Thank you Russell. You clearly explained what I was asking about.
While I was pondering with the situation, I came up with similar
solutions to yours including overloaded assignment operator which
doesn't assign the constant variable.
So is this the feature of the language which is meant to be like this?
I see that alternative solutions can be formed but what I am trying to
learn here the features of the language and possible situations where
the features can be useful. With regard to assignment of objects with
a constant data member,
I guess that's how it is designed to be.
For example,
class Person {
const unsigned reg_number;
char *name;
Date birthday;
};
If the application requires the registration number to be set just
once and to be made constant after that, the assignment operator
should not be allowed, I guess and the feature seems to make sense...
"Russell Hanneken" <rhanneken@pobox.com> wrote in message news:<n_v_a.3715$Nf3.779@newsread4.news.pas.earthl ink.net>...[color=blue]
> "CoolPint" <coolpint@yahoo.co.uk> wrote in message
> news:159a5c32.0308130958.4433f799@posting.google.c om...[color=green]
> > It seems to me that I cannot assign objects of a class which has a
> > constant data member since the data member cannot be changed once the
> > constructor calls are completed. Is this the way it is meant to be? Am I
> > not suppose not to have any constant data member if I am going to have the
> > assignment operator working for the class?
> >
> > Or am I missing something here and there is something I need to learn
> > about?[/color]
>
> You mean, you have a class like this
>
> class Foo
> {
> const int unchangeableValue;
> int changeableValue;
> public:
> Foo (int, int);
> };
>
> Foo::Foo (int value1, int value2) :
> unchangeableValue(value1), changeableValue(value2)
> {
> }
>
> and you're trying to do an assignment like this?
>
> Foo foo1(1, 2);
> Foo foo2(3, 4);
> foo1 = foo2;
>
> This is called "copy assignment," because both objects in the assignment are
> of the same type. The default method used for copy assignment is called
> "memberwise copy." In this method, each member of one object is simply
> assigned to the corresponding member of the other object. So this statement
>
> foo1 = foo2;
>
> leads to code that assigns foo2.unchangeableValue to foo1.unchangeableValue,
> and assigns foo2.changeableValue to foo1.changeableValue. The problem, of
> course, is that you can't assign foo2.unchangeableValue to
> foo1.unchangeableValue, because unchangeableValue is defined to be const.
>
> One solution would be to define unchangeableValue to be non-const. But that
> may not be what you want.
>
> Another solution might be to define unchangeableValue as a static member of
> Foo. Static members aren't involved in copy assignment. But the side
> effect would be that every instance of Foo would have the same
> unchangeableValue. This may or may not be acceptable to you.
>
> Another solution would be to "overload" the copy assignment operator for
> Foo. In other words, substitute your own copy-assignment operation for the
> default memberwise-copy operation.
>
> You overload the assignment operator by adding a new member function to Foo.
> The name of this function is "operator=":
>
> class Foo
> {
> int const unchangeableValue;
> int changeableValue;
> public:
> Foo (int, int);
> Foo &operator= (Foo const &);
> };
>
> // constructor
> Foo::Foo (int value1, int value2) :
> unchangeableValue(value1), changeableValue(value2)
> {
> }
>
> // copy assignment
> Foo &Foo::operator= (Foo const &other)
> {
> // we're ignoring unchangeableValue
> changeableValue = other.changeableValue;
> return *this;
> }
>
> Now this statement
>
> foo1 = foo2;
>
> leads to this function call:
>
> foo1.operator=(foo2);
>
> Since operator= doesn't attempt to assign to unchangeableValue, it's legal.
>
> Hope that helps. Let us know if you have further questions.
>
> Regards,
>
> Russell Hanneken
>
rhanneken@pobox.com[/color]
