type parameter variance in generic call signature is incorrectly bivariant

🔎 Search Terms

unsound generic function assignability type parameter variance generic call signature bivariant

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about generics and variance.

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=6.0.3#code/C4TwDgpgBAygFgewK4BsAmAhCBJAdgNwEMAnAS0N2AB4ARAPigF4oqAVOgCgA8AuKVgGQ0ANFBB9WASkZ1BNAFDyAxglwBnYFDUQVuNH3jJ0WPETIVqSXAGtcCAO64GzKN1EhJTBiGWqNUZGAAOSQAWwAjCGIDRFRMHAISckoqXDDI4mctHVU0eUCQiKiqK1sHJw4AFlEAcg1iGskAOmAEAFEuMFUISnIUDklFXX9SXEKMmKN40ySLVPSorO5PGVcuFvbO7t7CfslRLkHhzVGAVRs7R0m4k0TzFNLLpyYoUfGo+TOL8o464AbavVGoogA

💻 Code

type ShouldBeInvariant<D> = <T>(x: T&D, y: T)=>T&D

const second: ShouldBeInvariant<unknown> =  (x, y) => y
const outNumber: ShouldBeInvariant<number> = second
outNumber<unknown>(4, 'str').toExponential()

const inNumber: ShouldBeInvariant<number> = (x) => (x.toExponential(), x)
const inUnknown: ShouldBeInvariant<unknown> = inNumber
inUnknown('str', 'str')

🙁 Actual behavior

The code type checks, while it shouldn't because that is unsound, using ShouldBeInvariant covariantly or contravariantly both causes runtime errors.

🙂 Expected behavior

ShouldBeInvariant should be invariant against its parameter. Both assigns:

const outNumber: ShouldBeInvariant<number> = second

and

const inUnknown: ShouldBeInvariant<unknown> = inNumber

should report errors

Additional information about the issue

No response

Beginne mit dem verknüpften TypeScript Playground-Beispiel und reproduziere die unsounden Zuweisungen mit ShouldBeInvariant. Verfolge die Zuweisungsverträglichkeit generischer Aufrufsignaturen und die Varianzprüfung. Füge anschließend einen Regressionstest hinzu, der zeigt, dass beide Zuweisungen Fehler melden und dass die ungültigen Aufrufe nicht mehr typgeprüft werden.