copy constructor question

[Ilia Poliakov]

#include <iostream>

using namespace std;

class B
{
public:
B() {};
B(const B&)
{
cout << "Constructo r called for B" << endl;
};
~B() {};
};

struct A
{
B m_oB;

public:
A(B& rB)
{
cout << "test1" << endl;
m_oB = rB;

cout << "test2" << endl;

B oTestB = rB;
};

~A() {};
};

void main()
{
B b;
A a(b);
}

Output:

test1
test2
Constructor called for B

Why was not B::B(const B&) constructor called in test1 case?

PS. I work under MSVC7.0

[Ron Natalie]

Re: copy constructor question


"Ilia Poliakov" <ipoliak@allesk lar.de> wrote in message news:bidd4d$7kq 1h$1@ID-120622.news.uni-berlin.de...[color=blue][color=green]
>> A(B& rB)[/color]
> {
> cout << "test1" << endl;
> m_oB = rB;[/color]

This isn't construction it's assignment!
[color=blue]
> void main()
> {[/color]
Main must return int!

[color=blue]
> Why was not B::B(const B&) constructor called in test1 case?
>[/color]
Because you don't construct a B object from another B object.
You construct an A object from a B. That constructor does
an assigment of B to B which calls the implicitly-generated
copy-assignment operator after m_oB was derfault constructed.

If you want to copy construct the member in the other constructor
you need to do this.

A(B& rB) : m_oB(rB) {
}

[Kevin Saff]

Re: copy constructor question


"Ilia Poliakov" <ipoliak@allesk lar.de> wrote in message
news:bidd4d$7kq 1h$1@ID-120622.news.uni-berlin.de...[color=blue]
> struct A
> {
> B m_oB;
>
> public:
> A(B& rB)
> {
> cout << "test1" << endl;
> m_oB = rB;[/color]

// This calls operator=, not a constructor.
[color=blue]
>
> cout << "test2" << endl;
>
> B oTestB = rB;[/color]

// This calls the copy constructor, not operator=.
[color=blue]
> };
>
> ~A() {};
> };
>
> void main()
> {
> B b;
> A a(b);
> }
>
> Output:
>
> test1
> test2
> Constructor called for B
>
> Why was not B::B(const B&) constructor called in test1 case?[/color]

T newT = oldT;

// actually calls:

T newT(oldT);

// if available. However,

T newT;
newT = oldT;

// will call operator=.

// You should define operator= to do what you want.

[Liu Jian]

Re: copy constructor question


"Ilia Poliakov" <ipoliak@allesk lar.de> wrote in message
news:bidd4d$7kq 1h$1@ID-120622.news.uni-berlin.de...[color=blue]
> #include <iostream>
>
> using namespace std;
>
> class B
> {
> public:
> B() {};
> B(const B&)
> {
> cout << "Constructo r called for B" << endl;
> };
> ~B() {};
> };
>
> struct A
> {
> B m_oB;
>
> public:
> A(B& rB)
> {
> cout << "test1" << endl;
> m_oB = rB;[/color]

// Change it like following calls copy constractor of B
// B oB = rB;
// or B oB(rB);

[color=blue]
>
> cout << "test2" << endl;
>
> B oTestB = rB;
> };
>
> ~A() {};
> };
>
> void main()
> {
> B b;
> A a(b);
> }
>
> Output:
>
> test1
> test2
> Constructor called for B
>
> Why was not B::B(const B&) constructor called in test1 case?
>
> PS. I work under MSVC7.0
>
>[/color]

[Kevin Goodsell]

Re: copy constructor question

Ilia Poliakov wrote:
[color=blue]
> #include <iostream>
>
> using namespace std;
>
> class B
> {
> public:
> B() {};[/color]

Get rid of these semi-colons at the end of functions.
[color=blue]
> B(const B&)
> {
> cout << "Constructo r called for B" << endl;
> };
> ~B() {};
> };
>
> struct A
> {
> B m_oB;
>
> public:
> A(B& rB)
> {
> cout << "test1" << endl;
> m_oB = rB;[/color]

Copy assignment. No problem.
[color=blue]
>
> cout << "test2" << endl;
>
> B oTestB = rB;[/color]

Copy construction. No problem.
[color=blue]
> };
>
> ~A() {};
> };
>
> void main()[/color]

void is not and never has been an acceptable return type for main. main
has always been required to return int.
[color=blue]
> {
> B b;
> A a(b);
> }
>
> Output:
>
> test1
> test2
> Constructor called for B
>
> Why was not B::B(const B&) constructor called in test1 case?[/color]

It's not clear why you believe it should have been. No construction
occurred there. Only an assignment to an existing object.

-Kevin
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