Hi,
Considering the fact the given e.g. in LRM (page no. 282, e.g. d)
@(posedge clk) a ##1 @(posedge clk1) s1 |=> @(posedge clk2) s2;
will be parsed as:
@(posedge clk) a ##1 (@(posedge clk1) s1 |=> @(posedge clk2) s2);
So (@(posedge clk1) s1 |=> @(posedge clk2) s2) will become a property expression.
But as per assertion BNF, ## operator can not have property expression in RHS. So the e.g. given in LRM is wrong.
Please confirm.
-- Regards Surya.Received on Tue Mar 7 20:57:37 2006
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