Pete V. Pete V. - 10 months ago 183
TypeScript Question

Algebraic data types in TypeScript

Worded an other way:

How would you type the

DOM property in TypeScript?


declare var windowState: WindowState
type WindowState = 1 | 2 | 3 | 4
var windowState = 5 // Error -- Awesome!

Original question:

How do I
in TypeScript so that it describes an algebraic data type? The purpose of this is describing an existing API.

Say in Haskell it would be:

data Weather = 'sunny' | 'bad'

When I try the following, TypeScript obviously complains that a
type is expected

type Weather = 'sunny' | 'bad'

One idea I had is using a JavaScript 2015
, however TypeScript doesn't seem to know about these.

An other idea was using an
, however TypeScript complains that a
member initializer must be constant expression

const enum Weather {
sunny = 'sunny',
bad = 'bad',
windy = Symbol('windy')

I would have thought that a
constant is a constant expression.

Answer Source

TypeScript 2.0 has support for discriminated unions/algebraic data types. The documentation is here.

You can combine string literal types, union types, type guards, and type aliases to build an advanced pattern called discriminated unions, also known as tagged unions or algebraic data types. Discriminated unions are useful in functional programming. Some languages automatically discriminate unions for you; TypeScript instead builds on JavaScript patterns as they exist today. There are four ingredients:

  1. Types that have a common, string literal property — the discriminant.
  2. A type alias that takes the union of those types — the union.
  3. Type guards on the common property.

Let's start:

interface Square {
    kind: "square";
    size: number;
interface Rectangle {
    kind: "rectangle";
    width: number;
    height: number;
interface Circle {
    kind: "circle";
    radius: number;

First we declare the interfaces we will union. Each interface has a kind property with a different string literal type. The kind property is called the discriminant or tag. The other properties are specific to each interface. Notice that the interfaces are currently unrelated. Let's put them into a union:

type Shape = Square | Rectangle | Circle;

Now let's use the discriminated union:

function area(s: Shape) {
    switch (s.kind) {
        case "square": return s.size * s.size;
        case "rectangle": return s.height * s.width;
        case "circle": return Math.PI * s.radius ** 2;

In each of those branches, TypeScript will narrow down the type. If you try to use a case clause with a value that isn't present as any kind property, then TypeScript will error.