Extract<T, T> should be the same as T. But It does not behave the same way. #46794
Description
Bug Report
🔎 Search Terms
Extract
Extract is not pure
Extract behaviour
🕗 Version & Regression Information
v4.5.0-beta
⏯ Playground Link
Playground link with relevant code
💻 Code
type Input = number[] | string[];
type JoinArrays<T> = T extends (infer Element)[] ? Element[] : never;
type JoinArraysWithExtract<T> = Extract<T, T> extends (infer Element)[] ? Element[] : never;
type Output = JoinArrays<Input> // number[] | string[]
type OutputWithExtract = JoinArraysWithExtract<Input> // (string | number)[]
type Wanted = (number | string)[];
🙁 Actual behavior
type Extract<T, U> = T extends U ? T : never implies that replacing Extract<T, T> with T should return exactly the same type.
But replacing it in the example I have provided results in two different types.
🙂 Expected behavior
type JoinArrays<T> = T extends (infer Element)[] ? Element[] : never;
should be the same as
type JoinArraysWithExtract<T> = Extract<T, T> extends (infer Element)[] ? Element[] : never;