RE: [sv-ec] $root vs $unit

1. I've checked the LRM and I cannot find any statement forbidding $root.T
 
2.about $unit:: you claim that I can do for example :
 
1st compilation:
  typedef int T;
2nd compilation
  typedef reg T;
  T v; //this will be int?
  $unit::T v; //this will be reg?
 
2nd compilation to the same lib for sure override T from 1st one. What is
the sens of using $unit?
 
 
DANiel

  _____  

From: Bresticker, Shalom [mailto:shalom.bresticker@intel.com] 
Sent: Tuesday, November 06, 2007 12:57 PM
To: danielm; sv-ec@server.eda.org
Subject: RE: [sv-ec] $root vs $unit


I don't think $root.T is legal. 
T exists in $unit, not $root.
Consider what would happen if T were defined in two different compilation
units. $unit always refers to the local compilation unit, but what would
$root.T refer to? There are two different T's.
 
Shalom


  _____  

From: owner-sv-ec@server.eda.org [mailto:owner-sv-ec@server.eda.org] On
Behalf Of danielm
Sent: Tuesday, November 06, 2007 12:43 PM
To: sv-ec@server.eda.org
Subject: [sv-ec] $root vs $unit


What is the diff between $root and $unit ie in below case.
What should be the types of a,b,c.
 
typedef reg T;
module top;
 typedef int T;
 typedef $root.T T1;

 typedef $unit::T T2;
 
 T a;
 T1 b;
 T2 c;
endomdule
 
 
DANiel

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