问题
Is it permitable to design std::optional (currently std::experimental::optional) in such a way, that for trivially default constructible type T corresponding std::optional< T > is also trivially default constructible?
The same question regading std::variant and its integral discriminator.
My own answer is: "No, it cannot be designed in this way, because value of its integral discriminator obtained during default initialization will be indeterminate if the object has automatic storage duration or if it is reinterpret_cast-ed from non-zero-initialized storage." Requirement to the user to do value-initialization every time is not allowed on my mind.
回答1:
Your answer is correct: you cannot. The specification requires that its "initialized flag" is set to false upon default construction.
回答2:
As you explained yourself, you can't implement std::optional in such a way, because you would change its semantics (is_trivially_default_constructible is part of the class interface).
However, if you require this semantic for some reason in your code, there is no reason, why you couldn't implement a very similar optional class that is trivially default constructible. Then, when used, just zero initialize it via {} and - if that is what you want - treat zero as true in the bool operator.
来源：https://stackoverflow.com/questions/34061095/trivially-default-constructible-stdoptional-and-stdvariant