// --- Example: Functions. Arguments by reference
/*
    Demonstrates that arguments passed by reference
	are SHARED variables. Any change to the variable's
        value in the function affects the variable's
	value in main as well.  It's the SAME variable,
        even though the calling and called functions
        may (or may not) use different names for it.
     -----------------------------------------------
*/

#include <iostream.h>
int double_value ( int & i )
{
 /*   This function uses type conversion to add its arguments
    and return the sum. It illustrates the locality of the
    variables in its argument list.
  */

	cout << "\n ---------------------------- ";
	cout << "\n  Beginning function \"double_value( int& i )\"";
	cout << "\n ---------------------------- ";

	cout << "\n i = " << i  ;

        cout << "\n Doubling...\n";
        i += i;
	   
	cout << "\n Now i = " << i  ;
 
	cout << "\n ---------------------------- ";
        cout << "\n End function \"double_value()\"";
	cout << "\n ----------------------------\n ";
        return i;
}
int main(void) 
{
   int   imain = 5; 

	cout << "\n ----------------------------\n ";
	cout <<    " Beginning function \"main()\"";
	cout << "\n ---------------------------- ";

	cout << "\n imain = " << imain ;

//      cout << "\n ----------------------------\n "
	cout << " \n Calling function: double_value( imain )";
//       cout << "\n ---------------------------- ";


 	int jmain = double_value( imain );
	   
	cout << " ----------------------------\n "
	     <<    " Back in function \"main()\"";
	cout << "\n ---------------------------- ";

	cout << "\n returned value (2*imain)  = " << jmain;
	cout << "\n Now imain = " << imain ;
 
        cout << "\n Note: the value of imain is changed "
	     << "\n by the call to \"double_value().\" \n ";

        return 0;
}