Const function in class can change the member value?

Question

In the code below, why function SymmetricAxis can change x and y in p3? I think that const function will not allowed to change member's value. But it does, so I am confused. Besides, if I change p3 to const CPoint p3, the compiler do not allow me to do this. But if p3 is not const, the program can change the member in p3.

#include<iostream>
#include<math.h>
using namespace std;

class CPoint
{
private:
    double x; 
    double y; 
public:
    CPoint(double xx = 0, double yy = 0) : x(xx), y(yy) {};
    double Distance(CPoint p) const;  
    double Distance0() const;         
    CPoint SymmetricAxis(char style) const;
    void input();  
    void output(); 
};
void CPoint::input(){
    cout << "Please enter point location: x y" << endl;
    cin >> x >> y;
}
void CPoint::output(){
    cout << "X of point is: " << x << endl << "Y of point is: " << y << endl; 
}
CPoint CPoint::SymmetricAxis(char style) const{
    CPoint p1;
    switch (style){
        case 'x':
            p1.y = -y;
            break;
        case 'y':
            p1.x = -x;
        case '0':
            p1.x = -x;
            p1.y = -y;
            break;
    }
    return p1;
}
int main(){
    CPoint p1, p2(1, 10), p3(1,10);
    p1.input();
    p1.output();
    p3 = p1.SymmetricAxis('0');
    p3.output();
    return 0;
}

Answer 1

SymmetricAxis does not change the value of p3 . SymmetricAxis merely returns a new CPoint as an unnamed temporary value. (That temporary value is initialized by the local variable p1 in the body of SymmetricAxis .)

The copy-assignment operator copies this temporary value over the value of p3.

The const qualifier on SymmetricAxis only means that the call p1.SymmetricAxis('0') will not change p1 . It says nothing about what you assign the result of that call to.

_{(Implementation / optimization note: The compiler is allowed to optimize away one or more of these copies, but the meaning of const in this context presumes these copies happen.)}

Answer 2

You're changing variables on a local inside the function, not any of the member variables. If you for example write this->y = 0 , you will get a compile error. The const qualifier only promises not to change *this .

To clarify, *this refers to p1 (which you call the function on.) You create a local variable also called p1 which you are allowed to modify (because it's not the same this ). p3 does not come into play at all.