[feature] .toEqualUnorderedTypeOf

[gomain]

I have a use case where the order of items in the construction of a tuple type is non-deterministic.
To test I currently do:

  const tuple = expectTypeOf<TupleType>();
  tuple.toMatchTypeOf<[any, any, any]>(); // has length 3
  tuple.items.exclude<A | B>().toEqualTypeOf<C>(); // one of the items is C
  tuple.items.exclude<A | C>().toEqualTypeOf<B>(); // one of the items is B
  tuple.items.exclude<B | C>().toEqualTypeOf<A>(); // one of the items is A

But this only works when tuple items are indeed not union types. There is no way to test if one of the items type is a certain union type.

I propose a .toEqualUnorderedTypeOf:

  expectTypeOf<TestType>().toEqualUnorderedTypeOf<[A, B, A]>();
  /* any of these would pass, and fail otherwise
   * [A, B, A]
   * [B, A, A]
   * [A, A, B]
   */

Other matchers/combinators along this line are

• .toContainTypeOf // passes if at least one item has equal type
• .toContainItemsTypeOf // passes if tuple items is a superset of provided tuple
• .excludeItem // returns tuple with items of equal type removed, or fail if not found
• .excludeItems // returns tuple minus (in set sense) the provided tuple, or fail if can't

[mmkal]

These seems a little niche and probably complex to implement - could you say more about the use case? Maybe there's a way it could be made deterministic, it seems a bit strange for types to be non-deterministic.

[gomain]

There is this trick to implement transforming union types to tuples of constituents:

// must condition on T
type Contra<T> = T extends infer I ? (arg: I) => void : never;

// must wrap in tuple
type InferContra<F> = [F] extends [(arg: infer T) => void] ? T : never;

/*
 * Which one is picked is nondeteriminstic.
 * Recall that "a" | "b" == "b" | "a".
 */
type PickOne<T> = InferContra<InferContra<<Contra<<Contra<T>>>>;

type _Union2Tuple<ACC extends any[], T> = PickOne<T> extends infer U
  ? Exclude<T, U> extends never
    ? [T, ...ACC]
    : _Union2Tuple<[U, ...ACC], Exclude<T, U>>
  : never;

type Union2Tuple<T> = _Union2Tuple<[], T>;

/*
 * Because the order of constructing the tuple type is nondeterministic,
 * we can not assert with a tuple type. We can assert the length of the tuple
 * by _matching_ with tuples such as `[any, any]` to assert a length of 2.
 * Then exclude all but except one type from the tuple items type in all
 * combinations.
 */
{
  const tuple = expectTypeOf<Union2Tuple<"a" | 1>>();
  tuple.toMatchTypeOf<[any, any]>();
  tuple.items.exclude<"a">().toEqualTypeOf<1>();
  tuple.items.exclude<1>().toEqualTypeOf<"a">();
}

[gomain]

@mmkal I'm willing to hack on how one would be implemented. Names are open for bike-shedding.

[mmkal]

@gomain this is an old issue, but v0.20.0 added a UnionToTuple type for some internal stuff. And you are right, I have observed the ordering of the tuple does seem to be non-deterministic. I assumed it would just be unstable between typescript versions, but it can change between times you hover on a type in the IDE.

If you still want to try it out, and can find a not-too-complicated way to implement, I'd accept a PR.

[mmkal]

For fun I thought I'd try this. It's fairly easy to use the existing utility types to get something that will work for this use case:

type IsSubset<Small extends any[], Big extends any[]> = And<{
  [SmallKey in keyof Small]: Or<{
    [BigKey in keyof Big]: StrictEqualUsingTSInternalIdenticalToOperator<Small[SmallKey], Big[BigKey]>
  }>
}>

type TuplesHaveIdenticalItems<A extends any[], B extends any[]> = And<
  [Extends<A['length'], B['length']>, IsSubset<A, B>, IsSubset<B, A>]
>

tests

expectTypeOf<IsSubset<[1, 2, 3, 4], [1, 2, 3]>>().toEqualTypeOf<false>()
expectTypeOf<IsSubset<[1, 2, 3], [1, 2, 3, 4]>>().toEqualTypeOf<true>()

expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [1, 2, 3, 4]>>().toEqualTypeOf<false>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [1, 2, 3]>>().toEqualTypeOf<true>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [3, 2, 1]>>().toEqualTypeOf<true>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [3, 2, 1, 1]>>().toEqualTypeOf<false>()

expectTypeOf<TuplesHaveIdenticalItems<[1, 1, 2, 2, 3], [1, 1, 1, 2, 3]>>().toEqualTypeOf<true>()

That last one is a little sad, because while they do have the same number of items and the same items, it doesn't feel like [1, 1, 2, 2, 3] is "the same" as [1, 1, 1, 2, 3].

---

alternate approach: "count" the matches between A and B, and compare them to the count of the matches between A and A:

type CountTrues<Bools extends boolean[], Tally extends string = ''> = Bools extends []
  ? Tally
  : Bools extends [infer Head, ...infer Rest extends boolean[]]
    ? CountTrues<Rest, `${Tally}${Head extends true ? 'I' : ''}`>
    : never

type Matches<A extends any[], B extends any[]> = {
  [Akey in keyof A]: CountTrues<{
    [Bkey in keyof B]: StrictEqualUsingTSInternalIdenticalToOperator<A[Akey], B[Bkey]>
  }>
}
type TuplesHaveIdenticalItems<A extends any[], B extends any[]> = And<[
  Extends<A['length'], B['length']>,
  StrictEqualUsingTSInternalIdenticalToOperator<Matches<A, B>, Matches<A, A>>,
]>

Tests look better!

expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [1, 2, 3, 4]>>().toEqualTypeOf<false>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [1, 2, 3]>>().toEqualTypeOf<true>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [3, 2, 1]>>().toEqualTypeOf<true>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 2, 3], [3, 2, 1, 1]>>().toEqualTypeOf<false>()

expectTypeOf<TuplesHaveIdenticalItems<[1, 1, 2, 2, 3], [1, 1, 1, 2, 3]>>().toEqualTypeOf<false>()
expectTypeOf<TuplesHaveIdenticalItems<[1, 1, 2, 2, 3], [1, 1, 3, 2, 2]>>().toEqualTypeOf<true>()

I'm tempted to put this in think but this might be too niche for this library though!

Playground for those interested