Implementing TypeScript interface with bare function signature plus other fields

Question

How do I write a class that implements this TypeScript interface (and keeps the TypeScript compiler happy):

interface MyInterface {     (): string;     text2(content: string); } 

I saw this related answer: How to make a class implement a call signature in Typescript?

But that only works if the interface only has the bare function signature. It doesn't work if you have additional members (such as function text2) to be implemented.

Answer 1

A class cannot implement everything that is available in a typescript interface. Two prime examples are callable signatures and index operations e.g. : Implement an indexible interface

The reason is that an interface is primarily designed to describe anything that JavaScript objects can do. Therefore it needs to be really robust. A TypeScript class however is designed to represent specifically the prototype inheritance in a more OO conventional / easy to understand / easy to type way.

You can still create an object that follows that interface:

interface MyInterface {     (): string;     text2(content: string); }  var MyType = ((): MyInterface=>{   var x:any = function():string { // Notice the any        return "Some string"; // Dummy implementation    }   x.text2 = function(content:string){       console.log(content); // Dummy implementation    }   return x; } ); 

Answer 2

There's an easy and type-safe way to do this with ES6's Object.assign:

const foo: MyInterface = Object.assign(   // Callable signature implementation   () => 'hi',   {     // Additional properties     text2(content) { /* ... */ }   } ) 

Intersection types, which I don't think were available in TypeScript when this question was originally asked and answered, are the secret sauce to getting the typing right.