Comment by s28l

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_p...