Can I write a method which is an extension of a base class, but with a different return type, if it's supported by the shared interface, without inserting type declaration in class 'a'?

In practice classes a & b are in javascript files, and they're failing lint-ts-typing checks.

typescript definition file

interface x {
    abc(): string | number;
}

javascript file

class a implements x {
    abc() {
        return 234;
    }
}

class b extends a implements x {
    abc() {
        return '123';
    }
}

throws the following: Property 'abc' in type 'b' is not assignable to the same property in base type 'a'. Type '() => string' is not assignable to type '() => number'. Type 'string' is not assignable to type 'number'.

playground link

CodePudding user response：

TS doesn't force you to implement the interface exactly
You are implementing is as

class a implements x {
    abc(): number {
        return 234;
    }
}

If you want to keep the number | string type, you have to do id expliticly:

class a implements x {
    abc(): number | string {
        return 234;
    }
}

CodePudding user response：

In your class a, you have to specify the return type of abc(). Otherwise it will assume that it only returns number value. Below code will resolve your error -

interface x {
    abc(): string | number;
}

class a implements x {
    abc(): string | number {
        return 234;
    }
}

class b extends a implements x {
    abc() {
        return '123';
    }
}