[xsd-users] Re: Copying objects

[Mathew Benson]

I got it.  I think we were saying the same thing, you are just able to state it better.  The problem is when I created an object with the new operator, it didn't belong to the DOM.  I found a better solution.

Sent from my iPhone

> On Mar 24, 2014, at 9:43 AM, Boris Kolpackov <boris at codesynthesis.com> wrote:
> 
> Hi Mathew,
> 
> You really have a knack for overcomplicating things. I am typically
> lost at about the second sentence of your email ;-).
> 
> Here is how the ID/IDREF stuff works in C++/Tree. The resolution
> of the IDREF is dynamic. That is, every time you call get() or
> operator->(), the idref class searches the object model that it
> belongs to for the matching ID.
> 
> In particular, notice this phrase in the above sentence: "searches
> the object model that it belongs to". That's the important part
> to understand and that I believe is the key to your struggles.
> 
> Let me try to illustrate this with an example. Say we have type
> Root that has two elements of type Foo and Bar. Foo has the ID
> attribute (let's call it id) and Bar has the IDREF attribute
> (let's call it ref). Here is some code that illustrates what
> I am talking about:
> 
> Bar b;
> b.ref ("a");
> 
> // b.ref is NULL.
> 
> Foo f;
> f.id ("a");
> 
> // b.ref is still NULL since f and b don't belong to the same model.
> 
> Root r;
> r.bar (b);
> 
> // r.bar.ref is still NULL.
> 
> r.foo (f);
> 
> // r.bar.ref now points to r.foo. BUT: b.ref is still NULL.
> 
> Bar b1 (r.bar ());
> 
> // b1.ref is NULL since it does not belong to r (a copy).
> 
> Bar& b2 (r.bar ());
> 
> // b2.ref is not NULL.
> 
> r.foo ().id ("b");
> 
> // r.bar.ref is now NULL since there is no ID "a" in the object model.
> 
> Boris