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

从链接的 TypeScript Playground 示例开始，重现涉及 ShouldBeInvariant 的不健全赋值。跟踪泛型调用签名的可赋值性和方差检查，然后添加一个回归测试，表明两个赋值都会报告错误，并且无效调用不再通过类型检查。