From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S936377Ab3DHVam (ORCPT ); Mon, 8 Apr 2013 17:30:42 -0400 Received: from mail.linuxfoundation.org ([140.211.169.12]:50395 "EHLO mail.linuxfoundation.org" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S936268Ab3DHVai (ORCPT ); Mon, 8 Apr 2013 17:30:38 -0400 Date: Mon, 8 Apr 2013 14:30:37 -0700 From: Andrew Morton To: Chanho Min Cc: Nadia Yvette Chambers , Jiri Kosina , Guennadi Liakhovetski , linux-kernel@vger.kernel.org Subject: Re: [PATCH] bitmap: speedup in bitmap_find_free_region when order is 0 Message-Id: <20130408143037.5fe18306172315fc040329de@linux-foundation.org> In-Reply-To: <1365387828-24326-1-git-send-email-chanho.min@lge.com> References: <1365387828-24326-1-git-send-email-chanho.min@lge.com> X-Mailer: Sylpheed 3.2.0beta5 (GTK+ 2.24.10; x86_64-pc-linux-gnu) Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit Sender: linux-kernel-owner@vger.kernel.org List-ID: X-Mailing-List: linux-kernel@vger.kernel.org On Mon, 8 Apr 2013 11:23:48 +0900 Chanho Min wrote: > If bitmap_find_free_region() is called with order=0, We can reduce > for-loops to find 1 free bit. First, It scans bitmap array by the > increment of long type, then find 1 free bit within 1 long type value. That seems sensible. I assume-without-checking that single-bit is a common case? > In 32 bits system and 1024 bits size, in the worst case, We need 1024 > for-loops to find 1 free bit. But, If This is applied, it takes > 64 for-loops. Instead, if free bit is in the first index of the bitmaps, > It will be needed additional 1 for-loop. But from second index, It > will be speed up significantly. > > --- a/lib/bitmap.c > +++ b/lib/bitmap.c > @@ -1099,6 +1099,37 @@ done: > } > > /** > + * bitmap_find_free_one - find a mem region > + * @bitmap: array of unsigned longs corresponding to the bitmap > + * @bits: number of bits in the bitmap > + * > + * Find one of free (zero) bits in a @bitmap of @bits bits and > + * allocate them (set them to one). > + * > + * Return the bit offset in bitmap of the allocated region, > + * or -errno on failure. > + */ > +static int __bitmap_find_free_one(unsigned long *bitmap, int bits) > +{ > + int pos, i; > + unsigned long mask = (unsigned long)(~((unsigned long) 0)); That seems unnecessarily complicated. "unsigned long mask = -1;" works :) > + int nlongs_reg = BITS_TO_LONGS(bits); > + > + for (i = 0 ; i < nlongs_reg ; i++) { > + if ((bitmap[i] & mask) != mask) { But here we could just do "if (bitmap[i] != -1)". Or ~0UL. > + for (pos = 0 ; pos < BITS_PER_LONG ; pos++) { > + if (!__reg_op(&bitmap[i], pos, 0, > + REG_OP_ISFREE)) > + continue; > + __reg_op(&bitmap[i], pos, 0, REG_OP_ALLOC); > + return pos; > + } > + } > + } > + return -ENOMEM; > +} afacit the code is buggy - if `bits' is not an exact multiple of BITS_PER_LONG, this search will wander off the end of the specified region? > +/** > * bitmap_find_free_region - find a contiguous aligned mem region > * @bitmap: array of unsigned longs corresponding to the bitmap > * @bits: number of bits in the bitmap > @@ -1116,6 +1147,9 @@ int bitmap_find_free_region(unsigned long *bitmap, int bits, int order) > { > int pos, end; /* scans bitmap by regions of size order */ > > + if (order == 0) > + return __bitmap_find_free_one(bitmap, bits); > + > for (pos = 0 ; (end = pos + (1 << order)) <= bits; pos = end) { > if (!__reg_op(bitmap, pos, order, REG_OP_ISFREE)) > continue;