From mboxrd@z Thu Jan 1 00:00:00 1970 Received: from f03n07e.au.ibm.com by ausmtp01.au.ibm.com (IBM AP 1.0) with ESMTP id NAA76890 for ; Tue, 28 Mar 2000 13:54:05 +1000 From: pnilesh@in.ibm.com Received: from d73mta05.au.ibm.com (f06n05s [9.185.166.67]) by f03n07e.au.ibm.com (8.8.8m2/8.8.7) with SMTP id NAA25152 for ; Tue, 28 Mar 2000 13:58:57 +1000 Message-ID: Date: Tue, 28 Mar 2000 09:20:57 +0530 Mime-Version: 1.0 Content-type: text/plain; charset=us-ascii Content-Disposition: inline Sender: owner-linux-mm@kvack.org Return-Path: To: linux-mm@kvack.org List-ID: When a executable file runs there is only one copy of the text part in the memory. But I have some doubts as I am not able to figure how exactly this is done. Suppose a text page of an executable is mapped in the address space of 2 processes. The page count will be one. The page table entries of both the process will have entry for this page. But when the page is discarded only the page entry of only one process get cleared , this is what I have understood from the swap_out () function . But the page table entry of the other process is still pointing to the page which has been discarded. Can any body please clear my doubt. Q When a page of a file is in page hash queue, does this page have page table entry in any process ? Q Can this be discarded right away , if the need arises? Nilesh Patel -- To unsubscribe, send a message with 'unsubscribe linux-mm' in the body to majordomo@kvack.org. For more info on Linux MM, see: http://www.linux.eu.org/Linux-MM/