Re: A C++ question

1. Unions always align on the pickiest member of the union.  If you have union of char
and int, it will align on 2-byte.  Union of char, int, float on 4-byte.  Union of char, int, float, double on 8-byte.
2. Doubles are always 8-bytes and some architectures require 8-byte fetches to be fetched on 8-byte boundaries
(address divisible by 8 with no remainder).

Just a bit of parlor trickery.

At 09:52 AM 6/28/2006, you wrote:

>I saw some code in ACE:
>
>class ACE_Export ACE_U_LongLong
>{
>   ....
>   ....
>
>union
>      {
>        struct
>          {
>#     if defined (ACE_BIG_ENDIAN)
>            /// High 32 bits.
>            ACE_UINT32 hi_;
>
>            /// Low 32 bits.
>            ACE_UINT32 lo_;
>#     else
>            /// Low 32 bits.
>            ACE_UINT32 lo_;
>
>            /// High 32 bits.
>            ACE_UINT32 hi_;
>#     endif /* ! ACE_BIG_ENDIAN */
>
>          } data_;
>
>        /// To ensure alignment on 8-byte boundary.
>        /**
>         * @note "double" isn't usually usable with
>         *       ACE_LACKS_FLOATING_POINT, but this seems OK.
>         */
>        double for_alignment_;
>      };
>   ....
>};
>
>I don't understand why for_alignment_ can ensure aligment on 8-byte 
>boundary. Anyone can tell me?
>
>Thanks!
>
>ChinaGirl from Canada
>
>
>
>
>
>
> 
>Yahoo! Groups Links
>
>
>
> 
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>
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