RE: [sv-bc] stream to object hierarchy

By reconstruct, it means that the streaming operation will not re-create
the object collection. Using the previous example as a guide,
reconstruct means that the following would not work - the two contained
objects would *not* be created.

 

Pair r = new();

{<<{r}} = q;

 

Perhaps you think that this is already covered and the sentence is
redundant, but I can't come up with a better word than "re-construct".

 

            Arturo

 

From: Bresticker, Shalom [mailto:shalom.bresticker@intel.com] 
Sent: Monday, April 14, 2008 12:04 AM
To: Arturo Salz; sv-bc
Cc: sv-ec@eda.org
Subject: RE: [sv-bc] stream to object hierarchy

 

Well, originally I did not understand other parts of the sentence
either, but you answered those.

 

Regarding "reconstruct", again you wrote "recreate a collection of
objects that had been previously streamed".

But that is not precise.

In your example, you take the source object and stream it into a queue
of bits.

You could unpack it into a completely different collection of objects.

"Populated" would work for me, but I would still have a problem with
"repopulated".

 

Thanks,

Shalom

 

________________________________

	From: Arturo Salz [mailto:Arturo.Salz@synopsys.com] 
	Sent: Monday, April 14, 2008 9:50 AM
	To: Bresticker, Shalom; Arturo Salz; sv-bc
	Cc: sv-ec@eda.org
	Subject: RE: [sv-bc] stream to object hierarchy

	I now understand your issue.  The term "reconstruct" was used in
its general meaning, that is, to recreate a collection of objects that
had been previously streamed. The term construct was used because
classes have constructors.

	Perhaps re-populate would be a more accurate way to describe
what streaming does - versus construct.

	 

	            Arturo

	 

	From: Bresticker, Shalom [mailto:shalom.bresticker@intel.com] 
	Sent: Sunday, April 13, 2008 7:42 PM
	To: Arturo Salz; sv-bc
	Cc: sv-ec@eda.org
	Subject: RE: [sv-bc] stream to object hierarchy

	 

	Thanks.

	 

	But why is this called "reconstruction"?

	 

	The particular example you gave does a copy.

	 

	But in general, an unpack operation does not have to be doing a
copy.

	 

	Thanks,

	Shalom

		 

________________________________

		From: Arturo Salz [mailto:Arturo.Salz@synopsys.com] 
		Sent: Sunday, April 13, 2008 11:48 PM
		To: Bresticker, Shalom; sv-bc
		Cc: sv-ec@eda.org
		Subject: RE: [sv-bc] stream to object hierarchy

		Shalom,

		 

		That sentence is in the context of this paragraph: 

		When applied to a class handle (i.e., an object), the
streaming operator shall stream the contents of the object, and not the
handle itself. Class items are streamed in declaration order; extended
class items shall fol-low the items of their superclass. Embedded class
handles are streamed as other aggregate types: they are recursively
traversed in depth-first order until reaching integral types.  A null
class handle shall be ignored (not streamed), and a warning may be
issued. Null handles are skipped by both the pack and unpack opera-tors;
therefore, the unpack operation shall not create class objects.

		 

		Hence it conveys two concepts:

		1) Re-emphasizes that last sentence stating that
streaming operations do not create objects.

		2) A stream can be unpacked into a collection of
contained objects, provided the objects exist prior to streaming.

		 

		Here's an example:

		 

		class Number;

		  int  n;

		  function new(int x); n = x; endfunction

		endclass

		 

		class Pair;

		  Number  left;

		  Number  right;

		  int     avg;

		endclass

		 

		Pair p, r;

		bit q[$];

		p = new(); p.left = new(2); p.right = new(4); p.avg = 3;
// create and initialize Pair object

		 

		q = {<<{p}};      // stream object into queue

		 

		r = new(); r.left = new(0); r.right = new(0);
// create other Pair object

		{<<{r}} = q;            // stream queue onto new object
collection - r is a copy of p

		 

		In order for 'r' above to be a copy of p, the two
contained objects (r.left and r.right) must have been created (as done
above).

		 

		            Arturo

		 

		From: owner-sv-bc@eda.org [mailto:owner-sv-bc@eda.org]
On Behalf Of Bresticker, Shalom
		Sent: Sunday, April 13, 2008 1:20 AM
		To: sv-bc
		Subject: [sv-bc] stream to object hierarchy

		 

		Hi, 

		1.4.14 says about the streaming operators, 

		If a particular object hierarchy is to be reconstructed
from a stream, the object hierarchy into which the stream is unpacked
must be created before the streaming operator is applied.

		Can someone explain to me what this means and give an
example? 

		Thanks, 
		Shalom 

		Shalom Bresticker 
		Intel Jerusalem LAD DA 
		+972 2 589-6582 
		+972 54 721-1033 

	
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