Re: [sv-bc] confusion in determining the type of an self determined binary expression during evalution of type operator

Hi Brad,
               I cannot understanding your reply. I want to know what 
will be the  type of a  self determined binary  or  concatination 
expression when they are used in type operator. Please let me know  
whether  you understand my query or not .

Thanks,
Sumay
Brad Pierce wrote:

>Sumay,
>
>The LRM says that "The type operator applied to an expression shall
>represent the self-determined result type of that expression."  So you
>are asking about how to calculate the self-determined type of an
>expression.  The type operator is just one of many contexts where a
>self-determined context must be calculated.  Two common such contexts in
>Verilog are concatenations and parameter declarations.
>
>-- Brad
>
>-----Original Message-----
>From: owner-sv-bc@eda.org [mailto:owner-sv-bc@eda.org] On Behalf Of
>Sumay Guin
>Sent: Monday, October 15, 2007 2:53 AM
>To: sv-bc@eda-stds.org
>Subject: [sv-bc] confusion in determining the type of an self determined
>binary expression during evalution of type operator
>
>Hi,
>      Consider the usage of type operator,
>      int a,b;
>      byte  c ;
>      b = type(a + 6 )'(c);
>
>      so , what will  be  the type of self determined binary expression
>( here a+6 ) when we try to find out it's type using type operator.
>      Also in case if user specified concatination/multiple
>concatination expression in type operator then what will be the type of
>the
>      concatination/multiple concatination expression ? As LRM does not
>state it clearly, Can someone tell me what would be the type of
>      above expressions.
>
>Thanks ,
>Sumay
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>




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