std::move真的能避免拷贝吗？关于其实际作用的技术疑问

• 我知晓std::move(object)会返回该对象的右值引用。
• 但有观点称使用push_back(std::move(object))替代push_back(object)可避免拷贝，这点我难以理解，以下示例似乎与之矛盾：

#include <utility>      
#include <iostream>     
#include <vector>       
#include <string>       
#include <iomanip>
template <class T> void print(T const & object){
    for (auto var : object){std::cout << ' '<<var;}
    std::cout << '\n';
}
int main () {
  std::string foo = "foo-string";
  std::string bar = "bar-string";
  std::vector<std::string> myvector = { "super", "vip" , "pro" , "ok" }; // capcity=size=4;

  // this block is just to increase the capacity of the vector
  myvector.push_back( "rango" );      //NOW: capacity = 8 , size =5
  std::cout << &myvector[0] <<'\n';   //check the address of the first element of the vector

  // this block is the point of the problem
  myvector.push_back(foo);                  // copies -> YES!
  myvector.push_back(std::move(bar));       // moves !?!? 

  std::cout << "myvector contains:";
  print(myvector);

  std::cout << &myvector[0] << '\n';   // re-check the address of first element of the vector = still the same.
  std::cout << "the moved-from object: " << std::quoted(bar); 

  return 0;
}

• 调用push_back(std::move(object))前后，vector首元素地址未发生变化，这意味着新插入的末元素位置与首元素存在固定偏移。
• 既然如此，若不将对象拷贝到该固定位置，如何将对象的值存储到该位置？

内容的提问来源于stack exchange，提问作者Rango