Re: Am I misusing std::vector?

From:
"kanze" <kanze@gabi-soft.fr>
Newsgroups:
comp.lang.c++.moderated
Date:
11 May 2006 18:57:29 -0400
Message-ID:
<1147335995.386152.35690@j73g2000cwa.googlegroups.com>
loose AT astron DOT nl wrote:

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.


Correct. Using the copy constructor, of course.

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?


Yes, something fundamental. Your class A uses dynamic memory.
As such, it is almost certain that the default copy constructor,
assignment operator and destructor do not work correctly. You
must provide one which does. (The default copy constructor uses
member by member copy. Which means that it copies pointers, and
not the object pointed to.)

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

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


This class is simply incorrect. There are several ways to fix
it:

 -- The most obvious one is to declare a private copy
    constructor and assignment operator, and not to provide an
    implementation of them. This inhibits the generation of the
    compiler defaults (which don't work). Of course, it also
    means that you cannot have standard containers of the
    class:-).

 -- Failing that, you can provide a copy constructor and an
    assignment operator which does deep copy, e.g.:

        A::A( A const& other )
            : ip( new int( *other.ip )
        {
        }

        A&
        A::operator=( A const& other )
        {
            int* tmp = new int( *other.ip ) ;
            delete ip ;
            ip = tmp ;
            return *this ;
        }

    (Note that for reasons of exception safety, and to support
    assignment to self, you must allocate and construct the new
    object before deleting the old.)

 -- Since deep copy can be very expensive, you may prefer using
    some sort of shared representation. How to do this best
    depends largely on the application: if the pointed to object
    is immutable, boost::shared_ptr is indicated; if it rarely
    changes, then you can also use boost::shared_ptr to
    implement a copy on write strategy (but attention: this is
    trickier that it looks, especially in a multi-threaded
    environment). If most objects will end up getting changed,
    deep copy is probably a better solution, but in a more
    complex object, it might be possible to share some parts of
    the representation, and use deep copy for others.

 -- Finally, you can choose reference semantics -- in this case,
    just suppress your destructor, and if you're not using
    garbage collection, replace the pointer with
    boost::shared_ptr (but it is a lot easier if you use garbage
    collection). The problem with this solution is that C++
    value oriented objects don't normally use reference
    semantics; there would be a mismatch between what you are
    doing, and what most experienced C++ programmers would
    expect.

Note that whatever you do, you have to reflect it in the
destructor as well. Currently, your destructor supposes deep
copy. Since the copy constructor and the assignment operator
only provide shallow copy, your class is unusable. Even in the
case of something as simple as:

    {
        A a1, a2 ;
        a1 = a2 ;
    }

one of the allocated blocks is freed twice (undefined behavior
-- which may or may not work in simple code, but will almost
certainly cause problems is larger code), and the other one is
never freed.

int main()
{
  const unsigned sz = 1000000;
  vector<A> v;
  v.resize(sz);


Attention: this is semantically (and often literally) the same
as:
    v.resize( sz, A() ) ;
You construct one object, and copy it sz times, using the copy
constructor. In your case, you now have sz copies of the
pointer -- which obviously isn't what you want.

  *v[sz-1].ip = -77;


And of course, every single pointer in the array points to the
block you've just modified.

  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 of neglecting the copy constructor and the
assignment operator is a frequent one for beginners. It should
be covered in you C++ tutorial text. If it's not, or for that
matter, even if it is, I'd strongly recommend your getting a
copy of Scott Meyers' book "Effective C++". I think he's
updated it considerably since it first came out, but even if
not, I can't think of anything in it which wouldn't still hold
today. (If he hasn't updated it, of course, then it won't cover
the issues of exception safety -- nor the importance of fully
constructing all of the new subcomponents before deleting the
old ones in the assignment operator, since that is related to
exception safety. But it's still worth reading.)

--
James Kanze GABI Software
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