From owner-freebsd-stable Sun Oct 29 10:43:35 2000 Delivered-To: freebsd-stable@freebsd.org Received: from wall.polstra.com (rtrwan160.accessone.com [206.213.115.74]) by hub.freebsd.org (Postfix) with ESMTP id 6823E37B479 for ; Sun, 29 Oct 2000 10:43:32 -0800 (PST) Received: from vashon.polstra.com (vashon.polstra.com [206.213.73.13]) by wall.polstra.com (8.9.3/8.9.3) with ESMTP id KAA06173; Sun, 29 Oct 2000 10:43:20 -0800 (PST) (envelope-from jdp@wall.polstra.com) Received: (from jdp@localhost) by vashon.polstra.com (8.11.0/8.11.0) id e9TIhJG15929; Sun, 29 Oct 2000 10:43:19 -0800 (PST) (envelope-from jdp) Date: Sun, 29 Oct 2000 10:43:19 -0800 (PST) From: jdp@polstra.com Message-Id: <200010291843.e9TIhJG15929@vashon.polstra.com> To: stable@freebsd.org Reply-To: stable@freebsd.org Cc: Cy.Schubert@uumail.gov.bc.ca Subject: Re: F00F-HACK still necessary? In-Reply-To: <200010291602.e9TG25B01059@cwsys.cwsent.com> References: <200010291602.e9TG25B01059@cwsys.cwsent.com> Organization: Polstra & Co., Seattle, WA Sender: owner-freebsd-stable@FreeBSD.ORG Precedence: bulk X-Loop: FreeBSD.ORG In article <200010291602.e9TG25B01059@cwsys.cwsent.com>, Cy Schubert - ITSD Open Systems Group wrote: > NO_F00F_HACK is only effective with the original Pentium. If you > define i686_CPU, NO_F00F_HACK is implied. Close, but not quite right. If you _don't_ define I586_CPU then NO_F00F_HACK is implied. John -- John Polstra jdp@polstra.com John D. Polstra & Co., Inc. Seattle, Washington USA "Disappointment is a good sign of basic intelligence." -- Chögyam Trungpa To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-stable" in the body of the message