Understanding std::vector::push_back(std::move(v[i])) [duplicate]

Question

Beginner question.

In the following code, I was expecting v2[0]=1 would also change the value of v1, but it seems not.

push_back seems to receive T&& since C++11(ref), and std::move is equivalent to static_cast<T&&>(t); according to this answer. Also std::vector::operator[] returns reference(ref). So I thought v2.push_back(std::move(v1[0])); would make a reference to the same value.

What am I missing here? I thought the output would be 1 and 1.

#include <iostream>
#include <vector>

int main(){
    std::vector<int> v1{5}, v2;
    v2.push_back(std::move(v1[0])); 

    v2[0] = 1;

    std::cout << v1[0] << '\n';
    std::cout << v2[0] << '\n';

    // output:
    // 5
    // 1
}

Answer 1

So I thought v2.push_back(std::move(v1[0])); would make a reference to the same value.

v1[0] is an lvalue referring to the first element of the vector, and std::move(v1[0]) is an rvalue referring to that element. _{The move has little to do with the behaviour of the example.}

But the elements of v2 aren't references. They are integers. So, you initialise the non-reference second element of the vector using the value of the first element through the reference. It's analogous to following simpler example:

int a = 5;
int& ref = a;
int b = ref;  // b is not a reference
b = 1;        // this has no effect on a,
              // because it is a separate object from b

std::cout << a << '\n';
std::cout << b << '\n';

Answer 2

A vector like all containers can't store references. If you want v2 elements reference some elements of v1, you can use std::reference_wrapper<>.

#include <iostream>
#include <vector>
#include <functional>  // for std::reference_wrapper<>

int main() {
   std::vector<int>  v1{ 5 };
   std::vector<std::reference_wrapper<int>>  v2;
   v2.push_back( v1[0] );  // move has no sense, so illegal here

   // v2[0] = 1;   illegal a reference_wrapper is not exactly a reference
   // we must use : v2[0].get() = 1;
   // or            static_cast<int&>(v2[0]) = 1;
   v1[0] = 3;    // possible

   std::cout << v1[0] << ' ' << v2[0] << '\n';

   // output: 3 3
}

Note that referencing elements in vector v1 is not a good idea, because any modification in v1 will made all references invalid.