Streaming Multiple Objects into One Document
But what if you really only want one context? What if you want the second instance of
yellow_cat in the stream to simply refer to the first one, thus preserving the fact that they are really the same object. One way to do this is to make use of the same mechanism employed by the streams to create these contexts in the first place. You do this by using the guard classes
RWWithObjectOutputContext (for output) and
RWWithObjectInputContext (for input). By creating these guards, you effectively create your own master context for resolution of object references.
To get both instances of streaming yellow_cat in the same context, all you have to do is this:
// examples\serial\advanced\catanddog.cpp
{ // 1
RWWithObjectOutputContext context(out); // 2
out << yellow_cat << blue_dog << yellow_cat; // 3
}
out.openContext();
out << yellow_cat << blue_dog << yellow_cat;
out.closeContext();
But using the guard is a safer.
So now you have a single context containing all pets. The second time you stream out yellow_cat, the stream will contain a reference to the first yellow_cat rather than a whole new copy.
The code for input needs to match the code for output:
{
dog* blue_dog2;
cat* yellow_cat2;
cat* yellow_cat3;
RWWithObjectInputContext(in);
in << yellow_cat2 << blue_dog2 << yellow_cat3;
}
Now you have your pet collection. Both pointers yellow_cat2 and yellow_cat3 point to the same cat, as they should.