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Decompose a TypeScript union type into specific types

I wonder whether it's possible to “split” union types into the specific subtypes in TypeScript. This is the code I tried to use, it should be obvious what I'm trying to achieve from the snippet:

type SplitType<T> =
  T extends (infer A)|(infer B)
  ? Something<A, B>
  : T;

In this example Something<A, B> could be [A, B], or a completely different type. This would mean that SplitType<string> would just output a string, but SplitType<number|string> would mean [number, string].

Is something like that possible in TypeScript? And if not, is there a feature that will allow this in the future (eg. variadic types)?

like image 580
m93a Avatar asked Jun 05 '26 15:06

m93a


2 Answers

With Matt McCutchen's answer, and the recursive conditional types since v4.1:

type UnionToParm<U> = U extends any ? (k: U) => void : never;
type UnionToSect<U> = UnionToParm<U> extends ((k: infer I) => void) ? I : never;
type ExtractParm<F> = F extends { (a: infer A): void } ? A : never;

type SpliceOne<Union> = Exclude<Union, ExtractOne<Union>>;
type ExtractOne<Union> = ExtractParm<UnionToSect<UnionToParm<Union>>>;

type ToTuple<Union> = ToTupleRec<Union, []>;
type ToTupleRec<Union, Rslt extends any[]> = 
    SpliceOne<Union> extends never ? [ExtractOne<Union>, ...Rslt]
    : ToTupleRec<SpliceOne<Union>, [ExtractOne<Union>, ...Rslt]>
;

type test = ToTuple<5 | 6 | "l">;
like image 178
SE12938683 Avatar answered Jun 09 '26 22:06

SE12938683


For a fixed maximum number of union members, we can extract the union members in a single implementation-defined order by generating an intersection of call signatures and then matching it against a type with multiple call signatures. This version only works with strictFunctionTypes enabled.

// https://stackoverflow.com/a/50375286
type UnionToIntersection<U> = 
  (U extends any ? (k: U)=>void : never) extends ((k: infer I)=>void) ? I : never

type UnionToFunctions<U> =
    U extends unknown ? (k: U) => void : never;

type IntersectionOfFunctionsToType<F> =
    F extends { (a: infer A): void; (b: infer B): void; (c: infer C): void; } ? [A, B, C] :
    F extends { (a: infer A): void; (b: infer B): void; } ? [A, B] :
    F extends { (a: infer A): void } ? [A] :
    never;

type SplitType<T> =
    IntersectionOfFunctionsToType<UnionToIntersection<UnionToFunctions<T>>>;

type Test1 = SplitType<number>;                    // [number]
type Test2 = SplitType<number | string>;           // [string, number]
type Test3 = SplitType<number | string | symbol>;  // [string, number, symbol]
like image 43
Matt McCutchen Avatar answered Jun 09 '26 22:06

Matt McCutchen



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