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Thread: multilib




multilib
user name
2006-03-26 23:37:51
JimD posted <20060326113037.63d2f7f3keelie.localdomain>, excerpted below,
 on Sun, 26 Mar 2006 11:30:37 -0500:

> What does (-multilib) mean?  I tried to unmask it but I
cannot figure out
> how to get multilib turned on.
> 
> P.S.  I have also seen the (-FOO) syntax before and
cannot figure out how
> to enable a USE option that is in that format.

I see SS's pointing to the FAQ, and that's correct, but I
read the blurb
in the FAQ and don't find it worded as clearly as it could
be.

In general, a USE flag in parenthesis, as (-multilib),
indicates that USE
flag is set by the profile and can't be changed.  (Well, it
could be, but
you better be very sure you know what you are doing, and as
with many
things in life, if you are asking, you don't.)

Why would this be?  Simple.  The regular amd64 profiles
require multilib,
altho there are subprofiles available that have it disabled,
if you
prefer.  Thus, it's part of the profile, enabled or
disabled by the
profile you use, and not settable separately.  (This is a
fairly new
situation.  Old 2004 profiles had it as a settable USE flag,
which is why
it still exists as a USE flag, even if it can't be set as
such.)

Similarly with some other flags, including mmx, sse, etc. 
All amd64 based
processors, both those from AMD and those from Intel,
include these
extensions, so gcc knows they can be used already, and the
flags need not
be specified.  Additionally, as the flags were developed on
32-bit
processors, they often invoke the 32-bit specific assembly
language
extensions.  Trying to use these 32-bit specific extensions
in a 64-bit
program is a sure way to failure, so all around, it's
simply better to
force the USE flags off for 64-bit mode.  Again, gcc already
knows it can
use the 64-bit versions, and the USE flags apply the 32-bit
versions
which will break stuff in 64-bit mode, so leave them off,
and nobody gets
hurt.  <g>  Do note, however, that if you use a 32-bit
chroot, you can
safely use those flags there, and they'll work as intended,
because the
chroot will be using a 32-bit profile, not the usual 64-bit
profile.

Perhaps this is clearer than the way the FAQ says it?

-- 
Duncan - List replies preferred.   No HTML msgs.
"Every nonfree program has a lord, a master --
and if you use the program, he is your master." 
Richard Stallman in
http://www.linuxdevcenter.com/pub/a/linux
/2004/12/22/rms_interview.html


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