> 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 ? The rules for implicit conversions will give you the size and the signedness of the type. Since this may not match the type of any operand in the expression, I think you must always regard it as a new type synthesized from that size and signedness. It doesn't matter whether it is a binary operator or a multiple concatenation; it is a generic vector either way. Steven Sharp sharp@cadence.com -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.Received on Mon Oct 15 14:42:52 2007
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