Microsoft/TypeScript
Ver no GitHubConditional type eagerly calculated to be invariant breaks referential transparency
Open
#62.798 aberto em 23 de nov. de 2025
BugCursed?Domain: check: Variance RelationshipsHelp Wanted
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Description
🔎 Search Terms
Conditional types, Invariance, Referential transparency
🕗 Version & Regression Information
Tried with v5.9.3
⏯ Playground Link
💻 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.)