My understanding is /8 is private which means /8+x with x being any whole number is public So 10.0.0.0 /9-32 is public not private which is why you could not use /24 on the internet.

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Okay, you officially confused me. lol

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/8 = 255.0.0.0 right? /24 = 255.255.255.0 right? You asked if you could do 255.255.255.0 and it still be private, correct? The answer is ONLY /8 is private. If you use /24 its public so the answer is NO you can not use /24 and have it still be private ip. Make sense now?

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The three IP blocks mentioned in the Wikipedia article are all private address blocks and will not route on the public Internet. It does not matter how you slice and dice them, all IPs in those blocks are still private. 10.0.0.0/8 10.1.0.0/16 10.0.1.0/24 10.0.0.1/32 All private. See RFC 3330.

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That is exactly what I was wondering. If you slice up the default class A private address, does it prevent it from still being a private address? I was not able to find specific information the internet that stated this. Everything stated the default (10.0.0.0/8) scheme. I wanted to slice up the address to a 10.0.0.0/24 scheme and needed to make sure that it was still considered private. Thanks for the reply Fred.

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fredf take a look at that again. Note ONLY /8 is private. ALL 10.x.x.x are private? No way! What do you think DARPA is using?

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Wanderer: What about this line in the WIKI article that you linked: "For example, while 10.0.0.0/8 would be a single class A network, it is not uncommon for organisations to divide it into smaller /16 or /24 networks." Also, I found this on Microsofts Technet site: "The 10.0.0.0/8 private network has 24 host bits that a private organization can use for any subnetting scheme within the organization." According to that statement you should be able to borrow host bits to create multiple class A networks that are still considered private. At least that is what I gather from reading it. http://technet.microsoft.com/en-us/...

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wanderer, I'm totally amazed that with your knowledge/experience you'd make such a gross error. All addresses in the 10.0.0.0 – 10.255.255.255 range are private, no matter how much you subnet that range they're still private addresses. BTW, 10.0.1.0/8 10.0.2.0/8 10.0.3.0/8 10.0.4.0/8 Those are not 4 separate subnets. Those are 4 host addresses on the same subnet.

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chris30 yes you can subnet and use them INTERNALLY. From the same article. "Though discouraged, some enterprises have begun to use this address space internally for interconnecting private networks to eliminate the chance of address conflicts when using standards-based private ranges." Fishmonger, hey I never said I was perfect nor have I ever claimed to know it all. I offer what I know and believe I know for everyone to see and use as they see fit. If I wasn't open to correction or was afraid of being wrong I certainly wouldn't post. :-) You can use ANY ip addressing scheme you want on a lan/wan not connected to the internet. I have even seen 192.168.x.x ips in a tracert on the internet! Article refers to that as "leakage" The question was if you use /24 is it still considered private on the internet. My answer is no given the logic that 10.0.0.0 – 10.255.255.255 is the ENTIRE class A ip range. So what makes it private or public? What I read and find is that /8 in Class A is the ONLY range considered private which would make any other subnet range public. It is also my understanding that all of the public class A ip ranges were gobbled up in the very beginning by Cal Berkeley, Darpa, etc. I may be wrong but it certainly has resulted in an interesting conversation.

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What I read and find is that /8 in Class A is the ONLY range considered private which would make any other subnet range public. If you're talking about classfull addressing, then that is correct. However, now with the addition of CIDR, we can subnet that down as much as needed. To be a public address, first and foremost, it would need to be assigned out by IANA . Per RFC 1918, IANA has reserved 1 classfull address range in each class to be private so that individuals/companies can use those ranges without having IANA assigning them out. Additionally, those ranges are "not routable" meaning that border/edge routers are supposed to be configured to not route those addresses to the Internet. This has the benefit of have an unknown number of LANS to use the same private address range without the problem of address conflicts. As we all know this is not a perfect world and there are some routers that are miss-configured and will route traffic from private address to the Internet.

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chris30's question still remains. If you use something other than /8 with a class A ip range is that still considered PRIVATE despite if you only use it internally. Consensus of opinion?

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You already know my answer, but here it is a different form. A careful reading of the RFC's will show that all IP addresses starting from: 00001010.00000000.00000000.00000000 (10.0.0.0) and ending at: 00001010.11111111.11111111.11111111 (10.255.255.255) with a CIDR between 8 and 30 are private addresses in the class A range.

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OK. So that means IANA will never assign a class A address block to anyone ever. There are no reserved blocks. Correct?

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The entire class A address blocks are as follows: 0.0.0.0 - 127.255.255.255 (127.0.0.0/8, loopback) So, yes a class A block could be assigned to someone as a public address.

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chris30 that isn't what fishmonger is saying nor does the webpedia article support that with this range 10.0.0.0 – 10.255.255.255 which is the Entire Class A block Now that is very similiar to what I learned back in the early '90's when I started in IT. This apparently has changed with rfc 1918 in 1996 http://www.faqs.org/rfcs/rfc1918.html quoted from the rfc 3. Private Address Space "10.0.0.0 - 10.255.255.255 (10/8 prefix)" This is what Fishmonger is pointing out. When I read this in the past my conclusion was only the /8 was private. Fishmonger is saying the entire class A block is private as seen in post #18 no matter the prefix "with a CIDR between 8 and 30 are private addresses in the class A range" Though he does omit /1-7 and /31-32 [not much point in those being only one and two hosts] I am now questioning my understanding/interpretation of rfc 1918 and am hoping Fishmonger can add to this conversation.

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10.0.0.0 – 10.255.255.255 is Not THE Entire Class A block, it's a small portion of the Class A address range. Here's an excerpt from page 259 of TCP/IP Guide by Charles M. Kozierok, that might help. 1. If the first bit is a 0, it's a class A address, and you're done. (Half the address space has a 0 for the first bit, so this is why Class A takes up half the address space.) If it's a 1, continue to step 2. 2. If the second bit is a 0, it's a Class B address, and your done. (Half of the remaining non-Class A addresses, or one quarter of the total.) If it's a 1, continue to step 3. 3. if the third bit is a 0, then it's a Class C address and you're done. (Half again of what's left, or one-eighth of the total.) If it's a 1, continue to step 4. 4. If the fourth bit is a 0, it's a Class D address. (Half the remainder, or one-sixteenth of the address space.) if it's a 1, it's a Class E address. (The other half, one-sixteenth.) Class A: Lowest Binary value of 1st octet: 00000001 Highest Binary value of 1st octet: 01111110 Decimal Range: 1 to 126 Address Range: 1.0.0.0 to 126.255.255.255 Note, the Class A networks 0 and 127 are reserved for special purposes.

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