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 seen in [19.6.3.1/7] which states "bool(rhs) is unchanged."

This can also be seen by the following code:

#include <ios>
#include <iostream>
#include <optional>
#include <utility>

int main() {
  std::optional<int> foo{ 0 };
  std::optional<int> bar{ std::move(foo) };

  std::cout << std::boolalpha
            << foo.has_value() << '\n'  // true
            << bar.has_value() << '\n'; // true
}

This seems to contradict other instances of moving in the standard library such as with std::vector where the container moved from is usually reset in some way (in vector's case it is guaranteed to be empty afterwards) to "invalidate" it even if the objects contained within it themselves have been moved from already. Is there any reason for this or potential use case this decision is supposed to support such as perhaps trying to mimic the behavior of a non-optional version of the same type?

Mainly explained in the documentation: https://en.cppreference.com/w/cpp/utility/move — Phil1970, Aug 12 '18 at 02:18
"*in vector's case it is guaranteed to be empty afterwards*" Um... no it isn't. — Nicol Bolas, Aug 12 '18 at 04:42
@NicolBolas Well in most cases, cppreference at least says that the "After the move, other is guaranteed to be empty().", but in the case of using a custom allocator such is not true (but I wasn't talking about using a custom allocator so while technically correct that's not really what I meant). — Lemon Drop, Aug 12 '18 at 14:58
@LemonDrop: Well then cppreference is lying. The *standard* does not say that, and the standard is what matters. — Nicol Bolas, Aug 12 '18 at 15:01
@BenVoigt: The behavior of the move constructor/assignment operators are described in the table for `[container.requirements.general]/4`. Specifically, "*Postconditions*: `u` shall be equal to the value that `rv` had before this construction". It says nothing about the state of `rv`, so the general statement about moves for all standard library types applies. — Nicol Bolas, Aug 12 '18 at 18:36
@NicolBolas: Ah yes, missed that because it didn't contain the word "move". — Ben Voigt, Aug 12 '18 at 18:40
@NicolBolas Upon looking some more into it [someone suggested](https://stackoverflow.com/a/17735913/2085551) that since the complexity requirement of the move constructor is constant, the only way it can be implemented involves leaving the other vector in such a state, though I don't know how I feel about that not being stated explicitly. — Lemon Drop, Aug 12 '18 at 23:18
On the state of a moved-from `vector`: https://stackoverflow.com/a/17735913/576911 — Howard Hinnant, Aug 13 '18 at 01:25
@LemonDrop That is not true, however. E.g. a `vector` could be left with contents `{1,2,3,4}` with constant time complexity. — Arne Vogel, Aug 13 '18 at 06:10

M.M · Accepted Answer · 2018-08-13T01:58:56.807

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.

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.

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).

As an aside, letting a moved-from small string with active SSO retain its contents on being used for move-construction would mean branching. Simply constructing an empty string and swapping sight-unseen is more efficient. — Deduplicator, Jun 04 '19 at 12:50

Howard Hinnant · Answer 2 · 2018-08-13T01:22:26.920

32

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.

edited Aug 13 '18 at 01:22

answered Aug 12 '18 at 13:40

Howard Hinnant

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Problem is that most(all?) std implementations do the wrong thing with std::string so they give unreasonable expectations to users. :/ – NoSenseEtAl Aug 18 '18 at 10:14
2

@NoSenseEtAl: I can only speak for the libc++ `std::string`. It does indeed leave a moved-from `string` empty, but not because it is "wrong", but because that is the fastest thing for this design. The libc++ `string` was actually designed from the move members outward. It copies all of the bits of the `string` object, and then zeros all of the bits of the source `string` object. It does not check whether the `string` is long or short. This copy&zero algorithm is correct whether the string is in the long or short mode. It would be incorrect to avoid the zero when in the long mode. – Howard Hinnant Aug 18 '18 at 13:41
I find it weird that it is faster, since anyway destination string will need to branch on long or short because it needs to know if it is storing pointers or internal buffer, but I trust you. :) – NoSenseEtAl Aug 19 '18 at 03:44
1

Another nice thing about libc++ is that its source is so easily inspectable. ;-) https://github.com/llvm-mirror/libcxx/blob/master/include/string#L1846-L1862 – Howard Hinnant Aug 19 '18 at 13:42
well github has bad code navigation, but anyways you would need benchmarks, not just the source, but like I said I believe you, since I assume you did them when you implemented string rvr ctor. :) – NoSenseEtAl Aug 19 '18 at 14:23
2

The "moving from a string" case in libc++ was described (by Howard) here: https://stackoverflow.com/questions/52696413/unnecessary-emptying-of-moved-from-stdstring . – Marshall Clow Jun 03 '19 at 21:34

score 5 · Answer 3 · answered Aug 12 '18 at 01:57

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.

score 4 · Answer 4 · edited Jun 04 '19 at 11:22

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.

Why does moving std::optional not reset state

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