[C++] Returning values by reference in C++

A C++ program can be made easier to read and maintain by using references rather than pointers. A C++ function can return a reference in a similar way as it returns a pointer.

When a function returns a reference, it returns an implicit pointer to its return value. This way, a function can be used on the left side of an assignment statement. For example, consider this simple program:

#include <iostream>
#include <ctime>

using namespace std;

double vals[] = {10.1, 12.6, 33.1, 24.1, 50.0};

double& setValues( int i )
{
  return vals[i];   // return a reference to the ith element
}

// main function to call above defined function.
int main ()
{

   cout << "Value before change" << endl;
   for ( int i = ; i < ; i++ )
   {
       cout << "vals[" << i << "] = ";
       cout << vals[i] << endl;
   }

   setValues() = 20.23; // change 2nd element
   setValues() = 70.8;  // change 4th element

   cout << "Value after change" << endl;
   for ( int i = ; i < ; i++ )
   {
       cout << "vals[" << i << "] = ";
       cout << vals[i] << endl;
   }
   return ;
}

When the above code is compiled together and executed, it produces the following result:

Value before change
vals[] = 10.1
vals[] = 12.6
vals[] = 33.1
vals[] = 24.1
vals[] =
Value after change
vals[] = 10.1
vals[] = 20.23
vals[] = 33.1
vals[] = 70.8
vals[] = 

When returning a reference, be careful that the object being referred to does not go out of scope. So it is not legal to return a reference to local var. But you can always return a reference on a static variable.

int& func() {
   int q;
   //! return q; // Compile time error
   static int x;
   return x;     // Safe, x lives outside this scope
}