May I change the held type in a std::variant from within a call to std::visit

问题

Does the following code invoke undefined behaviour?

std::variant<A,B> v = ...;

std::visit([&v](auto& e){
  if constexpr (std::is_same_v<std::remove_reference_t<decltype(e)>,A>)
    e.some_modifying_operation_on_A();
  else {
    int i = e.some_accessor_of_B();
    v = some_function_returning_A(i); 
  }
}, v);

In particular, when the variant does not contain an A, this code re-assigns an A while still holding a reference to the previously held object of type B. However, because the reference is not used anymore after the assignment, I feel the code is fine. However, would a standard-library be free to implement std::visit in a way such that the above is undefined behaviour?

回答1:

The code is fine.

There is no requirement in the specification of std::visit that the visitor not change the alternative of any of the variants it is invoked on. The only requirement is:

Requires: For each valid pack m, e(m) shall be a valid expression. All such expressions shall be of the same type and value category; otherwise, the program is ill-formed.

Your visitor is a valid expression for each m and always returns void, so it satisfies the requirements and has well-defined behavior.

来源：https://stackoverflow.com/questions/55187548/may-i-change-the-held-type-in-a-stdvariant-from-within-a-call-to-stdvisit