STL vector: Moving all elements of a vector

std::move works fine. Here is the sample code for the same

    vector<int> v1 = {1,2,3,10,20,30,100,200,300,999};
    vector<int> v2;

    cout << "Size of v1 before move = " << v1.size() << endl;
    cout << "Capacity of v1 before move = " << v1.capacity() << endl;

    v2 = std::move(v1);

    cout << "Size of v2 after move = " << v2.size() << endl;
    cout << "Capacity of v2 after move = " << v2.capacity() << endl;

    cout << "Size of v1 after move = " << v1.size() << endl;
    cout << "Capacity of v1 after move = " << v1.capacity() << endl;

-----------Output-------------------------
Size of v1 before move = 10
Capacity of v1 before move = 10
Size of v2 after move = 10
Capacity of v2 after move = 10
Size of v1 after move = 0
Capacity of v1 after move = 0

I have two STL vectors A and B and I'd like to clear all elements of A and move all elements of B to A and then clear out B.

This can be done with a combination of swap. First swap A and B for the first half. Then swap an empty std::vector<> with B or call clear(). The difference is that clear() will not release the memory, but only destroy the objects:

std::vector<int> a, b; // initialize them somehow
swap(a,b);

// clear b without releasing the memory:
std::size_t capacity = b.capacity();
b.clear();
assert(b.capacity()==capacity);

// or release the memory
std::vector<int>().swap(b);
assert(b.capacity()==0);

Using C++11, it's as simple as:

A = std::move(B);

Now A contains the elements that were previously held by B, and B is now empty. This avoids copying: the internal representation is simply moved from B to A, so this is an O(1) solution.

As for C++03, as Prætorian states, you could swap the vectors. There is a specialization of the std::swap function, which takes std::vectors as its arguments. This effectively swaps the internal representation, so you end up avoiding creating copies of the elements held by them. This function works in O(1) complexity as well.

If you have a C++11 compiler you can move B into A.

A = std::move(B);

If you're working with an older compiler, just swap the two

A.swap(B);

In both cases, the only O(N) operation will be clearing the contents of A. In the first case the clearing will be done during the assignment itself, while in the second it will happen when B goes out of scope (since the contents were swapped).