Discriminated Union

If you have a class with a literal member (the literal TypeScript supports at the moment are string literals) then you can use that property to discriminate between union members.

As an example consider the union of a Square and Rectangle, here we have a member kind that exists on both union members and is of a particular literal type:

interface Square {
    kind: "square";
    size: number;
}

interface Rectangle {
    kind: "rectangle";
    width: number;
    height: number;
}
type Shape = Square | Rectangle;

If you use a type guard style check (==, ===, !=, !==) or switch on the discriminant property (here kind) TypeScript will realize that it means that the object must of the type that has that literal and do a type narrowing for you :)

function area(s: Shape) {
    if (s.kind === "square") {
        // Now TypeScript *knows* that `s` must a square ;)
        // So you can use its members safely :)
        return s.size * s.size;
    }
    else {
        // Wasn't a square? So TypeScript will figure out that it must be a Rectangle ;)
        // So you can use its members safely :)
        return s.width * s.height;
    }
}

Exhaustive Checks

Quite commonly you want to make sure that all members of a union have some code(action) against them.

interface Square {
    kind: "square";
    size: number;
}

interface Rectangle {
    kind: "rectangle";
    width: number;
    height: number;
}

// Someone just added this new `Circle` Type
// We would like to let TypeScript give an error at any place that *needs* to cater for this
interface Circle {
    kind: "circle";
    radius: number;
}

type Shape = Square | Rectangle | Circle;

As an example of where stuff goes bad:

function area(s: Shape) {
    if (s.kind === "square") {
        return s.size * s.size;
    }
    else if (s.kind === "rectangle") {
        return s.width * s.height;
    }
    // Would it be great if you could get TypeScript to give you an error?
}

You can do that by simply adding a fall through and making sure that the inferred type in that block is compatible with the never type. For example:

function area(s: Shape) {
    if (s.kind === "square") {
        return s.size * s.size;
    }
    else if (s.kind === "rectangle") {
        return s.width * s.height;
    }
    else {
        // ERROR : `Circle` is not assignable to `never`
        const _exhaustiveCheck: never = s;
    }
}

Switch

TIP: of course you can also do it in a switch statement:

function area(s: Shape) {
    switch (s.kind) {
        case "square": return s.size * s.size;
        case "rectangle": return s.width * s.height;
        case "circle": return Math.PI * s.radius * s.radius;
        default: const _exhaustiveCheck: never = s;
    }
}

strictNullChecks

If using strictNullChecks and doing exhaustive checks you should return the _exhaustiveCheck variable (of type never) as well, otherwise TypeScirpt infers a possible return of undefined. So:

function area(s: Shape) {
    switch (s.kind) {
        case "square": return s.size * s.size;
        case "rectangle": return s.width * s.height;
        case "circle": return Math.PI * s.radius * s.radius;
        default: 
          const _exhaustiveCheck: never = s;
          return _exhaustiveCheck;
    }
}