Check if variable is a specific interface type in a typescript union

Question

Is it possible to create a typeguard, or something else that accomplishes the same purpose, to check if a variable is a specific interface type in a typescript union?

Answer 1

In TypeScript 2 you can use Discriminated Unions like this:

interface Foo {
    kind: "foo";
    a:string;
}
interface Bar {
    kind: "bar";
    b:string;
}
type FooBar = Foo | Bar;
let thing: FooBar;

and then test object using if (thing.kind === "foo").

If you only have 2 fields like in the example I would probably go with combined interface as @ryan-cavanaugh mentioned and make both properties optional:

interface FooBar {
    a?: string;
    b?: string
}

Note that in original example testing the object using if (thing.a !== undefined) produces error Property 'a' does not exist on type 'Foo | Bar'.

And testing it using if (thing.hasOwnProperty('a')) doesn't narrow type to Foo inside the if statement.

@ryan-cavanaugh is there a better way in TypesScript 2.0 or 2.1?