From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S261189AbULAFbi (ORCPT ); Wed, 1 Dec 2004 00:31:38 -0500 Received: (majordomo@vger.kernel.org) by vger.kernel.org id S261226AbULAFbi (ORCPT ); Wed, 1 Dec 2004 00:31:38 -0500 Received: from ems.hclinsys.com ([203.90.70.242]:64527 "EHLO ems.hclinsys.com") by vger.kernel.org with ESMTP id S261189AbULAFbh (ORCPT ); Wed, 1 Dec 2004 00:31:37 -0500 Subject: Query regarding current macro From: Jagadeesh Bhaskar P To: LKML Content-Type: text/plain Message-Id: <1101879238.7423.13.camel@myLinux> Mime-Version: 1.0 X-Mailer: Ximian Evolution 1.4.5 (1.4.5-7) Date: Wed, 01 Dec 2004 11:03:58 +0530 Content-Transfer-Encoding: 7bit Sender: linux-kernel-owner@vger.kernel.org X-Mailing-List: linux-kernel@vger.kernel.org Hi all, I was going through the explanation of the current macro, and found out that it was got by masking 13-bits LSB the esp register. So is it that always the process descriptor will start at a location with the last 13 bits are 0?? I have read that both the kernel stack and process descriptor of a process is stored in together in an 8KB page. Now the offsets in the page should start from all bits 0, rite? So then why masking only the 13 bits LSB?? What is the significance of keeping that length at 13?? Please do help!! TIA -- With regards, Jagadeesh Bhaskar P