How to tell TypeScript that optional property in object exists and is set?

Question

I have the following code:

interface First
{
  propertyA: string;
}

// Here propertyA is optional
// Imagine that this interface came from external library.
interface Second
{
  propertyA ?: string;
}

function fn(arg: First)
{
  // ...
}

// I know that this object can be declared as type of First,
// but I really need set this as type of Second interface
let myVar: Second = {propertyA: 'some string'};

// I really need in this way to make the call.
fn(myVar); // Error

if(myVar.hasOwnProperty('propertyA'))
{
  fn(myVar); // Still same error
}

if(myVar.propertyA)
{
  fn(myVar); // Still same error
}

But TypeScript throw error:

Argument of type 'Second' is not assignable to parameter of type 'First'. Property 'propertyA' is optional in type 'Second' but required in type 'First'.

So, how to tell TypeScript that optional property propertyA in myVar exists and is set?

Answer 1

old question, but there's a very clean solution in newer versions of typescript

fn(myVar!);

In Typescript, what is the ! (exclamation mark / bang) operator when dereferencing a member?