Incorrect behavior of ArrayLike<infer T> | Iterable<infer T>?

Bug Report

Search Terms: assignability, extends, generics, array like, iterable

Version & Regression Information: This is the behavior in every version I tried

Code: Playground link with relevant code

// Non-distributive:
type Foo0 = ArrayLike<string> | Iterable<string> | number[] extends ArrayLike<any> | Iterable<any> ? true : false
//   ^?: type Foo0 = true
type Bar0 = ArrayLike<string> | Iterable<string> | number[] extends ArrayLike<infer T> | Iterable<infer T> ? T[] : never
//   ^?: type Bar0 = never
type Baz0 = ArrayLike<string> | Iterable<string> | number[] extends ArrayLike<infer T> | Iterable<infer U> ? T[] | U[] : never
//   ^?: type Baz0 = never

// Distributive:
type Foo1T<T> = T extends ArrayLike<any> | Iterable<any> ? true : false
type Foo1 = Foo1T<ArrayLike<string> | Iterable<string> | number[]>
//   ^?: type Foo1 = true
type Bar1T<T> = T extends ArrayLike<infer T> | Iterable<infer T> ? T[] : never
type Bar1 = Bar1T<ArrayLike<string> | Iterable<string> | number[]>
//   ^?: type Bar1 = number[] | string[]
type Baz1T<T> = T extends ArrayLike<infer T> | Iterable<infer U> ? T[] | U[] : never
type Baz1 = Baz1T<ArrayLike<string> | Iterable<string> | number[]>
//   ^?: type Baz1 = number[] | string[] | unknown[]

Expected Behavior: Types Bar0, Baz0 and Baz1 being number[] | string[].

This all looks correct to me. Can you make a more explicit case for why you consider this wrong?

Aside: it's very confusing to have same-named variables in the same small repro

I was digging into the behavior of the Array-related types when I was making up a good definition of Array.isArray for, again, uhyo/better-typescript-lib.

I've edited the comment. Hopefully it's clearer now.

For the first 3 types, if Foo0 holds, I suspect that Bar0 and Baz0 also hold.

For the last type, Baz1, I can't understand why it consists of an unknown[].

I see. I believe this misunderstanding is what infer does when the same type variable appears in multiple positions.

What you're expecting is that these utterances act as separate paths that the checker can take, but that's not how it works. When you have the same infer T, TS is trying to find a type that fits both instances at once. It's effectively the same as if you had written

function check<T>(x: ArrayLike<T> | Iterable<T>) { }
// Fails
check(null as any as ArrayLike<string> | Iterable<string> | number[]);

which fails because there are three candidates collected for T - string, string, and number.

Baz1 includes an unknown because when zero candidates get collected for T or U respectively, the resulting inference is unknown. An equivalent example would be

type Foo<T> = T extends { a: infer A } | { b: infer B } ? ({aa: A} | { bb: B }) : never;
type S = Foo<{a : string}>;

There's nothing to occur here which would exclude the { bb: B } part of the union.

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.

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Incorrect behavior of `ArrayLike<infer T> | Iterable<infer T>`? #49764

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Incorrect behavior of ArrayLike<infer T> | Iterable<infer T>? #49764

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RyanCavanaugh commented on Jul 5, 2022

graphemecluster commented on Jul 6, 2022

RyanCavanaugh commented on Jul 7, 2022

typescript-bot commented on Jul 10, 2022

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Incorrect behavior of `ArrayLike<infer T> | Iterable<infer T>`? #49764