Re: Why is it called sparc64?

r&lr=&ie=UTF-8&oe=utf-8&frame=right&th=7a22f4c0e87d112c&seekm=43125ccc.0209041413.13c8f31a%40posting.google.com#link2 With a little linux bashing at the end ;) Thanks, -- Bill Moseley mailto:[EMAIL PROTECTED]

Why is it called sparc64?

BTW -- The Ultra 10 on sourceforge say: $ uname -a Linux usf-cf-sparc-linux-1 2.4.18 #2 Thu Apr 11 14:37:17 EDT 2002 sparc64 unknown I gather from the Solaris list it's a SPARC v9, and a sparc64 is something else. -- Bill Moseley mailto:[EMAIL PROTECTED]

Re: Memory alignment of pointers for sparc64

At 02:16 PM 09/05/02 -0600, Jason Gunthorpe wrote: > >On Thu, 5 Sep 2002, Bill Moseley wrote: > >> >> Here's where it's blowing up: >> >> struct dev_ino *p; >> struct stat buf; >> ... >> // allocate a bit of memo

Re: Memory alignment of pointers for sparc64

ference is not worth worrying about. Thanks for your time. -- Bill Moseley mailto:[EMAIL PROTECTED]

Re: Memory alignment of pointers for sparc64

At 09:17 AM 09/05/02 -0400, Michael J. Saletnik wrote: >On September 4, 2002 at 23:06, Bill Moseley wrote: > > the allocated pointers based on the sizeof(void *). > >On September 5, 2002 at 08:13, Ben Collins wrote: > > What was wrong with the original usage of sizeof(long)? &

Re: Memory alignment of pointers for sparc64

On Thu, 5 Sep 2002, Ben Collins wrote: > On Wed, Sep 04, 2002 at 11:06:04PM -0700, Bill Moseley wrote: > > I'm pretty sure you mean sparc and not sparc64 (even if you are running > an ultra, it is still 32bit userspace). On sparc64, sizeof(void *) does > in fact equal 8bytes

Memory alignment of pointers for sparc64

suggestions are more than welcome. Thanks. -- Bill Moseley mailto:[EMAIL PROTECTED]