Re: Am I misusing std::vector?

From:
"Eugene Alterman" <eugalt@verizon.net>
Newsgroups:
comp.lang.c++.moderated
Date:
11 May 2006 08:43:41 -0400
Message-ID:
<8Et8g.85263$dW3.66439@newssvr21.news.prodigy.com>
"loose AT astron DOT nl" <loose@astron.nl> wrote in message
news:1147252936.701110.275280@j73g2000cwa.googlegroups.com...

Hi,

I was surprised by the output of the program below. From what I
understand from the C++ STL documentation, vector<T>::resize(N), will
create N *newly constructed* objects of T. So, I expected *v[0].ip to
be equal to 0, and *v[sz-1].ip equal to -77. However, I discovered,
using a few different compilers (two different versions of gcc, and
icc), that both return -77. It turns out that only *one* object is
being constructed and this one object seems to be bitwise copied to the
other members. See my code below and try it for yourself.
Am I missing something here?

Regards,

Marcel Loose.

<code>
#include <vector>
#include <iostream>
using namespace std;

struct A
{
 A(int i = 0) { ip = new int(i); }
 ~A() { delete ip; }
 int* ip;
};

int main()
{
 const unsigned sz = 1000000;
 vector<A> v;
 v.resize(sz);
 *v[sz-1].ip = -77;
 cout << "v.size() = " << v.size() << endl;
 cout << "v[0].ip = " << v[0].ip << "; " << "*v[0].ip = " <<
*v[0].ip << endl;
 cout << "v[sz-1].ip = " << v[sz-1].ip << "; " << "*v[sz-1].ip = "
<< *v[sz-1].ip << endl;
 return 0;
}
</code>


The problem is with your A struct that does not have user-defined copy
constructor implementing deep copy semantics.

'resize' member function has the following declaration:
void resize(size_type sz, T c = T())

The new vector elements are created as copies of the second (default)
argument. In your case this means that they all get the same value of the ip
pointer member.

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