Experiment short-circuiting of union distribution

[RyanCavanaugh]

Suggestion

🔍 Search Terms

union distribute short-circuit performance

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

⭐ Suggestion

@amcasey and I were discussing #47481 and he had an interesting observation. Consider this program:

type HugeUnion1 = "a" | "b" | "c" | "d" | ... [10,000 more];
type HugeUnion2 = "q" | "r" | "s" | "t" | ... [10,000 more];

type Square<T, U> = T extends U ? true : false;

type M = Square<T, U>;

Naively, when evaluating M, we need to check if every member of HugeUnion1 is present in HugeUnion2. However, let's say we saw this (in the sorted form):

type HugeUnion1 = "a" | "q" | ... [10,000 more];
type HugeUnion2 = "a" | ... [10,000 more which do not include 'q'];

After evaluating "a", we have a true. After evaluating "q", we have a false. We can skip the rest of the list since no subsequent result will change the answer (because T does not appear in the conditional result)

Moreover, if instead of false we had never, we could immediately stop after seeing true, since true | never is just true

[DanielRosenwasser]

Seems like this would use similar rules as @ahejlsberg's #46429, right? You can avoid distribution and just map to a set of possibilities for A extends B ? C : D when

• The check-type (A) appears in neither branch and
• The extends type (B) has no infer positions

With the possibilities never, C, D, C | D, and possibly any (?)

[jakebailey]

With #53192 merged, I think that this one is effectively "done"