Different behavior when using a type as is, or as a type alias

[ehmicky]

🔎 Search Terms

type alias
extends array
different behavior

🕗 Version & Regression Information

This was introduced by version 4.2, and continues as of today (version 5.3).

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.3.3#code/C4TwDgpgBAggPAFQHxQLxQN4EMBcUFQQAewEAdgCYDOUwATgK4QDaAulAPy2PR4BmWADZUIAXwDcAWABQM0JCgAhNLDj0mbJOKgB6HZlzcmoueGgBhFfABGAe1uCIWMpu16D-ISJPT50ACJWajyahCTk1FCWXOq8UALCEG76AAYJIikANFAA7gAWAJYAxnlQBTRFtnR0EEXApgoAokGxocSklDQ29o7OoTE8UJ6JyVApsVm5hSVlNAVkldW1wEA

💻 Code

type A<T> = {a: T extends true[] ? true : false};

type B = A<true[]>; // {a: true}
type C = A<boolean[]>; // {a: false}
type D = A<true[]> extends C ? true : false; // `false`, which is correct
type E = A<true[]> extends A<boolean[]> ? true : false; // `true`, which is incorrect

🙁 Actual behavior

The types D and E have different values.

🙂 Expected behavior

The types D and E should have the same value since they are identical, except for one using a type alias.

[fatcerberus]

Looks like a variance measurement issue (A is erroneously measured as covariant)

[ehmicky]

Thanks @fatcerberus. I can confirm this,as the following code works:

type A<in T> = {a: T extends true[] ? true : false};

[whzx5byb]

Duplicate of #48070

[ehmicky]

Thanks @whzx5byb and sorry for the duplicate. Closing in favor of #48070.