C++11 - what does it mean to initialize a variable to a reference

Question

#include<vector>
using namespace std;

class Foo
{
  private:
    vector<int> m_vec;
  public:
    vector<int> getFoo()
    {
      return m_vec; // returns a copy of m_vec
    }
    vector<int>& getFooRef()
    {
      return m_vec; // returns a reference of m_vec
    }
};

int main()
{
  Foo foo;
  vector<int> x = foo.getFoo(); // init x with a copy of m_vec
  vector<int>& y = foo.getFooRef(); // y is a reference to foo.m_vec - no new copy is created
  vector<int> z = foo.getFooRef(); // what happens here?? Is the copy constructor invoked?
  vector<int>& a = foo.getFoo(); // what happens here?? Is `a` a reference to a copy of m_vec?
  return 0;
}

Possible duplicate of [What are the differences between a pointer variable and a reference variable in C++?](http://stackoverflow.com/questions/57483/what-are-the-differences-between-a-pointer-variable-and-a-reference-variable-in) — tkausl, Jun 21 '16 at 00:50
As `z` is not a reference itself, it will be initialized with a copy of `m_vec`. Reference to `m_vec` will be returned from the method, but assigning it to `z` will execute the [copy assignment operator](http://en.cppreference.com/w/cpp/language/copy_assignment). — Jezor, Jun 21 '16 at 00:55

score 3 · Answer 1 · answered Jun 21 '16 at 00:56

When you do this

vector<int> z = foo.getFooRef();

the compiler uses the reference on the right of the = operator to perform initialization of variable z to the left of the = operator.

The nature of the actual operation of initialization is determined by the left side, i.e. z. Since z is not a reference, operator = copies the content of the vector obtained by reference into z.

C++11 - what does it mean to initialize a variable to a reference

1 Answers1