There is a big difference between "undefined" and "unspecified" behavior. In this case, the behavior of `unique_ptr(unique_ptr&&)` is in fact specified. [0] Ho

There is a big difference between "undefined" and "unspecified" behavior. In this case, the behavior of `unique_ptr(unique_ptr&&)` is in fact specified. [0] However, the bigger issue with that code is that it can easily stop working with a simple refactor. Consider: void foo(std::unique_ptr<int> ptr) {} void bar(std::unique_ptr<int>&& ptr) {} int main() { std::unique_ptr<int> p1{new int{1}}; std::unique_ptr<int> p2{new int{2}}; foo(std::move(p1)); assert(p1 == nullptr); bar(std::move(p2)); assert(p2 != nullptr); } Neither of the above asserts will fire, but from the calling site, they look exactly the same. In my opinion, the more explicit option would be to do something like `bar(std::exchange(p2, nullptr))` [0]: overload (5) https://en.cppreference.com/w/cpp/memory/unique_ptr/unique_ptr https://en.cppreference.com/w/cpp/memory/unique_ptr/unique_p...