Question

If I move-construct a from b, is it still necessary to destruct b, or can I get away without doing so?

This question crossed my mind during the implementation of an optional<T> template. Excerpt:

~optional()
{
    if (initialized)
    {
        reinterpret_cast<T*>(data)->~T();
    }
}

optional(optional&& o) : initialized(o.initialized)
{
    if (initialized)
    {
        new(data) T(std::move(*o));   // move from o.data
        o.initialized = false;        // o.data won't be destructed anymore!
    }
}

Of course, I could just replace the bool initialized with a three-valued enumeration that distinguishes between initialized, non-initialized and moved-from. I just want to know if this is strictly necessary.

Answer 1

Yes, it is still necessary to destruct b. A moved from object is a valid, constructed object. In some cases, it may even hold resources that still need to be disposed of. In generic code such as you show, T may not even have a move constructor. You may invoke a copy constructor instead in this case. So you can definitely not assume that ~T() is a no-op and can be elided.