Question

Say, i have a function which returns a reference and i want to make sure that the caller only gets it as a reference and should not receive it as a copy.
Is this possible in C++?

In order to be more clear. I have a class like this.

class A
{
private:
    std::vector<int>  m_value;
    A(A& a){ m_value = a.m_value; }

public:
    A() {}
    std::vector<int>& get_value() { return m_value; }
};

int main()
{
    A a;
    std::vector<int> x = a.get_value();
    x.push_back(-1);
    std::vector<int>& y = a.get_value();
    std::cout << y.size();

    return 0;
}

Thanks,
Gokul.

Answer 1

You can do what you want for your own classes by making the class non copyable.

You can make an class non copyable by putting the copy constructor and operator= as private or protected members.

class C
{
private:
  C(const C& other); 
  const C& operator=(const C&);

};

There is a good example of making a NonCopyable class here that you can derive from for your own types.

If you are using boost you can also use boost::noncopyable.

Alt solution:

Another solution is to have a void return type and make the caller pass their variable by reference. That way no copy will be made as you’re getting a reference to the caller’s object.