syntax to control over distributivness

from: https://github.com/Microsoft/TypeScript/issues/30569#issuecomment-476007534

so basically we need a way to say it clear and loud in the language whether we want:

Promise<A> | Promise<B>

or

Promise<A | B>

as a result of a type operation

[Jamesernator]

(Unexpected) Distributivity of conditional types is something that bit me fairly strongly today, the fact some features are just magically distributive seems fairly confusing design in my opinion.

Like the main reason distributivity is wanted is so that people can specify things like Boxed<a | b> === Box<a> | Box<b>, but that seems more of a failure of the language to provide a way to specify a list of types to be distributed later.

Consider arrays, Array<A | B> is not Array<A> | Array<B>, the fact some generics just magically distribute seems more of an confusing feature than a predictable one. It’d be more obvious if there was just a way to declare a distributed union for any generic, i.e. maybe Array<A || B> === Array<A> | Array<B>. People then wouldn’t even need to define the wrappers, they could just || wherever they want spreading over a type constructor.

[jack-williams]

Tuples do have special treatment in that they correctly create substitution types in the true branch, effectively doing type narrowing—an object type will not do that. This is why the tuple method is proposed as the canonical way of doing this.

type OnlyNumber<N extends number> = [N];
type NoDistribution<T> = { o: T } extends { o: number } ? OnlyNumber<T> : "N" // error
type NoDistribution2<T> = [T] extends [number] ? OnlyNumber<T> : "N" // no error

I would also be reluctant to assign any special meaning to () in conditionals. () can already be used in types for grouping and they do not add any meaning they just change precedence (as they do in most other context): type A = (readonly number[])[].

Very much agree.