Re: Does push_back() do a copy?

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 28 Jun 2007 13:31:10 +0200

Message-ID:

<f6066l$j7l$1@murdoch.acc.Virginia.EDU>

Angus wrote:

I have a socket class CTestClientSocket which I am using to simulate
load testing.

I create multiple instances of the client like this:

for (int i = 0; i < 5; i++)
{
CTestClientSocket* pTemp = new CTestClientSocket(this, ip, port);
pTemp->Connect();
m_collClients.push_back(pTemp);
// delete pTemp;
}

Note I have commented out the delete pTemp

Then I send some data to a server like this:

// iterate through collection of clients
for (std::vector<CTestClientSocket*>::iterator it =
m_collClients.begin();
it != m_collClients.end(); it++)
{
(*it)->Send(byData);
}

This works if delete pTemp on the Connect is commented out. But I
thought that push_back would make a copy of pTemp.

Yes.

and so I could then delete pTemp.

No: pTemp is a pointer and that is all that is copied. The pointee is _not_
copied. Thus, if you delete pTemp, you loose the pointee and your program
is in deep undefined waters.

But if I delete pTemp I get an memory access
problem in the (*it)->Send(byData);

Right.

And I have to cleanup at the end. Does push_back NOT do a copy? I am
confused, can someone explain.

With pointers, you need to distinguish between the pointer and the pointee.
Both are objects. When used in standard containers, the former is copied,
the later is not.

Best

Kai-Uwe Bux