copy constructor of parent clas

[Kench]

Sorry if this becomes a repost. I posted this to comp.lang.c++.moderated 1
hour ago still it does not show up there so posting this here.

Hi,
Consider class A & B both of which implement a copy constructor.
class B inherits from A.
When copy constructor of B is called, first the constructor of A gets called
My question is that, why the copy constructor for A not called.
Is there a way to have it call the copy constructor of class A?

#include <iostream>
using namespace std;


class A
{
public:
A() { cout << "A constructor called\n";};
A(A& a) { cout << "Copy A constructor called\n";};
~A() { cout << "Destructor A called\n";};
};
class B: public A
{
public:
B() { cout << "B constructor called\n";};
B(B& b) { cout << "Copy B constructor called\n"; };
~B() { cout << "Destructor B called\n";};
};
int main(void)
{
B b;
B bb = B(b);
return 0;
}

 

[Victor Bazarov]

Sorry if this becomes a repost. I posted this to comp.lang.c++.moderated 1
hour ago still it does not show up there so posting this here.

It takes moderators up to a day to get through all posts in a moderated
newsgroup, usually. You should not expect a chat room performance from
the Usenet.

Consider class A & B both of which implement a copy constructor.
class B inherits from A.
When copy constructor of B is called, first the constructor of A gets called

Not unless you specifically omit initialising the base class object.

My question is that, why the copy constructor for A not called.

You didn't initialise the base class. It calls the default c-tor.

Is there a way to have it call the copy constructor of class A?

Of course. See below.

#include <iostream>
using namespace std;


class A
{
public:
A() { cout << "A constructor called\n";};

First of all, lose all the semicolons after function bodies. It
just looks so amateurish and sloppy...

A(A& a) { cout << "Copy A constructor called\n";};
~A() { cout << "Destructor A called\n";};
};
class B: public A
{
public:
B() { cout << "B constructor called\n";};
B(B& b) { cout << "Copy B constructor called\n"; };

You need to say

B(B& b) : A(b) { ...

to make sure that the 'A' part is copy-constructed.

~B() { cout << "Destructor B called\n";};
};
int main(void)
{
B b;
B bb = B(b);
return 0;
}

Victor

 

[Mark A. Gibbs]

Consider class A & B both of which implement a copy constructor.
class B inherits from A.
When copy constructor of B is called, first the constructor of A gets called
My question is that, why the copy constructor for A not called.
Is there a way to have it call the copy constructor of class A?

#include <iostream>
using namespace std;


class A
{
public:
A() { cout << "A constructor called\n";};
A(A& a) { cout << "Copy A constructor called\n";};
~A() { cout << "Destructor A called\n";};
};
class B: public A
{
public:
B() { cout << "B constructor called\n";};
B(B& b) { cout << "Copy B constructor called\n"; };

B(B& b) : A(b) { cout << "Copy B constructor called\n"; };

~B() { cout << "Destructor B called\n";};
};
int main(void)
{
B b;
B bb = B(b);
return 0;
}

mark

 

[Mark A. Gibbs]

Incidently, the "B bb = B(b)" is a little ugly (and tough to say :/). Go
with "B bb = b", or better yet "B bb(b)".

mark

 

[Kench]

It takes moderators up to a day to get through all posts in a moderated
newsgroup, usually. You should not expect a chat room performance from
the Usenet.
called

Not unless you specifically omit initialising the base class object.


You didn't initialise the base class. It calls the default c-tor.


Of course. See below.


First of all, lose all the semicolons after function bodies. It
just looks so amateurish and sloppy...


You need to say

B(B& b) : A(b) { ...

to make sure that the 'A' part is copy-constructed.


Victor

that one worked..thanks!