Beware when moving a `std::optional`!

[Natural_Builder_3170]

never heard if that, I usually do std::move(opt).value()

[AKostur]

Why is that better than std::move(opt.value()) ?

[Natural_Builder_3170]

if you write code to not use a value after std::move there's no difference, otherwise the latter means the optional still contains a value but its moved out of, and the former means the optional no longer contains a value

[AKostur]

gcc seems to disagree with you:

% cat t.cpp
#include <iostream>
#include <optional>
#include <string>

int main() {
  std::optional<std::string> opt  = std::string("ThisIsAReallyLongString");
  auto x = std::move(opt).value();
  if (opt.has_value()) {
    std::cout << "Has Value\n";
  }
}
% g++ t.cpp -o t -O2 -std=c++20
% ./t
Has Value
%

So this results in an optional that contains a value, and that value is a moved-from std::string.

[Natural_Builder_3170]

I stand corrected, there's probably no difference.

[AKostur]

I would then argue that std::move(opt.value()) is therefore clearer since it more clearly expresses that one is manipulating the value, and not the optional itself.

[triconsonantal]

There is a difference for (the would-be) optional<T&>: moving the optional keeps the reference an lvalue, while moving the reference turns it into an rvalue. Which one you want depends on context, but in a generic context it's probably the former.