Transform union type to intersection type

Is there a way to transform a union type into an intersection type :

type FunctionUnion = () => void | (p: string) => void
type FunctionIntersection = () => void & (p: string) => void

I would like to apply a transformation to FunctionUnion to get FunctionIntersection

Solution 1:

You want union to intersection? Distributive conditional types and inference from conditional types can do that. (Don't think it's possible to do intersection-to-union though, sorry) Here's the evil magic:

type UnionToIntersection<U> = 
  (U extends any ? (k: U)=>void : never) extends ((k: infer I)=>void) ? I : never

That distributes the union U and repackages it into a new union where all the consitutents are in contravariant position. That allows the type to be inferred as an intersection I, as mentioned in the handbook:

Likewise, multiple candidates for the same type variable in contra-variant positions causes an intersection type to be inferred.

Let's see if it works.

First let me parenthesize your FunctionUnion and FunctionIntersection because TypeScript seems to bind the union/intersection more tightly than function return:

type FunctionUnion = (() => void) | ((p: string) => void);
type FunctionIntersection = (() => void) & ((p: string) => void);

Testing:

type SynthesizedFunctionIntersection = UnionToIntersection<FunctionUnion>
// inspects as 
// type SynthesizedFunctionIntersection = (() => void) & ((p: string) => void)

Looks good!

Be careful that in general UnionToIntersection<> exposes some details of what TypeScript thinks is an actual union. For example, boolean is apparently internally represented as true | false, so

type Weird = UnionToIntersection<string | number | boolean>

becomes

type Weird = string & number & true & false

which in TS3.6+ gets eagerly reduced to

type Weird = never

because it's impossible to have a value which is string and number and true and false.

Hope that helps. Good luck!

Solution 2:

There is also a very related problem when you would like an intersection of several types, but not necessarily convert unions to intersections. There is just no way to get right to intersections without resorting to temporary unions!

The problem is that types we would like to get an intersection of might have unions inside, which will be converted to intersections too. Guards to the rescue:

// union to intersection converter by @jcalz
// Intersect<{ a: 1 } | { b: 2 }> = { a: 1 } & { b: 2 }
type Intersect<T> = (T extends any ? ((x: T) => 0) : never) extends ((x: infer R) => 0) ? R : never

// get keys of tuple
// TupleKeys<[string, string, string]> = 0 | 1 | 2
type TupleKeys<T extends any[]> = Exclude<keyof T, keyof []>

// apply { foo: ... } to every type in tuple
// Foo<[1, 2]> = { 0: { foo: 1 }, 1: { foo: 2 } }
type Foo<T extends any[]> = {
    [K in TupleKeys<T>]: {foo: T[K]}
}

// get union of field types of an object (another answer by @jcalz again, I guess)
// Values<{ a: string, b: number }> = string | number
type Values<T> = T[keyof T]

// TS won't believe the result will always have a field "foo"
// so we have to check for it with a conditional first
type Unfoo<T> = T extends { foo: any } ? T["foo"] : never

// combine three helpers to get an intersection of all the item types
type IntersectItems<T extends any[]> = Unfoo<Intersect<Values<Foo<T>>>>

type Test = [
    { a: 1 } | { b: 2 },
    { c: 3 },
]

// this is what we wanted
type X = IntersectItems<Test> // { a: 1, c: 3 } | { b: 2, c: 3 }

// this is not what we wanted
type Y = Intersect<Test[number]> // { a: 1, b: 2, c: 3 }

The execution in the given example goes like this

IntersectItems<[{ a: 1 } | { b: 2 }, { c: 3 }]> =
Unfoo<Intersect<Values<Foo<[{ a: 1 } | { b: 2 }, { c: 3 }]>>>> =
Unfoo<Intersect<Values<{0: { foo: { a: 1 } | { b: 2 } }, 1: { foo: { c: 3 } }}>>> =
Unfoo<Intersect<{ foo: { a: 1 } | { b: 2 } } | { foo: { c: 3 } }>> =
Unfoo<(({ foo: { a: 1 } | { b: 2 } } | { foo: { c: 3 } }) extends any ? ((x: T) => 0) : never) extends ((x: infer R) => 0) ? R : never> =
Unfoo<(({ foo: { a: 1 } | { b: 2 } } extends any ? ((x: T) => 0) : never) | ({ foo: { c: 3 } } extends any ? ((x: T) => 0) : never)) extends ((x: infer R) => 0) ? R : never> =
Unfoo<(((x: { foo: { a: 1 } | { b: 2 } }) => 0) | ((x: { foo: { c: 3 } }) => 0)) extends ((x: infer R) => 0) ? R : never> =
Unfoo<{ foo: { a: 1 } | { b: 2 } } & { foo: { c: 3 } }> =
({ foo: { a: 1 } | { b: 2 } } & { foo: { c: 3 } })["foo"] =
({ a: 1 } | { b: 2 }) & { c: 3 } =
{ a: 1 } & { c: 3 } | { b: 2 } & { c: 3 }

Hopefully this also shows some other useful techniques.