Conditional type eagerly calculated to be invariant breaks referential transparency

[devanshj]

🔎 Search Terms

Conditional types, Invariance, Referential transparency

🕗 Version & Regression Information

Tried with v5.9.3

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.9.3#code/C4TwDgpgBAYgPAFQHxQLxQN5QG4EMA2ArhAFxQJQQAewEAdgCYDOUA9gEYBWEAxsFAH4oAawghWAM3JQyFAL4AoBaEhQASgEY0sOBjkpqtRi3hZcZJsABOASzoBzKPsFQARAAtW+Zq5luGrBAsdKz8nt6uCgD0UVBxAHoCSirQMLjapvrK4KnsGbpQ5lCWtg5OSMk56gBMGemG9MyweUIeXj5+rgFBUCFh7ZExcVCJCkA

💻 Code

type F<T> = { value: T extends object ? keyof T : T }

type R1 = F<{}> extends F<{ a: string }> ? "holds" : "does not hold"
//   ^? "does not hold"

type Fa = F<{}>
type Fb = F<{ a: string }>

type R2 = Fa extends Fb ? "holds" : "does not hold"
//   ^? "holds"

🙁 Actual behavior

R1 computes to "does not hold"

🙂 Expected behavior

R1 computes to "holds"

Additional information about the issue

I'm assuming typescript eagerly calculates F to be invariant and then instead of checking the subtype relation on resolved types it calculates subtype relation of the parameters to F. (This would perhaps have made sense if F was explicitly annotated with in out T.)

[MartinJohns]

This sounds like a duplicate of #48070.

[RyanCavanaugh]

#55161 (comment) is also closely related, the variance we should have measured on F is "unreliable" but instead it looks like we're getting invariant (I haven't confirmed this).

This is a longstanding problem and straightforward attempts to fix it have generally made a lot of things worse and very few things better. Open to a PR but this might be effectively a duplicate of "We need a way to disable variance measurement on a type".

[Andarist]

I can only confirm that F is measured to be invariant here and that the other example ends up comparing different type aliases so it goes straight~ to the structural comparison.