copy constructor question

[Ilia Poliakov]

#include <iostream>

using namespace std;

class B
{
public:
B() {};
B(const B&)
{
cout << "Constructor 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]

"Ilia Poliakov" <ip*****@allesklar.de> wrote in message news:bi************@ID-120622.news.uni-berlin.de...

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

This isn't construction it's assignment!
void main()
{ Main must return int!

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

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]

"Ilia Poliakov" <ip*****@allesklar.de> wrote in message
news:bi************@ID-120622.news.uni-berlin.de...

struct A
{
B m_oB;

public:
A(B& rB)
{
cout << "test1" << endl;
m_oB = rB;
// This calls operator=, not a constructor.

cout << "test2" << endl;

B oTestB = rB;
// This calls the copy constructor, not operator=.
};

~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?

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]

"Ilia Poliakov" <ip*****@allesklar.de> wrote in message
news:bi************@ID-120622.news.uni-berlin.de...

#include <iostream>

using namespace std;

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

struct A
{
B m_oB;

public:
A(B& rB)
{
cout << "test1" << endl;
m_oB = rB;
// Change it like following calls copy constractor of B
// B oB = rB;
// or B 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

[Kevin Goodsell]

Ilia Poliakov wrote:

#include <iostream>

using namespace std;

class B
{
public:
B() {};
Get rid of these semi-colons at the end of functions.
B(const B&)
{
cout << "Constructor called for B" << endl;
};
~B() {};
};

struct A
{
B m_oB;

public:
A(B& rB)
{
cout << "test1" << endl;
m_oB = rB;
Copy assignment. No problem.

cout << "test2" << endl;

B oTestB = rB;
Copy construction. No problem.
};

~A() {};
};

void main()
void is not and never has been an acceptable return type for main. main
has always been required to return int.
{
B b;
A a(b);
}

Output:

test1
test2
Constructor called for B

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

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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