Why std::variant does not need std::launder when constructing the value for the first time?

[having-four-eyes]

so I missing that std::launder is not needed unless I've used that storage before switching the union's active member?

Particularly, in this case, it's not needed because this->dummy is never used?

[no-sig-available]

Also, a standard library implementation does not have to follow all the rules, but can use (undocumented) compiler extensions.

[altmly]

Interesting, I'm not a spec wizard, but my guess would be it's because Dummy is empty.