Why doesn't move vector trigger moving internal elements?

Question

For the following code, I am confused about p3 vector, why move operators are not triggered for Test? For p1 and p2, I can understand that all the elements need to copied be over.

Does that mean when we move a vector, it modifies the vector object so the elements of the vector will leave untouched. Whereas when we copy a vector, we have to copy every single element. Is there a vector operation where the move operator of an element will be trigger?

#include <iostream>
#include <vector> 

using namespace std;

class Test {
  public: 
    Test(int a) {
        std::cout << " default " << std::endl;
    }
    
    Test(const Test& o) {
        std::cout << " copy ctor " << std::endl;
    }
    
    Test& operator=(const Test& o) {
        std::cout << " copy assign " << std::endl;
        return *this;
    }
    
    Test(Test&& o) {
        std::cout << " move ctor" << std::endl;
    }
    
    Test& operator=(Test&& o) {
        std::cout << " move assign " << std::endl;
        return *this;
    }
};

int main()
{
  std::cout << " p: " << std::endl;
  std::vector<Test> p = {Test(0), Test(0)}; 
  std::cout << std::endl;
   
  std::cout << " vec p1 " << std::endl;
  std::vector<Test> p1 = p;
  std::cout << std::endl;
    
  std::cout << " vec p2 " << std::endl;
  std::vector<Test> p2;
  p2.reserve(2);
  p2.emplace_back(0);
  p2.emplace_back(0);
  std::cout << std::endl;
   
  std::cout << " vec p3 " << std::endl;
  std::vector<Test> p3 = std::move(p);
}

Output:

p: 
default 
default 
copy ctor 
copy ctor 

vec p1 
copy ctor 
copy ctor 

vec p2 
default 
default 

vec p3 

Answer 1

A vector is essentially a pointer to an array. Moving the vector simply swaps the array pointers between the two vectors, the arrays themselves are unchanged.