> On m68k, some bits of memory must be aligned on a 2-byte boundary. The > above makes that impossible.
I notice all the other architectures handled this code. This make me wonder if this could be seen as a bug in the compiler on m68k. Why does it not calculate the required 2-byte alignment if it need the data to be aligned on 2-byte boundaries? Note, I am not against changing the code of insserv to get it working on m68k, but I want to understand why the correct fix is in that code, and not in the compiler on m68k, when the code work on all other architectures. What is the value of sizeof(regex_t) on m68k? What is the alignment calculated for the struct? Happy hacking, -- Petter Reinholdtsen -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of "unsubscribe". Trouble? Contact [EMAIL PROTECTED]
