I am working on const-correctness of my code and just wondered why this code compiles:
class X
{
int x;
int& y;
public:
X(int& _y):y(_y)
{
}
void f(int& newY) const
{
//x = 3; would not work, that's fine
y = newY; //does compile. Why?
}
};
int main(int argc, char **argv)
{
int i1=0, i2=0;
X myX(i1);
myX.f(i2);
...
}
As far as I understand, f() is changing the object myX, although it says to be const. How can I ensure my compiler complains when I do assign to y? (Visual C++ 2008)
Thank a lot!
1
Because you are not changing any variable in X. Actually, you are changing _y which is an outsider with respect to your class. Don’t forget that:
y = newY;
Is assigning the value of newY to the variable pointed by y, but not the references them selves. Only on initialization the references are considered.
4
The situation is similar to pointer members. In a const member function, the const applies to the pointer itself, not the pointee.
It’s the difference between:
X* const //this is how the const applies: you can modify the pointee
const X*
Except X& const isn’t valid syntax, since the reference can’t be made to refer to another object in the first place (they are implicitly always const). In conclusion: const on methods has no effect on member references.
As additional information for the accepted answer, I want to say, in fact one can
change variables in X.
Because you are not changing any variable in X.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
class Bar {
public:
void funcA() const {
c++;
}
int b = 3;
int &c = b;
};
int main()
{
Bar b;
b.funcA();
cout << b.b << endl; //4
}
So the main idea for this problem is:
It modifies what the member refers to, it does not modify the member.
This also applies for pointer members.
See here Why can reference members be modified by const member functions?