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How to check if string is member of union type
There's some ongoing debate whether
enum in TypeScript should be used: The Dangers of TypeScript EnumsA TypeScript
enum is compiled to an object at runtime:enum Suits {
Hearts = "hearts",
Diamonds = "diamonds",
Spages = "spades",
Clubs = "clubs"
}var Suits;
(function (Suits) {
Suits["Hearts"] = "hearts";
Suits["Diamonds"] = "diamonds";
Suits["Spages"] = "spades";
Suits["Clubs"] = "clubs";
})(Suits || (Suits = {}));One benefit of this approach is being able to check at runtime whether a string is a member of the
enum:function isSuit(value: string): value is Suits {
return Object.keys(Suits).includes(value)
}The alternative for
enum is using a union type:type Suit = "hearts" | "diamonds" | "spades" | "clubs"Type information doesn't exist at runtime, how can we check at runtime if a string is a member of the union type?
function isSuit(value: string): value is Suit {
return ["hearts", "diamonds", "spades", "clubs"].includes(value as Suit)
}This works but there is an issue with this approach... When you add or remove members, we need to update the array. This is prone to errors because we might forget to update the array or the type. How can we do this automatically?
const ALL_SUITS = ['hearts', 'diamonds', 'spades', 'clubs'] as const;
type SuitTuple = typeof ALL_SUITS;
type Suit = SuitTuple[number];
function isSuit(value: string): value is Suit {
return ALL_SUITS.includes(value as Suit)
}We define the possible options using an array with a const assertion. A const assertion tells the compiler to infer the narrowest or most specific type it can for an expression. If you leave it off, the compiler will use its default type inference behavior, which will possibly result in a wider or more general type.
This gets compiled to
const ALL_SUITS = ['hearts', 'diamonds', 'spades', 'clubs'];
function isSuit(suit) {
return ALL_SUITS.includes(suit);
}Which is a lot nicer than using an
enum! 🤩Let me know if this blogpost was useful! 😊
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