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

Commencez par l’exemple TypeScript Playground lié et reproduisez les affectations non sûres impliquant ShouldBeInvariant. Suivez l’assignabilité des signatures d’appel génériques et la vérification de variance, puis ajoutez un test de régression montrant que les deux affectations signalent des erreurs et que les appels invalides ne passent plus la vérification des types.