Why does moving std::optional not reset state

Question

I was rather surprised to learn that the move constructor (and assignment for that matter) of std::optional does not reset the optional moved from, as can be se

Answer 1

While it might be reasonable to expect that std::optional behaves similarly to std::unique_ptr which resets state of moved-from object, there are reasons not to demand such behavior. I think one of them is that std::optional of a trivial type should be a trivially copyable type. As such, it cannot have a non-defaulted move constructor and cannot reset its has_value flag.

Having std::optional for a non-trivial type behave differently from std::optional for a trivial type is a rather bad idea.

Answer 2

What that paragraph says is that if that optional had a value, it still has a value. Since that value has been moved from (to the newly constructed object), it could be a different value than what it had before the move. This allows you to access the moved-from optional object the same way as a moved-from non-optional object, so the behavior of a T vs. optional<T> (when it contains an object) when accessed after the move is the same.

Also, the overall effect of a move from an optional depends on how the contained type T handles a move. Other classes (like vector) do not have this dependency.

Answer 3

In a word: Performance.

One of the chief motivating reasons for move semantics to exist in the first place is performance. So the special operations move construction and move assignment should be as fast as possible for all types.

In order to assist this goal, it is standard practice that moved-from objects be left in a valid but unspecified state. So the very minimum that optional move construction/assignment need to do is to move from the source argument. To specify setting the source to not have a value after the move is equivalent to saying:

After you move, do some extra, unnecessary work.

No matter how small that extra work is, it is non-zero. Some (and I dare say many) clients will not need that extra work, and should not have to pay for it. Clients who do need it can easily add x.reset() after the move, putting the moved-from optional into a well-specified state.

Answer 4

Unless otherwise specified, a moved-from object of class type is left in a valid but unspecified state. Not necessarily a "reset state", and definitely not "invalidated".

For primitive types , moving is the same as copying, i.e. the source is unchanged.

The defaulted move-constructor for a class type with primitive members will move each member, i.e. leave the primitive members unchanged; a user-defined move constructor might or might not "reset" them.

A moved-from vector may or may not still have elements in it. We would expect it not to, since that's efficient, but it cannot be relied on.

A moved-from std::string may still have elements in it, because of Small String Optimization.

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move on std::optional is actually specified by the standard (C++17 [optional.ctor]/7). It is defined as doing move on the contained type, if present. It does not turn a valued optional into a valueless optional.

So it is actually expected that your code outputs true true, and the actual contained value in foo should stay the same too.

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Regarding the question of why std::optional 's move-constructor is defined this way: I can't say for sure; but an optional is not like a vector with max size of 1. It's more like a variable with a validity flag tacked on. Accordingly, it makes sense for moving an optional to be like moving the variable.

If moving an optional left the old one "empty", then a = std::move(b); would invoke the destructor of b's managed object, which would be unexpected (to me, at least).