From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S1750915AbbIWEDJ (ORCPT ); Wed, 23 Sep 2015 00:03:09 -0400 Received: from mail-bn1on0143.outbound.protection.outlook.com ([157.56.110.143]:2337 "EHLO na01-bn1-obe.outbound.protection.outlook.com" rhost-flags-OK-OK-OK-FAIL) by vger.kernel.org with ESMTP id S1750703AbbIWEDH (ORCPT ); Wed, 23 Sep 2015 00:03:07 -0400 Authentication-Results: spf=none (sender IP is ) smtp.mailfrom=scottwood@freescale.com; Message-ID: <1442980978.19102.321.camel@freescale.com> Subject: Re: [PATCH v10 3/5] CPM/QE: use genalloc to manage CPM/QE muram From: Scott Wood To: Zhao Qiang-B45475 CC: "linux-kernel@vger.kernel.org" , "linuxppc-dev@lists.ozlabs.org" , "lauraa@codeaurora.org" , Xie Xiaobo-R63061 , "benh@kernel.crashing.org" , Li Yang-Leo-R58472 , "paulus@samba.org" Date: Tue, 22 Sep 2015 23:02:58 -0500 In-Reply-To: References: <1442560521-19354-1-git-send-email-qiang.zhao@freescale.com> <1442560521-19354-3-git-send-email-qiang.zhao@freescale.com> <20150921224720.GA12709@home.buserror.net> <1442967567.19102.318.camel@freescale.com> Content-Type: text/plain; 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> > > > > + > > > > > + unsigned long start; > > > > > unsigned long flags; > > > > > + unsigned long size_alloc = size; struct muram_block *entry; int > > > > > + end_bit; int order = muram_pool->min_alloc_order; > > > > > > > > > > spin_lock_irqsave(&cpm_muram_lock, flags); > > > > > - ret = rh_free(&cpm_muram_info, offset); > > > > > + end_bit = (offset >> order) + ((size + (1UL << order) - 1) >> > > > > order); > > > > > + if ((offset + size) > (end_bit << order)) > > > > > + size_alloc = size + (1UL << order); > > > > > > > > Why do you need to do all these calculations here? > > > > > > So do it in gen_pool_fixed_alloc? > > > > Could you explain why they're needed at all? > > Why it does the calculations? > If the min block of gen_pool is 8 bytes, and I want to allocate a > Region with offset=7, size=8bytes, I actually need block 0 and block 1, > And the allocation will give me block 0. How can you have offset 7 if the minimum order is 2 bytes? -Scott