From owner-freebsd-net@FreeBSD.ORG Fri Aug 9 23:17:06 2013 Return-Path: Delivered-To: freebsd-net@freebsd.org Received: from mx1.freebsd.org (mx1.freebsd.org [8.8.178.115]) (using TLSv1 with cipher ADH-AES256-SHA (256/256 bits)) (No client certificate requested) by hub.freebsd.org (Postfix) with ESMTP id ED572A4A for ; Fri, 9 Aug 2013 23:17:06 +0000 (UTC) (envelope-from peter@wemm.org) Received: from mail-vb0-x232.google.com (mail-vb0-x232.google.com [IPv6:2607:f8b0:400c:c02::232]) (using TLSv1 with cipher ECDHE-RSA-RC4-SHA (128/128 bits)) (No client certificate requested) by mx1.freebsd.org (Postfix) with ESMTPS id A322228E6 for ; Fri, 9 Aug 2013 23:17:06 +0000 (UTC) Received: by mail-vb0-f50.google.com with SMTP id x14so4548856vbb.9 for ; Fri, 09 Aug 2013 16:17:05 -0700 (PDT) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=wemm.org; s=google; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :cc:content-type; bh=W4dxAWQEgK2jILIbLfFVymdEY4a8zdUICCSrEj9UAXY=; b=hYiMPiatUr4RDEE4wGg/mMc2JbPLP3zBZaosAH9oPa8qbN9Q0u6Jm6DIk9Ou3q45sy 3ys1u0o4/EKEjhJIGIZDsDgGtRf5JjihdhAbbU4UGBpzdPk8iGYch1mbnic6hZ+XXJXu oOYtotBZpvYG224EMsv6xZgjbRHecfvBo4MF0= X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=google.com; s=20120113; h=x-gm-message-state:mime-version:in-reply-to:references:date :message-id:subject:from:to:cc:content-type; bh=W4dxAWQEgK2jILIbLfFVymdEY4a8zdUICCSrEj9UAXY=; b=P91yN/HRTrRFUro4QWjwuUtYw+iwuGZh5419wA09cBPnUUOcfuNQzQzztdkq99eAMB JJjZd4cna/V2Mh8wORarV5a+ei3FWzKA1Jw3APLZ7Tf7DSRMGIw64/nGNdzdt09aGxg3 rzvCnLPPgZxzkO6daq3csVEWgUD9VkZQ6oV/I+HUtMSEBoEZW4Tcv5PFa6RqntKP8ble tC/JK2auwg+jKPXMiRW4wIZVkYcQfjItavkqQh1aE2FeraBCWcZ/3OaRwwcRFE3o8wBY trLNN36nY2mNgIxUJSlv3b9tPv9Rb3xWlK+aw99uITS9djTHEXaHOml6Ksc9CEnmqgCd halQ== X-Gm-Message-State: ALoCoQn3JMKCguE751lzi1ddyo3ZbXnPuRnEsA5jl2ya0hOSh14VuQ7lCrQrTCYzh0cELdG7WmOn MIME-Version: 1.0 X-Received: by 10.52.30.18 with SMTP id o18mr1439025vdh.114.1376090225724; Fri, 09 Aug 2013 16:17:05 -0700 (PDT) Received: by 10.220.167.74 with HTTP; Fri, 9 Aug 2013 16:17:05 -0700 (PDT) In-Reply-To: References: <8B53C542-5CC3-45E6-AA62-B9F52A735EE5@my.gd> Date: Fri, 9 Aug 2013 16:17:05 -0700 Message-ID: Subject: Re: how calculate the number of ip addresses in a range? From: Peter Wemm To: Kimmo Paasiala Content-Type: text/plain; charset=ISO-8859-1 Cc: FreeBSD Net , Fleuriot Damien , s m X-BeenThere: freebsd-net@freebsd.org X-Mailman-Version: 2.1.14 Precedence: list List-Id: Networking and TCP/IP with FreeBSD List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Fri, 09 Aug 2013 23:17:07 -0000 On Fri, Aug 9, 2013 at 4:07 PM, Kimmo Paasiala wrote: > On Sat, Aug 10, 2013 at 1:44 AM, Peter Wemm wrote: >> On Fri, Aug 9, 2013 at 9:34 AM, Fleuriot Damien wrote: >>> >>> On Aug 8, 2013, at 10:27 AM, Peter Wemm wrote: >>> >>>> On Thu, Aug 8, 2013 at 12:04 AM, s m wrote: >>>>> hello guys, >>>>> >>>>> i have a question about ip addresses. i know my question is not related to >>>>> freebsd but i googled a lot and found nothing useful and don't know where i >>>>> should ask my question. >>>>> >>>>> i want to know how can i calculate the number of ip addresses in a range? >>>>> for example if i have 192.0.0.1 192.100.255.254 with mask 8, how many ip >>>>> addresses are available in this range? is there any formula to calculate >>>>> the number of ip addresses for any range? >>>>> >>>>> i'm confusing about it. please help me to clear my mind. >>>>> thanks in advance, >>>> >>>> My immediate reaction is.. is this a homework / classwork / assignment? >>>> >>>> Anyway, you can think of it by converting your start and end addresses >>>> to an integer. Over simplified: >>>> >>>> $ cat homework.c >>>> main() >>>> { >>>> int start = (192 << 24) | (0 << 16) | (0 << 8) | 1; >>>> int end = (192 << 24) | (100 << 16) | (255 << 8) | 254; >>>> printf("start %d end %d range %d\n", start, end, (end - start) + 1); >>>> } >>>> $ ./homework >>>> start -1073741823 end -1067122690 range 6619134 >>>> >>>> The +1 is correcting for base zero. 192.0.0.1 - 192.0.0.2 is two >>>> usable addresses. >>>> >>>> I'm not sure what you want to do with the mask of 8. >>>> >>>> You can also do it with ntohl(inet_addr("address")) as well and a >>>> multitude of other ways. >>> >>> >>> Hold on a second, why would you correct the base zero ? >>> It can be a valid IP address. >> >> There is one usable address in a range of 10.0.0.1 - 10.0.0.1. >> Converting to an integer and subtracting would be zero. Hence +1. >> >> -- > > To elaborate on this, for every subnet regardless of the address/mask > combination there are two unusable addresses: The first address aka > the "network address" and the last address aka the "broadcast > address". There may be usable address in between the two that end in > one of more zeros but those addresses are still valid. Some operating > systems got this horribly wrong and marked any address ending with a > single zero as invalid, windows 2000 was one of them. > > -Kimmo If we go back to the orignal question: "if i have 192.0.0.1 192.100.255.254 how many ip addresses are available in this range?" They're all in the same 192.0.0.0/8. Broadcast or sink addresses don't factor into it. -- Peter Wemm - peter@wemm.org; peter@FreeBSD.org; peter@yahoo-inc.com; KI6FJV UTF-8: for when a ' just won\342\200\231t do. ZFS must be the bacon of file systems. "everything's better with ZFS"