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

"part selects, cast operators, assignment operators, assignment patterns
and maybe others"

An assignment pattern doesn't have a self-determined type.  See 10.9 in
Draft 4.

-- Brad

-----Original Message-----
From: owner-sv-bc@eda.org [mailto:owner-sv-bc@eda.org] On Behalf Of
Feldman, Yulik
Sent: Tuesday, October 16, 2007 2:47 AM
To: Steven Sharp; Greg.Jaxon@synopsys.COM
Cc: sv-bc@eda-stds.org; gordonv@model.com
Subject: RE: [sv-bc] confusion in determining the type of an self
determined binary expression during evalution of type operator

I would say that you need to normalize only if the types of then- and
else- expressions do not match. If the types are matching (or "same"),
it would be more useful to let the conditional operator return this
type, rather than to artificially normalize it. This is especially true
for non-integral types.

Aside conditional operators and identifiers, mentioned in this mail
thread, other expressions that may (should be defined as able to) have a
type other than a simple scalar or a "normalized" one-dimensional vector
are part selects, cast operators, assignment operators, assignment
patterns and maybe others.

--Yulik.

-----Original Message-----
From: owner-sv-bc@server.eda.org [mailto:owner-sv-bc@server.eda.org] On
Behalf Of Steven Sharp
Sent: Tuesday, October 16, 2007 2:17 AM
To: sharp@cadence.com; Greg.Jaxon@synopsys.com
Cc: sv-bc@server.eda-stds.org; gordonv@model.com
Subject: Re: [sv-bc] confusion in determining the type of an self
determined binary expression during evalution of type operator


>There are lots of hard cases hereabouts:  e.g.,  c ? a : a[4:5]

Since you have to apply the LRM rules to determine the width of the
resulting value of ?:, and convert both operands to that width, it seems
clear to me that you have to normalize.  The result of (1 ? a : b) may
not have the type of a.

Steven Sharp
sharp@cadence.com


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