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TypeScript: Union to intersection type

Stefan Baumgartner

Written by @ddprrt

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Recently, I had to convert a union type into an intersection type. Working on a helper type UnionToIntersection<T> has taught me a ton of things on conditional types and strict function types, which I want to share with you.

I really like working with non-discriminated union types when I try to model a type where at least one property needs to be set, making all other properties optional. Like in this example:

type Format320 = { urls: { format320p: string } }
type Format480 = { urls: { format480p: string } }
type Format720 = { urls: { format720p: string } }
type Format1080 = { urls: { format1080p: string } }

type Video = BasicVideoData & (
Format320 | Format480 | Format720 | Format1080
)

const video1: Video = {
// ...
urls: {
format320p: 'https://...'
}
} // ✅

const video2: Video = {
// ...
urls: {
format320p: 'https://...',
format480p: 'https://...',
}
} // ✅

const video3: Video = {
// ...
urls: {
format1080p: 'https://...',
}
} // ✅

However, putting them in a union has some side-effects when you need e.g. all available keys:

// FormatKeys = never
type FormatKeys = keyof Video["urls"]

// But I need a string representation of all possible
// Video formats here!
declare function selectFormat(format: FormatKeys): void

In the example above, FormatKeys is never, because there are no common, intersecting keys within this type. Since I don’t want to maintain extra types (that might be error-prone), I need to somehow transform the union of my video formats to an intersection of video formats. The intersection means that all keys need to be available, which allows the keyof operator to create a union of all my formats.

So how do we do that? The answer can be found in the academic description of conditional types that have been released with TypeScript 2.8. There is a lot of jargon, so let’s go over this piece by piece to make sense out of it.

The solution #

I’ll start by presenting the solution. If you don’t want to know how this works underneath, just see this as a TL/DR.

type UnionToIntersection<T> = 
(T extends any ? (x: T) => any : never) extends
(x: infer R) => any ? R : never

Still here? Good! There is a lot to unpack here. There’s a conditional type nested within a conditional type, we use the infer keyword and everything looks like it’s way too much work that does nothing at all. But it does, because there are a couple of key pieces TypeScript treats special. First, the naked type.

The naked type #

If you look at the first conditional within UnionToIntersection<T>, you can see that we use the generic type argument as a naked type.

type UnionToIntersection<T> = 
(T extends any ? (x: T) => any : never) //...

This means that we check if T is in a sub-type condition without wrapping it in something.

type Naked<T> = 
T extends ... // naked!

type NotNaked<T> =
{ o: T } extends ... // not naked!

Naked types in conditional types have a certain feature. If T is a union, they run the conditional type for each constituent of the union. So with a naked type, a conditional of union types becomes a union of conditional types. For example:

type WrapNaked<T> = 
T extends any ? { o: T } : never

type Foo = WrapNaked<string | number | boolean>

// A naked type, so this equals to

type Foo =
WrapNaked<string> |
WrapNaked<number> |
WrapNaked<boolean>

// equals to

type Foo =
string extends any ? { o: string } : never |
number extends any ? { o: number } : never |
boolean extends any ? { o: boolean } : never

type Foo =
{ o: string } | { o: number } | { o: boolean }

As compared to the non-naked version:

type WrapNaked<T> = 
{ o: T } extends any ? { o: T } : never

type Foo = WrapNaked<string | number | boolean>

// A non Naked type, so this equals to

type Foo =
{ o: string | number | boolean } extends any ?
{ o: string | number | boolean } : never

type Foo =
{ o: string | number | boolean }

Subtle, but considerably different for complex types!

So, back in our example, we use the naked type and ask if it extends any (which it always does, any is the allow-it-all top type).

type UnionToIntersection<T> = 
(T extends any ? (x: T) => any : never) //...

Since this condition is always true, we wrap our generic type in a function, where T is the type of the function’s parameter. But why are we doing that?

Contra-variant type positions #

This leads me to the second condition:

type UnionToIntersection<T> = 
(T extends any ? (x: T) => any : never) extends
(x: infer R) => any ? R : never

As the first condition always yields true, meaning that we wrap our type in a function type, the other condition also always yields true. We are basically checking if the type we just created is a subtype of itself. But instead of passing through T, we infer a new type R, and return the inferred type.

So what we do is wrap, and unwrap type T via a function type.

Doing this via function arguments brings the new inferred type R in a contra-variant position. I will explain contra-variance in a later post. For now, it’s important to know that it means that you can’t assign a sub-type to a super-type when dealing with function arguments.

For example, this works:

declare let b: string
declare let c: string | number

c = b // ✅

string is a sub-type of string | number, all elements of string appear in string | number, so we can assign b to c. c still behaves as we originally intended it. This is co-variance.

This on the other hand, won’t work:

type Fun<X> = (...args: X[]) => void

declare let f: Fun<string>
declare let g: Fun<string | number>

g = f // 💥 this cannot be assigned

And if you think about it, this is also clear. When assigning f to g, we suddenly can’t call g with numbers anymore! We miss part of the contract of g. This is contra-variance, and it effectively works like an intersection.

This is what happens when we put contra-variant positions in a conditional type: TypeScript creates an intersection out of it. Meaning that since we infer from a function argument, TypeScript knows that we have to fulfill the complete contract. Creating an intersection of all constituents in the union.

Basically, union to intersection.

How the solution works #

Let’s run it through.

type UnionToIntersection<T> = 
(T extends any ? (x: T) => any : never) extends
(x: infer R) => any ? R : never

type Intersected = UnionToIntersection<Video["urls"]>

// equals to

type Intersected = UnionToIntersection<
{ format320p: string } |
{ format480p: string } |
{ format720p: string } |
{ format1080p: string }
>

// we have a naked type, this means we can do
// a union of conditionals:

type Intersected =
UnionToIntersection<{ format320p: string }> |
UnionToIntersection<{ format480p: string }> |
UnionToIntersection<{ format720p: string }> |
UnionToIntersection<{ format1080p: string }>

// expand it...

type Intersected =
({ format320p: string } extends any ?
(x: { format320p: string }) => any : never) extends
(x: infer R) => any ? R : never |
({ format480p: string } extends any ?
(x: { format480p: string }) => any : never) extends
(x: infer R) => any ? R : never |
({ format720p: string } extends any ?
(x: { format720p: string }) => any : never) extends
(x: infer R) => any ? R : never |
({ format1080p: string } extends any ?
(x: { format1080p: string }) => any : never) extends
(x: infer R) => any ? R : never

// conditional one!

type Intersected =
(x: { format320p: string }) => any extends
(x: infer R) => any ? R : never |
(x: { format480p: string }) => any extends
(x: infer R) => any ? R : never |
(x: { format720p: string }) => any extends
(x: infer R) => any ? R : never |
(x: { format1080p: string }) => any extends
(x: infer R) => any ? R : never

// conditional two!, inferring R!
type Intersected =
{ format320p: string } |
{ format480p: string } |
{ format720p: string } |
{ format1080p: string }

// But wait! `R` is inferred from a contra-variant position
// I have to make an intersection, otherwise I lose type compatibility

type Intersected =
{ format320p: string } &
{ format480p: string } &
{ format720p: string } &
{ format1080p: string }

And that’s what we have been looking for! So applied to our original example:

type FormatKeys = keyof UnionToIntersection<Video["urls"]>

FormatKeys is now "format320p" | "format480p" | "format720p" | "format1080p". Whenever we add another format to the original union, the FormatKeys type gets updated automatically. Maintain once, use everywhere.

Further reading #

I came to this solution after digging into what contra-variant positions are and what they mean in TypeScript. Next to type system jargon, it tells us effectively that we need to provide all constituents of a generic union if used as a function argument. And this works as an intersection during the assignment.

If you want to read more on this subject, I suggest catching up on the following articles.

More articles on TypeScript

Tidy TypeScript: Avoid traditional OOP patterns

Tidy TypeScript: Prefer type aliases over interfaces

Tidy TypeScript: Prefer union types over enums

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