280

What does Record<K, T> mean in Typescript?

Typescript 2.1 introduced the Record type, describing it in an example:

// For every properties K of type T, transform it to U
function mapObject<K extends string, T, U>(obj: Record<K, T>, f: (x: T) => U): Record<K, U>

see Typescript 2.1

And the Advanced Types page mentions Record under the Mapped Types heading alongside Readonly, Partial, and Pick, in what appears to be its definition:

type Record<K extends string, T> = {
    [P in K]: T;
}

Readonly, Partial and Pick are homomorphic whereas Record is not. One clue that Record is not homomorphic is that it doesn’t take an input type to copy properties from:

type ThreeStringProps = Record<'prop1' | 'prop2' | 'prop3', string>

And that's it. Besides the above quotes, there is no other mention of Record on typescriptlang.org.

Questions

  1. Can someone give a simple definition of what Record is?

  2. Is Record<K,T> merely a way of saying "all properties on this object will have type T"? Probably not all properties, since K has some purpose...

  3. Does the K generic forbid additional keys on the object that are not K, or does it allow them and just indicate that their properties are not transformed to T?

  4. With the given example:

    type ThreeStringProps = Record<'prop1' | 'prop2' | 'prop3', string>
    

    Is it exactly the same as this?:

    type ThreeStringProps = {prop1: string, prop2: string, prop3: string}
    
Matthias
  • 8,845
  • 7
  • 33
  • 52
  • 11
    The answer to 4. is pretty much "yes", so that should probably answer your other questions. – jcalz Aug 20 '18 at 18:30

2 Answers2

328
  1. Can someone give a simple definition of what Record is?

A Record<K, T> is an object type whose property keys are K and whose property values are T. That is, keyof Record<K, T> is equivalent to K, and Record<K, T>[K] is (basically) equivalent to T.

  1. Is Record<K,T> merely a way of saying "all properties on this object will have type T"? Probably not all objects, since K has some purpose...

As you note, K has a purpose... to limit the property keys to particular values. If you want to accept all possible string-valued keys, you could do something like Record<string, T>, but the idiomatic way of doing that is to use an index signature like { [k: string]: T }.

  1. Does the K generic forbid additional keys on the object that are not K, or does it allow them and just indicate that their properties are not transformed to T?

It doesn't exactly "forbid" additional keys: after all, a value is generally allowed to have properties not explicitly mentioned in its type... but it wouldn't recognize that such properties exist:

declare const x: Record<"a", string>;
x.b; // error, Property 'b' does not exist on type 'Record<"a", string>'

and it would treat them as excess properties which are sometimes rejected:

declare function acceptR(x: Record<"a", string>): void;
acceptR({a: "hey", b: "you"}); // error, Object literal may only specify known properties

and sometimes accepted:

const y = {a: "hey", b: "you"};
acceptR(y); // okay
  1. With the given example:

    type ThreeStringProps = Record<'prop1' | 'prop2' | 'prop3', string>
    

    Is it exactly the same as this?:

    type ThreeStringProps = {prop1: string, prop2: string, prop3: string}
    

Yes!

Hope that helps. Good luck!

jcalz
  • 125,133
  • 11
  • 145
  • 170
  • 4
    Very much learned and a question why "the idiomatic way of doing that is to use an index signature" not a Record one ? I don't find any related information about this "idiomatic way". – legend80s Oct 28 '19 at 06:55
  • 7
    You can use `Record` to mean `{[x: string]: V}` if you want; I've probably even done this myself. The index signature version is more direct: they are the same type, but the former is a type alias of a mapped type which evaluates to an index signature, while the latter is just the index signature directly. All else being equal, I'd recommend the latter. Similarly I wouldn't use `Record` in place of `{a: string}` unless there were some other compelling contextual reason do to so. – jcalz Oct 28 '19 at 13:58
  • 3
    "_All else being equal, I'd recommend the latter._" Why is that? My pre-Typescript self agrees, but I know the former will be more, um, self-commenting for people coming from the C# side, for instance, and no worse for JavaScript-to-Typescripters. Are you just interested in skipping the transpilation step for those constructs? – ruffin Nov 14 '19 at 14:08
  • 4
    Just my opinion: the behavior of `Record` only makes sense if you already know how index signatures work in TypeScript. E.g., given `x: Record`, `x.foo` will apparently be a `string` at compile time, but in actuality is likely to be `string | undefined`. This is a gap in how `--strictNullChecks` works (see [#13778](https://github.com/microsoft/TypeScript/issues/13778)). I'd rather have newcomers deal with `{[x: string]: V}` directly instead of expecting them to follow the chain from `Record` through `{[P in string]: V}` to the index signature behavior. – jcalz Nov 14 '19 at 14:27
  • I wanted to point out that logically a type can be defined as a set of all values contained within the type. given this interpretation I think Record is reasonable as an abstraction to simplify the code instead of laying out all possible values. It is similar to the example in the utility types documentation: Record where type T = 'a' | 'b' | 'c'. Never doing Record is not a good counter example, because it does not follow the same pattern. It also does not add to reuse or simplify the code through abstraction as the other examples do. – Scott Leonard Mar 13 '20 at 17:23
  • 1
    Can Record be used as C# dictionaries? Do Records have add(), remove() methods? – Kishan Vaishnav Jun 22 '20 at 09:04
  • 1
    @KishanVaishnav no, it's just a plain object type with keys an values. Sounds like you'd rather use something like [`Map`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map) instead. – jcalz Jun 22 '20 at 15:18
  • See [superluminary's answer](https://stackoverflow.com/a/56473984/3499424) for a compelling contextual reason to prefer `Record` over an index signature. When the set of keys is finite _and_ you want to use it in more than one place (e.g. constraining a string parameter in addition to associating with typed data), `Record` allows you to use a set of keys defined independently. – John Neuhaus Apr 23 '21 at 15:55
123

A Record lets you create a new type from a Union. The values in the Union are used as attributes of the new type.

For example, say I have a Union like this:

type CatNames = "miffy" | "boris" | "mordred";

Now I want to create an object that contains information about all the cats, I can create a new type using the values in the CatName Union as keys.

type CatList = Record<CatNames, {age: number}>

If I want to satisfy this CatList, I must create an object like this:

const cats:CatList = {
  miffy: { age:99 },
  boris: { age:16 },
  mordred: { age:600 }
}

You get very strong type safety:

  • If I forget a cat, I get an error.
  • If I add a cat that's not allowed, I get an error.
  • If I later change CatNames, I get an error. This is especially useful because CatNames is likely imported from another file, and likely used in many places.

Real-world React example.

I used this recently to create a Status component. The component would receive a status prop, and then render an icon. I've simplified the code quite a lot here for illustrative purposes

I had a union like this:

type Statuses = "failed" | "complete";

I used this to create an object like this:

const icons: Record<
  Statuses,
  { iconType: IconTypes; iconColor: IconColors }
> = {
  failed: {
    iconType: "warning",
    iconColor: "red"
  },
  complete: {
    iconType: "check",
    iconColor: "green"
  };

I could then render by destructuring an element from the object into props, like so:

const Status = ({status}) => <Icon {...icons[status]} />

If the Statuses union is later extended or changed, I know my Status component will fail to compile and I'll get an error that I can fix immediately. This allows me to add additional error states to the app.

Note that the actual app had dozens of error states that were referenced in multiple places, so this type safety was extremely useful.

superluminary
  • 38,944
  • 21
  • 142
  • 143
  • I assume most of the time `type Statuses` lives in typings NOT defined by you? Otherwise I can see something like an interface with an enum being a better fit right? – Victorio Berra Jun 17 '19 at 14:56
  • Hi @victorio, I'm not sure how an enum would solve the problem, you don't get an error in an enum if you miss a key. It's just a mapping between keys and values. – superluminary Jun 18 '19 at 08:37
  • 2
    I see what you mean now. Coming from C# we do not have clever ways to do that. The closest thing would be a dictionary of `Dictionary`. The Record type is a great way to represent that enum + metadata pattern. – Victorio Berra Jun 18 '19 at 16:52
  • Is there a way to initialize a record that has several keys mapped to a single value? Let's say I have 10 different statuses but in the UI all of them should be shown with a warning icon. If I go with record type, I'll have to duplicate the same 10 times, which I would be able to avoid if I use something like a switch statement. – Ε Г И І И О May 07 '21 at 13:48