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?

CodePudding user response：

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.

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.

CodePudding user response：

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.