From: Steven Sharp (sharp@cadence.com)
Date: Wed Jan 19 2000 - 17:03:33 PST
> What drive strength gets used in this case?
> Is (strong1, strong0) assumed?
Yes, just as in any continuous assignment without a drive strength specified.
> If so, then is there any value in declaring the net a trireg, since all of
> the charge strengths are weaker than strong?
Yes. In fact, what doesn't make sense is declaring a drive strength weaker
than the charge strength, since such a driver could not affect the trireg's
value. The trireg has value because the driver could go to z. E.g.
trireg y = en ? a : 1'bz;
> What if the net also has a continuous assignment on it?
You mean another continuous assignment? It already has one continuous
assignment, since that's what the assignment on the declaration is. If
there is another continuous assignment, the result will be produced by
a resolution function that takes in all of the driver outputs and the
trireg capacitive value.
> This started out as an innocuous observation about the BNF. Mostly, I
> wanted to know if this was an error in the BNF or something which was
> intentionally added to it, since it isn't in the 1995 BNF. Perhaps we
> can discuss this during the BTF telecon.
I provided Cliff with some feedback on the BNF based on what is actually
accepted by Verilog-XL. A trireg declaration with an assignment can set
a drive strength, but not a charge strength. A trireg declaration without
an assignment can set a charge strength and doesn't need a drive strength.
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