RE: [sv-ac] question on subroutines on empty sequences

Hi Lisa,
 
In the formal semantics appendix
 
(R, v = e) is derived as (R ##0 (1, v = e))
 
Although this rule applies to local var assignment, I would extend it
for match items too.
 
(R, match_item_statement) is derived as (R ##0 (1,
match_item_statement))
 
So if R is empty, match_item_statement should not be executed since 
it becomes (empty ##0 (1, match_item_statement)) which does not result
in match.
 
Tej


________________________________

	From: owner-sv-ac@server.eda.org
[mailto:owner-sv-ac@server.eda.org] On Behalf Of Lisa Piper
	Sent: Wednesday, January 03, 2007 3:41 PM
	To: sv-ac@server.eda-stds.org
	Subject: [sv-ac] question on subroutines on empty sequences
	
	

	I have a question. If there is a "sequence_match_item"
associated with a potentially empty sequence, does it get executed when
the sequence is empty?  Given the example in the std, there could be a
sequence_match_item associated with the potentially empty seq
(potentially empty meaning  a repetition range like [*0:3])  and also a
sequence_match_item associated with the "seq".  Are both executed
simultaneously?

	 

	Starting on page 293 of the document, page 259 of the pdf,  it
states:

	 

	" Using 0 as the repetition number, an empty sequence results,
as follows:

	a [*0]

	 

	An empty sequence is one that does not match over any positive
number of clock ticks. The following rules

	apply for concatenating sequences with empty sequences. An empty
sequence is denoted as empty, and a

	sequence is denoted as seq.

	- (empty ##0 seq) does not result in a match.

	- (seq ##0 empty) does not result in a match.

	- (empty ##n seq), where n is greater than 0, is equivalent to
(##(n-1) seq).

	- (seq ##n empty), where n is greater than 0, is equivalent to
(seq ##(n-1) 'true).

	 

	For example:

	b ##1 ( a[*0] ##0 c)

	produces no match of the sequence.

	 

	b ##1 a[*0:1] ##2 c

	is equivalent to

	(b ##2 c) or (b ##1 a ##2 c)

	 

	Lisa

	 


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