CODEWITHSUNDEEP
javascripttypescript
mantsevich
mantsevich

Reputation: 41

a reference to the constructor function

How can I get reference to the constructor function for any object in typescript?

Example in JavaScript:

var anyObject = new this.constructor(options);

or

var anyObject = new someObj.constructor(options);

or

class Greeter { 
  greeting: string; 
  constructor(message: string, options: any) { 
       this.options = options; 
       this.greeting = message; 
  } greet() { 
     return "Hello, " + this.greeting; 
  } createAnyObj(){  
     return new this.constructor(this.options); 
  } 
} 

 var t= new Greeter('mes',{param1: 'val1'});
 var b=t.createAnyObj();

Thanks.

More examples:

>>>Link to typescript playground

Upvotes: 1

Views: 506

Answers (3)

mantsevich
mantsevich

Reputation: 41

This problem was solved with next solution:

module Example{
    // Thank you, JavaScript!
    export function getConstructor(obj){
        return obj.constructor;
    }
    export class A{
        constructor(private options: any){}
        dublicate(){
            return new (getConstructor(this))(this.options);
        }
        someMethod(){
            return 'This is class A';
        }
    }
    export class B extends A{
        someMethod(){
            return 'This is class B';
        }
    }
}

// Class A
var objA = new Example.A({key1: 'val1'});
console.log(objA.someMethod(), objA); //=> This is class A
var objA2 = objA.dublicate();
console.log(objA2.someMethod(), objA2); //=> This is class A

// Class B
var objB = new Example.B({key2: 'val2'});
console.log(objB.someMethod(), objB); //=> This is class B
var objB2 = objB.dublicate();
console.log(objB2.someMethod(), objB2); //=> This is class B

Live example => Open playground

Upvotes: 1

Fenton
Fenton

Reputation: 251082

You could achieve what you need using the following code:

class A {
    constructor(public options: any) {

    }

    duplicate() {
        return new A(this.options);
    }

    someMethod() {
        return 'Hello! I am class A!';
    }
}

class B extends A {
    constructor(options: any) {
        super(options);
    }

    someMethod() {
        return 'This is class B!';
    }

    duplicate() : A {
        return new B(this.options);
    }
}

var objA = new A({key1: 'value1'});
var objWhithOptionsObjA = objA.duplicate(); //=> Instaceof class A
alert(objWhithOptionsObjA.someMethod());

var objB = new B({key2: 'value2'});
var objWhithOptionsObjB = objB.duplicate(); //=> Instaceof class B
alert(objWhithOptionsObjB.someMethod());

You need a constructor on B that just passes options to the super class. You override the duplicate method to return a B, which is a sub-type of A.

Upvotes: 2

Rajeesh
Rajeesh

Reputation: 4495

Use the new operator to instantiate the object with the required parameter in the constructor. This is similar to how you do it in other Object Oriented programming languages, below is an example from the typescript official website 

class Greeter {
    constructor(public message: string, public options?: any) {          
     }

     greet() {
      return "Hello, " + this.message;
     }

     createAnyObj(){  
        return new Greeter(this.message, this.options);         
     } 
}

var greeter = new Greeter("world");

Upvotes: 0

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