Override method with different argument types in extended class - Typescript

Question

I want to override a method and pass different argument types to it:

class Base {
    public myMethod(myString: string): undefined {
        return;
    }
}

class Child extends Base {
    public myMethod(myNumber: number): undefined {
        return super.myMethod(String(myNumber));
    }
}

Yet this yields a typescript error:

Property 'myMethod' in type 'Child' is not assignable to the same property in base type 'Base'. Type '(myNumber: number) => undefined' is not assignable to type '(myString: string) => undefined'. Types of parameters 'myNumber' and 'myString' are incompatible. Type 'string' is not assignable to type 'number'.

Is there a way to do this without creating a typescript error?

Answer 1

There is not a way to do this*. In TypeScript, inheritance implies subtyping, so you can't inherit from a base class while not being a valid subtype of it.

_{* Technically not true but I'm not going to mention them because they're gross hacks that you shouldn't do.}

Answer 2

As mentioned by others this is not a good idea because you break Liskov substitution.

What you can easily do, is provide an override that takes both string and number. This allows your class to still be used wherever the base class is expected.

class Base {
     public myMethod(myString: string): undefined {
         return;
     }
 }

class Child extends Base {
    public myMethod(myNumberOrString: number | string): undefined {
        if (typeof myNumberOrString === 'number') {
            return super.myMethod(String(myNumberOrString));
        } else {
            return super.myMethod(myNumberOrString);
        }
    }
}