Re: Providing a no-overhead way for a contained class to access its container?

From:
PeteOlcott <PeteOlcott@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 17 Jun 2008 11:03:38 -0700 (PDT)
Message-ID:
<2cd8a3eb-0696-46ce-bec5-7cbcbde888e9@8g2000hse.googlegroups.com>
On Jun 17, 9:45 am, Puppet_Sock <puppet_s...@hotmail.com> wrote:

On Jun 17, 8:30 am, PeteOlcott <PeteOlc...@gmail.com> wrote:
[snips]

What about the case where the contained class must store its data in
its container?


It's not even a little clear what you mean.

Do you mean: The objects in the container stick copies
of themselves into the same container? If so, then you
have a pathological design that should be refactored.

Do you mean: Actual copies of the objects are stored in
the container, as opposed to pointers to the objects?
If that's the issue, and you are concerned about such
things as excess effort involved in swapping copies
instead of pointers, there are many ways of proceeding.

Or do you mean something else?
Socks


Here is what I mean. For this specific use it is about the highest
performance (space and time) possible. A Large number of Contained
elements can be read in and written to disk as a single block read or
block write. Also all of the extra memory allocation overhead that
would normally be associated with the individual Contained elements
has been reduced to making two large allocations.

#define uint8 unsigned char
#define uint32 unsigned int

ContainedClass {
  uint32 Offset;
  bool operator<(const ContainedClass& CC);
}

ContainerClass {
   uint32 Length; // All ContainedClass elements are the same Length
   std::vector<uint8> Bytes;
   std::vector<ContainedClass> Contained;
   uint8& operator[](uint32 N){ return Bytes[N]; };
   void sort(){ std::sort(Contained.begin(), Contained.end()) };
} Container;

bool ContainedClass::operator<(const ContainedClass& CC)
{
  uint32 Last = this->Offset + Container.Length;
  for (int N = this->Offset, M = CC.Offset; N < Last; N++, M++)
    if (Container[N] < Container[M])
      return true;
    else if (Container[N] > Container[M])
      return false;
return false; // They must be Equal, thus Not(LessThan)
}

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