To demonstrate my problem let's look at the following sample:
#include "stdafx.h"
#include <iostream>
using namespace std;
class MyClass {
public:
long double x, y;
MyClass(const long double &xx = 0, const long double &yy = 0);
long double distance(const MyClass &b) {
return sqrt((x - b.x)*(x - b.x) + (y - b.y)*(y - b.y));
}
};
MyClass::MyClass(const long double &xx, const long double &yy) {
x = xx; y = yy;
}
void WriteDistance(const MyClass &a, const MyClass &b) {
cout << a.distance(b) << endl;
}
int main()
{
MyClass a = MyClass(2., 3.);
MyClass b = MyClass(3., 4.);
cout << a.distance(b) << endl;
return 0;
}
There is a class MyClass and there is a class function distance which takes one MyClass variable and returns the distance between an existing point and the argument point.
The problem is: in the main(), the function works (gives no error). However there are errors in the WriteDistance() function, that reads: the object has type qualifiers that are not compatible with the member function "MyClass::distance" and 'long double MyClass::distance(const MyClass &)': cannot convert 'this' pointer from 'const MyClass' to 'MyClass &'.
If I overload the distance function (to take not only one MyClass object, but perhaps also two long doubles, just for the comfort of availability), the error reads: no instance of overloaded function "MyClass::distance" matches the argument list and object (the object has type qualifiers that prevent a match) and 'MyClass::distance': 2 overloads have no legal conversion for 'this' pointer.
The question is: why this error occurs and how to prevent it? I found out that not making the MyClass &a const (so deleting "const") gets rid of the error. But why? Members of this site told me countless times to always pass by const reference to prevent copying of objects. If I cannot pass by const reference, does that mean that my function WriteDistance changes the a object in some way in the process? Is there some workaround to being able to having it really const?
