C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  adding alignment support n2140 & n2165 inconsistent alignmen          C++Talk.NET Forum Index -> C++ language, library and standards View previous topic :: View next topic   Author Message Larry Guest Posted: Tue May 08, 2007 3:12 am    Post subject: adding alignment support n2140 & n2165 inconsistent alignmen ===================================== MODERATOR'S COMMENT: Apologies for the delay in moderation; this appears to be caused by my ISP (Cox) dropping and/or blocking e-mail. ===================================== END OF MODERATOR'S COMMENT Sorry if the following appears more than once. I sent it but got no response in 24 hrs; so, I'm resubmitting. --- original message: --- According to the table on: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/#mailing2007-03 the document: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2006/n2140.pdf is the predecessor to the document: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2165.pdf According to n2165, page. 10, aligned_storage::alignment_value is: the least common multiple of all Alignments However, according to n2140, page 5, aligned_storage::alignment_value is the "largest digit that is set" in "the binary-ors all Alignments". I can understand n2165; however, I don't understand n2140. The largest bit set is a number from 0 to 31 on a 32 bit machine. So, if the only element in an aligned_union is an int, with alignment 4, then the largest digit set of binary-or of 4 is 2, and 2 is the wrong alignment for an int. OTOH, if n2140 meant 2 raised to the power of the largest digit that is set, then it would make more sense, but this, AFAICT, would not be the same as the least common multiple. So, is n2140 wrong? In addition, if all alignments are 2 raised to some unsigned power (as required by paragraph 4 on page 5 of n2140) then wouldn't the max of all alignments give the correct alignment for the union? OTOH, if the extended alignments (page 3 of n2165) were allowed, then least common multiple instead of max would be the correct method for calculating the composite (union) alignment. So, is lcm instead of max used on page 10 to accommodate extended alignments? TIA. --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top jg Guest Posted: Thu May 10, 2007 10:57 pm    Post subject: Re: adding alignment support n2140 & n2165 inconsistent alig ===================================== MODERATOR'S COMMENT: ===================================== END OF MODERATOR'S COMMENT On May 7, 8:12 pm, Larry <cppljev...@suddenlink.net> wrote: Quote: ===================================== MODERATOR'S COMMENT: Apologies for the delay in moderation; this appears to be caused by my ISP (Cox) dropping and/or blocking e-mail. ===================================== END OF MODERATOR'S COMMENT Sorry if the following appears more than once. I sent it but got no response in 24 hrs; so, I'm resubmitting. --- original message: --- According to the table on: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/#mailing2007-03 the document: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2006/n2140.pdf is the predecessor to the document: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2165.pdf According to n2165, page. 10, aligned_storage::alignment_value is: the least common multiple of all Alignments However, according to n2140, page 5, aligned_storage::alignment_value is the "largest digit that is set" in "the binary-ors all Alignments". I can understand n2165; however, I don't understand n2140. The largest bit set is a number from 0 to 31 on a 32 bit machine. So, if the only element in an aligned_union is an int, with alignment 4, then the largest digit set of binary-or of 4 is 2, and 2 is the wrong alignment for an int. OTOH, if n2140 meant 2 raised to the power of the largest digit that is set, then it would make more sense, but this, AFAICT, would not be the same as the least common multiple. So, I didn't read the document, instead of answering your question, I am just saying mine. If all the alignment is 2's power, I think the lcm is the same as largest digit that is set" in "the binary-ors all Alignments". (assume n2140 means 2's power.) In practice, is there any alignment that is not 2's power ? JG --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Larry Evans Guest Posted: Sun May 13, 2007 2:47 am    Post subject: Re: adding alignment support n2140 & n2165 inconsistent alig On 05/10/2007 05:57 PM, jg wrote: [snip] Quote: In practice, is there any alignment that is not 2's power ? If boost::alignment_of is correct, then apparently so: http://archives.free.net.ph/message/20051026.011808.725ad7bb.en.html --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. 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