Question

94

Trying to override and extend method signature in child class in Typescript

rated 0 times [ 101] [ 7] / answers: 1 / hits: 15871 / 6 Years ago, thu, april 19, 2018, 12:00:00

I have a base class that I am trying to extend:

export class BaseClass<T extends SomeOtherClass> {

  constructor(param: ParamType) {

  }



  doSomething(param1: Param1Type): BaseClass<T> {

    // do something with param1;

    return this;

  } 

}

My class:

export class MyClass<T extends SomeOtherClass> extends BaseClass<T> {



  constructor(param: ParamType) {

    super(param);

  }



  doSomething(param1: Param1Type, param2: Param2Type): MyClass<T> {

    // super.doSomething(param1);

    // do something with param2;

    return this;

  }

}

but I'm getting a warning:

Property 'doSomething' in type 'MyClass<T>' is not assignable to the same property in base type 'BaseClass<T>'.

  Type '(param1: Param1Type, param2: Param2Type) => MyClass<T>' is not assignable to type '(param1: Param1Type) => BaseClass<T>'.

Is it not possible to extend method signatures in typescript? How do I extend the capabilities of the BaseClass if I need to add a parameter to the overridden method and is this the correct way of calling the parent method in es6 syntax. I'm aware that prior to es6 I could have called BaseClass.prototype.doSomething.call(this, param1).

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eliezerc

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answered 6 Years ago tiffany · Accepted Answer

The problem as pointed out by others is that if param2 is required it breaks polymorphism:

// We should be able to do this assignment 

let baseRef: BaseClass<SomeOtherClass> = new MyClass<SomeOtherClass>(); 

baseRef.doSomething() // param2 is not required by the base class so MyClass will not receive it even though it NEEDS it

One solution, is to make the second parameter optional, so the call baseRef.doSomething() is valid for the derived type as well :

export class MyClass<T extends SomeOtherClass> extends BaseClass<T> {



    constructor(param: string) {

        super(param);

    }



    doSomething(param1: string, param2?: string): MyClass<T> {

        super.doSomething(param1);

        return this;

    }

}

A second solution, if we only want to share code between the classes, is to disallow the assignment let baseRef: BaseClass<SomeOtherClass> = new MyClass<SomeOtherClass>(); by not really inheriting BaseClass but rather inherit a class that excludes the doSomething method:

type PartialBaseClass = new <T> (param: string)  => { [P in Exclude<keyof BaseClass<T>, 'doSomething'>] : BaseClass<T>[P] }

const PartialBaseClass:PartialBaseClass = BaseClass



export class MyClass<T extends SomeOtherClass> extends PartialBaseClass<T> {



    constructor(param: string) {

        super(param);

    }



    doSomething(param1: string, param2: string): MyClass<T> {

        BaseClass.prototype.doSomething.call(this, param1);

        return this;

    }

}

// This is now invalid ! 

let baseRef: BaseClass<SomeOtherClass> = new MyClass<SomeOtherClass>()    ;