Demystifying TypeScript Discriminated Unions

TypeScript is a wonderful tool for writing JavaScript that scales. It’s more or less the de facto standard for the web when it comes to large JavaScript projects. As outstanding as it is, there are some tricky pieces for the unaccustomed. One such area is TypeScript discriminated unions.

Specifically, given this code:

interface Cat { weight: number; whiskers: number;
}
interface Dog { weight: number; friendly: boolean;
}
let animal: Dog | Cat;

…many developers are surprised (and maybe even angry) to discover that when they do animal., only the weight property is valid, and not whiskers or friendly. By the end of this post, this will make perfect sense.

Before we dive in, let’s do a quick (and necessary) review of structural typing, and how it differs from nominal typing. This will set up our discussion of TypeScript’s discriminated unions nicely.

Structural typing

The best way to introduce structural typing is to compare it to what it’s not. Most typed languages you’ve probably used are nominally typed. Consider this C# code (Java or C++ would look similar):

class Foo { public int x;
}
class Blah { public int x;
}

Even though Foo and Blah are structured exactly the same, they cannot be assigned to one another. The following code:

Blah b = new Foo();

…generates this error:

Cannot implicitly convert type 'Foo' to 'Blah'

The structure of these classes is irrelevant. A variable of type Foo can only be assigned to instances of the Foo class (or subclasses thereof).

TypeScript operates the opposite way. TypeScript considers types to be compatible if they have the same structure—hence the name, structural typing. Get it?

So, the following runs without error:

class Foo { x: number = 0;
}
class Blah { x: number = 0;
}
let f: Foo = new Blah();
let b: Blah = new Foo();

Types as sets of matching values

Let’s hammer this home. Given this code:

class Foo { x: number = 0;
} let f: Foo;

f is a variable holding any object that matches the structure of instances created by the Foo class which, in this case, means an x property that represents a number. That means even a plain JavaScript object will be accepted.

let f: Foo;
f = { x: 0
}

Unions

Thanks for sticking with me so far. Let’s get back to the code from the beginning:

interface Cat { weight: number; whiskers: number;
}
interface Dog { weight: number; friendly: boolean;
}

We know that this:

let animal: Dog;

…makes animal any object that has the same structure as the Dog interface. So what does the following mean?

let animal: Dog | Cat;

This types animal as any object that matches the Dog interface, or any object that matches the Cat interface.

So why does animal—as it exists now—only allow us to access the weight property? To put it simply, it’s because TypeScript does not know which type it is. TypeScript knows that animal has to be either a Dog or Cat, but it could be either (or both at the same time, but let’s keep it simple). We’d likely get runtime errors if we were allowed to access the friendly property, but the instance wound up being a Cat instead of a Dog. Likewise for the whiskers property if the object wound up being a Dog.

Type unions are unions of valid values rather than unions of properties. Developers often write something like this:

let animal: Dog | Cat;

…and expect animal to have the union of Dog and Cat properties. But again, that’s a mistake. This specifies animal as having a value that matches the union of valid Dog values and valid Cat values. But TypeScript will only allow you to access properties it knows are there. For now, that means properties on all the types in the union.

Narrowing

Right now, we have this:

let animal: Dog | Cat;

How do we properly treat animal as a Dog when it’s a Dog, and access properties on the Dog interface, and likewise when it’s a Cat? For now, we can use the in operator. This is an old-school JavaScript operator you probably don’t see very often, but it essentially allows us to test if a property is in an object. Like this:

let o = { a: 12 }; "a" in o; // true "x" in o; // false

It turns out TypeScript is deeply integrated with the in operator. Let’s see how:

let animal: Dog | Cat = {} as any; if ("friendly" in animal) { console.log(animal.friendly);
} else { console.log(animal.whiskers);
}

This code produces no errors. When inside the if block, TypeScript knows there’s a friendly property, and therefore casts animal as a Dog. And when inside the else block, TypeScript similarly treats animal as a Cat. You can even see this if you hover over the animal object inside these blocks in your code editor:

Showing a tooltip open on top of a a TypeScript discriminated union example that shows `let animal: Cat`.

Discriminated unions

You might expect the blog post to end here but, unfortunately, narrowing type unions by checking for the existence of properties is incredibly limited. It worked well for our trivial Dog and Cat types, but things can easily get more complicated, and more fragile, when we have more types, as well as more overlap between those types.

This is where discriminated unions come in handy. We’ll keep everything the same from before, except add a property to each type whose only job is to distinguish (or “discriminate”) between the types:

interface Cat { weight: number; whiskers: number; ANIMAL_TYPE: "CAT";
}
interface Dog { weight: number; friendly: boolean; ANIMAL_TYPE: "DOG";
}

Note the ANIMAL_TYPE property on both types. Don’t mistake this as a string with two different values; this is a literal type. ANIMAL_TYPE: "CAT"; means a type that holds exactly the string "CAT", and nothing else.

And now our check becomes a bit more reliable:

let animal: Dog | Cat = {} as any; if (animal.ANIMAL_TYPE === "DOG") { console.log(animal.friendly);
} else { console.log(animal.whiskers);
}

Assuming each type participating in the union has a distinct value for the ANIMAL_TYPE property, this check becomes foolproof.

The only downside is that you now have a new property to deal with. Any time you create an instance of a Dog or a Cat, you have to supply the single correct value for the ANIMAL_TYPE. But don’t worry about forgetting because TypeScript will remind you. 🙂