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Beginner's Question: Assignment of Objects with a constant data member possible?

P: n/a
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?

Clear, easy to understand explanation would be very much appreciated as I am
just beginning to learn C++ features. Thank you in advance!
Jul 19 '05 #1
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5 Replies


P: n/a
"CoolPint" <co******@yahoo.co.uk> wrote...
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?
You can write your own assignment operator and do whatever you think
is appropriate. I guess you need to learn about operator overloading.
Clear, easy to understand explanation would be very much appreciated as I am just beginning to learn C++ features. Thank you in advance!


What book are you reading? Does it have a chapter on operators? Does
it talk about operator overloading for your custom types? If not,
discard it and find a better one.

class HasConstData
{
int const a;
int b;
public:
HasConstData(int b) : a(42), b(b) {}
HasConstData& operator=(HasConstData const& r)
{
b = r.b;
return *this;
}
};

int main()
{
HasConstData hcd1(1), hcd2(2);
hcd1 = hcd2;
}

Victor
Jul 19 '05 #2

P: n/a
CoolPint wrote:
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?

Clear, easy to understand explanation would be very much appreciated as I am
just beginning to learn C++ features. Thank you in advance!


You need to explicitly write the assignment operator and copy
constructor methods for your class.

<example>
class X {
private:
int a_;
const int b_;
public:
// ctor
X(int a, int b) : a_(a), b_(b) { }
// copy ctor
X(const X& x) : a_(x.a_), b_(x.b_) { }
// assignment
X & operator=(const X& x) {
a_ = x.a_;
return *this;
}
};

int main()
{
X x1(1,2), x2(3,4);
X x3(x1); // copy ctor
// x3.{a_==1, b_==2}

// x2.{a_==3, b_==4}
x2 = x3; // assignment
// x2.{a_==1, b_==4}
}
</example>

--
Jim

To reply by email, remove "link" and change "now.here" to "yahoo"
jfischer_link5809{at}now.here.com
Jul 19 '05 #3

P: n/a
"CoolPint" <co******@yahoo.co.uk> wrote in message
news:15**************************@posting.google.c om...
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?


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
rh*******@pobox.com
Jul 19 '05 #4

P: n/a
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" <rh*******@pobox.com> wrote in message news:<n_****************@newsread4.news.pas.earthl ink.net>...
"CoolPint" <co******@yahoo.co.uk> wrote in message
news:15**************************@posting.google.c om...
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?


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
rh*******@pobox.com

Jul 19 '05 #5

P: n/a
CoolPint writes:
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?
Clear, easy to understand explanation would be very much appreciated as I am just beginning to learn C++ features. Thank you in advance!


Based on the amount of time that has elapsed since you made your post, I
think most others are in the same boat I am, they don't understand your
question. How about posting some code that illustrates your
problem/concern/misapprehension?
Jul 19 '05 #6

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