55 - Union to Intersection

[githubxiaowen]

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

https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-8.html#type-inference-in-conditional-types

[Sociosarbis]

type UnionToIntersection<U> = (U extends any ? (arg: U) => any : never) extends ((arg: infer I) => void) ? I : never
type I = UnionToIntersection<'foo' | 42 | true> // never

type A = 'foo' | 42 | true

type B = A extends any ? (arg: A) => any : never // (arg: A) => any 

type C = B extends ((arg: infer I) => void) ? I : never //  'foo' | 42 | true

type D = 'foo' & 2 & true // never

Hi, @githubxiaowen ,I try your solution in my editor.It works when all statements combined as a type template.But When I separate those statements into three statements in order to figure out what mechanism your solution is based on, it didn't work as expect.

Could you give me more hints to understand the underlying magic of the solution. (I have read the link you attached above.)

BTW, do you know the reason why the three separated statement doesn't work the same as the template type ?

[jo32]

type UnionToIntersection<U> = (U extends any ? (arg: U) => any : never) extends ((arg: infer I) => void) ? I : never
type I = UnionToIntersection<'foo' | 42 | true> // never

type A = 'foo' | 42 | true

type B = A extends any ? (arg: A) => any : never // (arg: A) => any 

type C = B extends ((arg: infer I) => void) ? I : never //  'foo' | 42 | true

type D = 'foo' & 2 & true // never

Hi, @githubxiaowen ,I try your solution in my editor.It works when all statements combined as a type template.But When I separate those statements into three statements in order to figure out what mechanism your solution is based on, it didn't work as expect.

Could you give me more hints to understand the underlying magic of the solution. (I have read the link you attached above.)

BTW, do you know the reason why the three separated statement doesn't work the same as the template type ?

@Sociosarbis It is because 'foo' & 42 & true is logically never -- you can never find a type which is 42 and true at the same time.

the magic of this solution is given a type like (arg: U) => any | (arg: V) => any, if you want to find an argument(using keyword infer) satisfying U and V, the type of it must be U & V;

[swnb]

amazing!

[vogler]

If the intersection for a field is never, you can't provide a value for the object anymore.
So when using UnionToIntersection on objects out of your control the below StripNever may be useful.

type t = UnionToIntersection<{a: number, b: number} | {a: number, b: string}>;
const t1: t = {a: 1}; //  Property 'b' is missing in type '{ a: number; }' but required in type '{ a: number; b: number; }'
const t2: t = {a: 1, b: 1}; // Type 'number' is not assignable to type 'never'.

type NonNeverKeys<T> = { [K in keyof T]: T[K] extends never ? never : K }[keyof T];
type StripNever<T> = T extends object ? { [K in NonNeverKeys<T>]: StripNever<T[K]> } : T;

const t3: StripNever<t> = {a: 1}; // ok

[Jackie1210]

[wine-fall]

The solution is amazing, however, i think write as below may be understood more easily

type UnionToIntersection<U, T = U> =  (U extends T ? (arg: U) => any : never) extends (arg: infer I) => void ? I : never;

And what does the U extends U mean? (when U is an Union Type) you can check link: #614

[kstratis]

type UnionToIntersection<U> = (U extends any ? (arg: U) => any : never) extends ((arg: infer I) => void) ? I : never
type I = UnionToIntersection<'foo' | 42 | true> // never

type A = 'foo' | 42 | true

type B = A extends any ? (arg: A) => any : never // (arg: A) => any 

type C = B extends ((arg: infer I) => void) ? I : never //  'foo' | 42 | true

type D = 'foo' & 2 & true // never

Hi, @githubxiaowen ,I try your solution in my editor.It works when all statements combined as a type template.But When I separate those statements into three statements in order to figure out what mechanism your solution is based on, it didn't work as expect.

Could you give me more hints to understand the underlying magic of the solution. (I have read the link you attached above.)

BTW, do you know the reason why the three separated statement doesn't work the same as the template type ?

I was wondering the exact same thing, so I turned to SO.
Turns out your break-down is missing generics:
https://stackoverflow.com/a/74003461/178728

[Sociosarbis]

type UnionToIntersection<U> = (U extends any ? (arg: U) => any : never) extends ((arg: infer I) => void) ? I : never
type I = UnionToIntersection<'foo' | 42 | true> // never

type A = 'foo' | 42 | true

type B = A extends any ? (arg: A) => any : never // (arg: A) => any 

type C = B extends ((arg: infer I) => void) ? I : never //  'foo' | 42 | true

type D = 'foo' & 2 & true // never

Hi, @githubxiaowen ,I try your solution in my editor.It works when all statements combined as a type template.But When I separate those statements into three statements in order to figure out what mechanism your solution is based on, it didn't work as expect.
Could you give me more hints to understand the underlying magic of the solution. (I have read the link you attached above.)
BTW, do you know the reason why the three separated statement doesn't work the same as the template type ?

I was wondering the exact same thing, so I turned to SO. Turns out your break-down is missing generics: https://stackoverflow.com/a/74003461/178728

an answer with clear explanations. thx @kstratis