RE: [sv-bc] 6.9.2 Equivalent types - question

OK, let me clarify my comments.

The structure of each sub-section of 6.9 is to say that each level of
compatibility includes everything included in the preceding level, plus
a list of additional cases.

In particular, 6.9.2 (1) says, "If two types match, they are
equivalent.".

This IMPLIES, or at least, gives the impression, that 6.9.2 (2)
describes a case not included in (1), i.e., NON-matching.

This implication/impression is strengthened by the fact that the example
in 6.9.1 (3) is of PACKED structs, whereas the text of 6.9.2 (2) and its
example are of UNPACKED structs.

The combination of these facts gives a strong impression that 6.9.1(3)
was intended to relate only to packed structs and unions.

Since both Francoise and Greg seem to agree that 6.9.1(3) does indeed
apply to unpacked as well as packed structs and unions,
6.9.2(2) should be deleted and the example of 6.9.1(3) should be of
unpacked structs.

Thanks,
Shalom 

>-----Original Message-----
>From: Greg Jaxon [mailto:Greg.Jaxon@synopsys.com]
>Sent: Monday, November 21, 2005 8:27 PM
>To: Bresticker, Shalom
>Cc: sv-bc@eda.org
>Subject: Re: [sv-bc] 6.9.2 Equivalent types - question
>
>The relationship between Matching and Equivalence here could
>be made clearer by changing 6.9.2(2) to say
>
>"2) Anonymous enum, unpacked struct, or unpacked union types
>     are Equivalent if and only if they are Matching."
>
>Shalom is right that for these cases, Equivalence is /not/
>weaker
>than Matching.  These are the types for which our goal was to
>provide "strong type checking", so for these, equivalence makes
>no compromises and becomes as strong as exact matching.
>
>Greg Jaxon
>Synopsys DCSV
>
>Bresticker, Shalom wrote:
>> Hi,
>>
>> I have a question on Matching types vs. Equivalent types.
>>
>> 6.9.1(3) says,
>>
>> "3) An anonymous enum, struct, or union type matches itself
>among data objects declared within the same declaration
>statement and no other data types.
>>
>> struct packed {int A; int B;} AB1, AB2;// AB1, AB2 have
>matching types
>> struct packed {int A; int B;} AB3;     // the type of AB3
>does not match the type of AB1"
>>
>>
>> 6.9.2(2) says,
>>
>> "2) An anonymous enum, unpacked struct, or unpacked union
>type is equivalent to itself among data objects declared within
>the same declaration statement and no other data types.
>>
>> struct {int A; int B;} AB1, AB2; // AB1, AB2 have equivalent
>types
>> struct {int A; int B;} AB3;      // AB3 is not type
>equivalent to AB1"
>>
>>
>> If we look at the two texts, it seems that the case described
>by 6.9.2(2) is included in 6.9.1(3). 6.9.2(2) relates only to
>unpacked structs and unions (ignoring enums for now), whereas
>6.9.1(3) makes no distinction between unpacked and packed
>structs and unions.
>>
>> The example of 6.9.1(3) is of packed structs, but the text
>makes no such restriction.
>>
>> Since matching is stronger than equivalence, this seems
>strange.
>>
>> Comments?
>>
>> Thanks,
>> Shalom
>>
>>
>> Shalom Bresticker
>> Intel Jerusalem LAD DA
>> +972 2 589-6852
>> +972 54 721-1033
>> I don't represent Intel
>>
>>