C / C++ / MFC

22-Sep-08 19:37

Hello everyone,

Just want to confirm a basic concept when reading a couple of documents about diamond problem and virtual inheritance.

1. Diamond problem covers both ambiguity methods and data member variable?

2. Virtual inheritance only solves ambiguity data member issue (making data member one copy)?

3. So, it is not correct to say virtual inheritance solves diamond problem 100% since ambiguity member methods is not covered and solved?

thanks in advance,
George

tony_Udz22-Sep-08 20:07

tony_Udz

22-Sep-08 20:07

Hi,

Try the following code it might help you:

==========================================================================
#include "iostream.h"
#include "stdio.h"

class A
{
public:
int m_nAInt;
A(int n = 1):m_nAInt(n)
{
cout << "Class A Constructor called" << endl;
}

Display()
{
cout << "A::Display()" << endl;
}
};

class B: public virtual A
{
public:
B()
{
cout << "Class B Constructor called" << endl;
}

};

class C: public virtual A
{
public:
C()
{
cout << "Class C Constructor called" << endl;
}

};

class D: public B, C
{
public:
D()
{
cout << "Class D Constructor called" << endl;
}

};

int main(int argc, char* argv[])
{
D dObj;
dObj.Display();
return 0;
}
==========================================================================

George_George22-Sep-08 20:11

22-Sep-08 20:11

What is the issue, tony_udz? Could you clarify please?

You code runs ok without any compile/build warning/errors.

Here is the output.

Class A Constructor called
Class B Constructor called
Class C Constructor called
Class D Constructor called
A::Display()

regards,
George

SandipG 22-Sep-08 20:30

22-Sep-08 20:30

Hi George,

George_George wrote:
You code runs ok without any compile/build warning/errors.

That means virtual resolved methods as well as data member ambiguity so that does not clear your doubt.

I hope it helps..

Regards,
Sandip.

George_George22-Sep-08 20:36

22-Sep-08 20:36

Thanks, Sandip!

So, you mean virtual inheritance makes one copy of both methods and data?

regards,
George

SandipG 22-Sep-08 20:41

22-Sep-08 20:41

George_George wrote:
So, you mean virtual inheritance makes one copy of both methods and data?

Yes unless you do not override the common method in both derived classes.
In above case if you override Display method in both derived classes you will get ambiguity error.

Regards,
Sandip.

George_George22-Sep-08 20:49

22-Sep-08 20:49

Thanks Sandip,

I have tried that when overriding in both classes, there will be compile error regarding the ambiguity issue. But when overriding only in one class, there is no compile error regarding the ambiguity issue, why??

regards,
George

SandipG 22-Sep-08 21:07

22-Sep-08 21:07

George_George wrote:
But when overriding only in one class, there is no compile error regarding the ambiguity issue, why??

In that case i think preference is given to overridden method.
and you can always use scope resolution operator to execute desired method Smile | :)

Regards,
Sandip.

George_George22-Sep-08 21:26

22-Sep-08 21:26

Thanks Sandip,

"In that case i think preference is given to overridden method." -- you mean when calling D.Display, there are two choices,

1. D::B::Display;
2. D::C::A::Display.

But since (1) is "more" overridden than (2), (1) is preferred, but if we derive Display in both B and C, D will have no choice and there will be ambiguity compile error?

regards,
George

SandipG 22-Sep-08 21:30

22-Sep-08 21:30

George_George wrote:
"In that case i think preference is given to overridden method." -- you mean when calling D.Display, there are two choices,

1. D::B::Display;
2. D::C::A::Display.

But since (1) is "more" overridden than (2), (1) is preferred,

Yes from the behavior it seems like that.
Even i did not come across any documents that states above behavior.

Regards,
Sandip.

George_George22-Sep-08 21:34

22-Sep-08 21:34

Thanks Sandip,

Here is what I learned from you -- virtual inheritance solves both member method and member data issue -- so we can say virtual inheritance solves all diamond problem issues (both ambiguity member method and data member). Could you review and confirm my understanding now is correct please? Smile | :)

regards,
George

SandipG 22-Sep-08 21:37

22-Sep-08 21:37

Yes unless there is no overriding of methods from the common ancestor as i said earlier.

Regards,
Sandip.

George_George22-Sep-08 21:39

22-Sep-08 21:39

I agree, thanks Sandip!

regards,
George

Is there any memory leak in this code?

laksh220422-Sep-08 23:03

laksh2204

22-Sep-08 23:03

I have some queries here..

1. By making the inheritance virtual how the ambiguity is actually resolved?
I mean, what this virtual inheritance does internally? Does it create another vptr of another vtable?
if it is so how does the new vtable looks like??

2. And as the diamond problem says that there is an ambiguity that whether D will have path
A->B->D
or
A->C->D

A
/ \
B C
\ /
D

So after removing ambiguity which path is taken?

SRKSHOME22-Sep-08 19:33

SRKSHOME

22-Sep-08 19:33

Hi,
Is there any memory leak in the following code? I cannot see any one. But bound checker telling there is a leak? Please advice me. Thanks in advance.

class test<br />
{<br />
	char* m_name;<br />
public:<br />
	test():m_name(NULL)<br />
	{}<br />
	void Set(const char* name)<br />
	{<br />
		<br />
		if(name!=NULL && strlen(name))<br />
		{<br />
			m_name = new char[strlen(name)+1];<br />
			strcpy(m_name,name);<br />
		}<br />
		else<br />
			delete [] m_name;<br />
	}<br />
	~test()<br />
	{<br />
		Set(NULL);<br />
	}<br />
	void display()<br />
	{<br />
		cout<<m_name<<endl;<br />
	}<br />
};<br />
int main(int argc, char* argv[])<br />
{       test tst;<br />
	tst.Set("123456");<br />
	tst.display();	<br />
	return 0;<br />
}<br />

User 21559722-Sep-08 19:53

User 215597

22-Sep-08 19:53

SRKSHOME wrote:
void Set(const char* name)
{

if(name!=NULL && strlen(name))
{
m_name = new char[strlen(name)+1];
strcpy(m_name,name);
}
else
delete [] m_name;
}

Assuming name is valid
On first call to Set method, m_name is NULL and you allocate memory to it.
On second call to Set method, m_name had valid allocated memory from first call. But prior to allocating new memory, the previous allocated memory is not deallocated

modified 21-Apr-21 21:01pm.

SRKSHOME22-Sep-08 20:49

SRKSHOME

22-Sep-08 20:49

Thanks for your reply. But if you look at the main function I have called set("123456"). I have never called set(..) for second time. Second time this functin is called from destructor and NULL is passed to Set(). Hence I cannot see memory leak here. Please correct me if I am wrong.

User 21559722-Sep-08 21:46

User 215597

22-Sep-08 21:46

I see the above as obvious memory leak in code. Also Sarath (in the reply below) pointed to the same problem. Try fixing the problem and then see if there is still memory leak.

You may also want to re-think about how m_name is deallocated when object lifetime ends.

modified 21-Apr-21 21:01pm.

Sarath C22-Sep-08 20:46

Sarath C

22-Sep-08 20:46

The program will be having memory leak, if you call set twice. You can rewrite the code better.

class test
{
	char* m_name;
public:
	test():m_name(NULL)
	{}
	void Set(const char* name)
	{
		// Leave the data untouched on invalid input
		if( name == NULL || 0 == strlen(name))
			return;

		if( m_name )
			delete []m_name;

		// Validate the pointer after allocating if necessary or put try/catch if exception enabled
		m_name = new char[strlen(name)+1]; 
		strcpy(m_name,name);
	}
	~test()
	{
		if( m_name ) 
                     delete [] m_name;
	}
	void display()
	{
		cout<<m_name<<endl;
	}
};

-Sarath.
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