Move a std::map vs move all the elements of a std::map

Question

std::map<int, Obj> mp;
// insert elements into mp

// case 1
std::map<int, Obj> mp2;
mp2 = std::move(mp);

// case 2
std::map<int, Obj> mp3;
std::move(std::begin(mp), std::end(mp), std::inserter(mp3, std::end(mp3));

I am confused by the two cases. Are they exactly the same?

Answer 1

No, they are not the same.

• Case 1 moves the content of the whole map at once. The map's internal pointer(s) are "moved" to mp2 - none of the pairs in the map are affected.
• Case 2 moves the individual pair's in the map, one by one. Note that map Key s are const so they can't be moved but will instead be copied. mp will still contain as many elements as before - but with values in an indeterminable state.

Answer 2

Are they exactly the same?

No, They are not!

The first one invokes the move constructor of the std::map⁴ and the move operation will be done at class/ data structure level.

[...]

4. Move constructor. After container move construction (overload (4)), references, pointers, and iterators (other than the end iterator) to other remain valid, but refer to elements that are now in *this. The current standard makes this guarantee via the blanket statement in container.requirements.general, and a more direct guarantee is under consideration via LWG 2321

Complexity

4) Constant. If alloc is given and alloc != other.get_allocator(), then linear.

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The second std::move is from <algorithm> header, which does element wise(i.e. key value pairs) movement to the other map.

1. Moves the elements in the range [first, last), to another range beginning at d_first, starting from first and proceeding to last - 1. After this operation the elements in the moved-from range will still contain valid values of the appropriate type, but not necessarily the same values as before the move.

Complexity

Exactly last - first move assignments.