Shalom, It is true that matching is stronger than equivalence. All packed types are equivalent to each other this is the reason why they do not appear in section 6.9.2 (2). The ex only illustrates unpacked structs. The rule for matching types is the same for anonymous unpacked or packed structs or enums. The ex only illustrates packed anonymous structs in 6.9.1 (3) Francoise ' -----Original Message----- From: owner-sv-bc@eda.org [mailto:owner-sv-bc@eda.org] On Behalf Of Bresticker, Shalom Sent: Monday, November 21, 2005 10:10 AM To: sv-bc@eda.org Subject: [sv-bc] 6.9.2 Equivalent types - question 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 IntelReceived on Mon Nov 21 07:45:23 2005
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