Call an abstract method with prototype

Question

I have an abstract class with an abstract method. I discovered that Typescript won't throw any error if I called the method with prototype and call. For exmaple:

abstract class Human{
    age: string;
    constructor(age: string){
        this.age = age;
    }
    abstract name(): void
}

class Human1 extends Human{
    constructor(age: string){
        super(age);
        this.name()
        
    }
name(): void {
    Human.prototype.name.call(this); // this is allowed despite that name is undefined/abstract
    let a = "do something";
    }
}

let a = new Human1("23")

Is there a way to make Typescript throw an error here?

Answer 1

I discovered that Typescript won't throw any error if I called the method with prototype and call.

Yes, TypeScript is pretty bad at typing prototype objects. It thinks that Human.prototype has the type Human¹, and for that it only considers the public interface. The abstract modifier is only affecting subclass implementations.

Is there a way to make Typescript throw an error here?

Yes, don't use .call to call the parent method. There's the super keyword for that:

name(): void {
    super.name(); // this errors
    let a = "do something";
}

^{(playground demo)}

This does not compile but throws the error "Abstract method 'name' in class 'Human' cannot be accessed via super expression. (ts 2513)" as expected.

_{1: Notice also that it thinks Human.prototype.age is a number, and that Human.prototype.constructor is any Function and not typeof Human.}