I have object of type any and I assign object of type myResponse like below

    public obj: any;
    public set Result() {
        obj = myResponse;
    }

I have another function where I want to cast any type to my specific type like below

public myFunction(){
    let x: MyResponse = (MyResponse) obj;
    conosle.log(x.somePropoerty);
} 

I have tried different methods which I got online like angular brackets for casting, Object.assign but it did not work.

MyResponse class is like below

export class MyResponse{
    public property1: string;
    public property2: number;
    //some other code
}

I have object of type any and I assign object of type myResponse like below

    public obj: any;
    public set Result() {
        obj = myResponse;
    }

I have another function where I want to cast any type to my specific type like below

public myFunction(){
    let x: MyResponse = (MyResponse) obj;
    conosle.log(x.somePropoerty);
} 

I have tried different methods which I got online like angular brackets for casting, Object.assign but it did not work.

MyResponse class is like below

export class MyResponse{
    public property1: string;
    public property2: number;
    //some other code
}

• javascript
• angular
• typescript

Share Improve this question Follow edited Jul 22, 2017 at 9:35 Maciej Treder 12.4k55 gold badges5656 silver badges7777 bronze badges asked Jul 22, 2017 at 7:32 ViruViru 2,24622 gold badges1919 silver badges2929 bronze badges 6

• What do you mean? Is MyResponse a class or an interface? How it is defined? Does the content of obj at least match the same interface? Explain. – Neil Lunn Commented Jul 22, 2017 at 7:46
• MyResponse is class...class has some property..Yes content obj should match the same class...As I assigned instance of same class to obj and now I am trying to convert it back to MyResponse class. – Viru Commented Jul 22, 2017 at 7:51
• So how do you usually get a class instance from some provided data? Pretty sure you do that on a regular basis, and then why would you think this would be any different. So think! What does new Date(0) actually do? – Neil Lunn Commented Jul 22, 2017 at 7:53
• obj = myResponse; should be this.obj = myResponse as MyResponse; – Hitmands Commented Jul 22, 2017 at 8:15
• Have you tried this let x: MyResponse = Object.assign({},myResponse ) and x['someProperty'] to get the value – Sreemat Commented Jul 22, 2017 at 8:58

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3 Answers 3

Sorted by: Reset to default 2

There is no casting in TypeScript, only type assertions. You can assert that obj is of type MyResponse by doing:

let x: MyResponse = obj as MyResponse;

Do note that this is for pile-time checking only. If your obj is not a MyResponse instance at runtime then this won't work.

Casting (Microsoft call it Type assertions) in TypeScript can be done in two ways.

Let's say we got following object definition:

class MyClass {
    public someMethod:void() {
         console.log("method invoked");
    }
    public static staticMethod(object: MyClass) {
         console.log("static method");
         object.someMethod();
    }
}

The first way of casting is very similar to Java style. You need to use <> signs.

let someVariable: any = new MyClass();
someVariable.someMethod(); //won't pile
MyClass.staticMethod(someVariable); //won't pile

(<MyClass> someVariable).someMethod(); //will pile
MyClass.staticMethod(<MyClass> someVariable); //will pile

Second way is as @Saravana showed (using as keyword):

//all below lines piles
let someVariable: any = new MyClass();
let another: MyClass = someVariable as MyClass;
(someVariable as MyClass).someMethod(); 

MyClass.staticMethod(someVariable as MyClass);

Check out this link for more informations: https://www.typescriptlang/docs/handbook/basic-types.html#type-assertions

To properly cast into "Type", you basically need to change the prototype of the object. That can be done by creating a new instance using the Constructor function, then copying your existing properties.

Otherwise type assertion will only provide a pile-time check! During runtime the "someMethod" will be undefined