition based on T (thus allowing various signatures including overloads).
Please note that direct reference by Base to CallSignature is not permissible, as it serves as a distinct interface for each subclass (hence being passed as F).
Furthermore, the essence of this inquiry lies in having Base dictate the interface that Foo must adhere to, even if Foo introduces a generic that alters this requirement in some manner.