I'm learning C++ from C++ Primer Plus, 5th Edition, and have come across the following problem.

The following class declaration is given as part of the question:

/*
 Name: move.h
 Copyright: 
 Author: Steven Taylor
 Date: 31/01/08 15:05
 Description: Class 'Move' header file. C++ Primer Plus, Ch.10,Prog. Ex 6.
*/

#ifndef MOVE_H_
#define MOVE_H_

class Move
{
  private:
    double x;
    double y;
  public:
    Move(double a = 0, double b = 0); // sets x, y to a, b
    void showmove() const;          // shows current x, y values
    Move add(const Move &m) const;        
// this function adds x of m to x of invoking object to get new x,
// adds y of m to y of invoking object to get new y, creates a new
// move object initialized to new x, y values and returns it
    void reset(double a = 0, double b = 0);      // resets x,y to a, b
};
#endif

The question continues. Create member function definitions and a program that exercises the class.

The gist of my question surrounds the prototype/function Move add(const Move &m) const; . I keep getting a compiler error that indicates 29 ...move.cpp assignment of data-member `Move::x' in read-only structure . When I remove the const reference at the end of the add function, the program compiles successfully and the function changes the private variables. If leaving that const reference in place, as per my book, is there another way to modify the private variables.

This is what I've done so far:

/*
  Name: move.cpp
  Copyright: 
  Author: Steven Taylor
  Date: 31/01/08 15:14
  Description: Class methods for move. C++ Primer Plus, Ch.10. PE 5. Page 498
*/
#include <iostream>
#include "move.h"
using std::cout;
using std::endl;
Move::Move(double a, double b)
{
    x = a;
    y = b;
}
void Move::showmove() const
{
    cout << "x = " << x << "  y = " << y << endl;
}
Move Move::add(const Move &m) const 
{
    //double c, d;
    //c = x + m.x;
    //d = y + m.y;
    //x = c;
    //y = d;
        
    x += m.x;
    y += m.y;
    Move *temp = new Move(x, y);
    return *temp;
}
void Move::reset(double a, double b)
{
    x = y = 0;
}

and the main program;

/*
  Name: main.cpp
  Copyright: 
  Author: Steven Taylor
  Date: 31/01/08 15:27
  Description: C++ Primer Plus, Ch. 10, Prog. Ex. 6, page 498.
*/

#include <iostream>
#include "move.h"

int main()
{
    using std::cout;
    using std::cin;
    
    cout << "Move one\n";
    Move one;
    one.showmove();
    
    cout << "\n\nMove two\n";
    Move two(101.5,55.6);
    two.showmove();
    
    cout << "\n\nMove three\n";
    Move three = one.add(two);
    three.showmove();
    
    cout << "\n\nMove one - after 'one.add(two)'\n";
    one.showmove();
    cout << "\n\nMove one - reset\n";
    one.reset();
    one.showmove();
    cout << "\n\nMove two - reset\n";
    two.reset();
    two.showmove();
    cout << "\nMove three - reset\n";
    three.reset();
    three.showmove();
     
    // exit routine
    cout << "\n\n...Press ENTER to Exit System...";
    cin.get();
    return 0;
}

Any advice appreciated.

when you declare a function as const, it implies that you cannot change the state of the object inside that function. that means you cannot change the value of any member variable inside the function and you cannot make a call to any non-const function either. that is y its not allowing you to modify 'x' inside add, however when your remove the const your are able to modify it. it has nothing to do with the variable being private, even if it was public it would not have allowed you to modify it.

Comments
Very helpful.

when you declare a function as const, it implies that you cannot change the state of the object inside that function. that means you cannot change the value of any member variable inside the function and you cannot make a call to any non-const function either. that is y its not allowing you to modify 'x' inside add, however when your remove the const your are able to modify it. it has nothing to do with the variable being private, even if it was public it would not have allowed you to modify it.

That is what I thought. To recap:

Move add(const Move &m) const;        
// this function adds x of m to x of invoking object to get new x,
// adds y of m to y of invoking object to get new y, creates a new
// move object initialized to new x, y values and returns it

The above is part of the question. As you can see the function is required to do two things, 1. set x & y of the invoking object and 2. return a new object. With the prototype as provided it is therefore impossible to alter x & y of the invoking object. Therefore I'm left to assume that the question in the first place is a little suspect.

Thank you all. This is now resolved. I fell victim to not reading the question properly.

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