As Source Address is always an individual address (Unicast), the least
significant bit of first byte is always 0.
I saw this exp. on geeksforgeeks.com, so I want to ask that whether a source address has to be unicast or not. I understand that a source address is an individual address but what about broadcasting? Even if it has an individual MAC address, it is not unicasting. As a 2nd question, why is the least significant bit of first byte is always 0 and what does it have to do with source address being always an individual address? Thanks.
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When define a subnet, such as 1.2.3.0/24, it means 255 hosts in the subnet start from 1.2.3.0 to 1.2.3.255.
But what if I wrongly define the subnet with 1.2.3.64/24? Does it then mean the range is start from 1.2.3.64 to 1.2.3.255?
I could not find a formal document about this.
EDIT: in Java ipaddress lib (5.2.1), the 1.2.3.64/24 will be treated as range from 1.2.3.64 to 1.2.3.255, instead of 1.2.3.0 to 1.2.3.255!. The code is: new IPAddressString("1.2.3.64/24").getSequentialRange()
EDIT: My apology: the above description of ipaddress lib's behavior is wrong. It was other things that caused me think the the range become 1.2.3.64 to 1.2.3.255.
Indeed, the
new IPAddressString("1.2.3.64/24").getSequentialRange().getLower()
is 1.2.3.64,
and
new IPAddressString("1.2.3.64/24").getSequentialRange().getUpper()
is also 1.2.3.64,
so I think this is a reasonable implementation!. I have no problem of that.
The
new IPAddressString("1.2.3.64/24").getAddress().toPrefixBlock().getLower()`
will be the expect one: 1.2.3.0.
And the `.getUpper()` is 1.2.3.255.
so this is the perfect way to get correct ip range.
There is no formal specification indicating that setting the un-fixed bits (the host bits) to 0 is a convention. However, it is common practice when describing a subnet as opposed to a single address, that you leave the host bits as zero. It is an informal convention. If you are describing a subnet, why would you set those bits to anything other than zero if they are intended to be ignored? So that is what all network engineers and most other people do, they leave the bits as zero when they are describing the full subnet, when the host can take on any value.
The specification for something like 1.2.3.64/24 is that the network part of the address is the first 24 bits, and the host part of the address is the remaining 8 bits. So, the network is 1.2.3.0 and the host is 0.0.0.64. The specification says nothing else.
Likewise, 1.2.3.0/24 indicates the network is 1.2.3.0 and the host is 0.0.0.0. However, like I said, when the host is 0, that is commonly used to refer to the entire subnet, meaning the host can take on any value. When people write down a string for that subnet of 255 addresses, they will write 1.2.3.0/24.
In fact, it is unusual to see 1.2.3.0 used as a single address because many routers do not accept a host of 0.
How you parse those things is specific to any library. Some libraries parse 1.2.3.64/24 as the whole subnet, throwing out the 0.0.0.64. Some libraries parse 1.2.3.64/24 as 1.2.3.64. Some libraries have separate methods to do one or the other.
With the IPAddress library, the same parser parses both addresses and subnets. When it parses 1.2.3.64/24 or 1.2.3.0/24, the parser needs to decide what you mean by each. For the former, it interprets 1.2.3.64/24 as the address 1.2.3.64 with prefix length 24, because why else is the 0.0.0.64 there at all if you do not intend that 64 to mean something? Why throw it out?
For 1.2.3.0/24, it interprets that as the whole subnet, because, like I said, a host of 0 is commonly used to refer to the entire subnet, and it is rare to use 1.2.3.0 as an address.
However, if you wish to choose some other meaning when parsing those things, the library provides other options.
There is no formal specification for this. In fact, there is no formal specification for a lot of the different aspects of IP addressing, a lot of it is informal and simply evolved over the years to the common practices in use.
new IPAddressString("1.2.3.64/24").getSequentialRange() is parsed as the range "1.2.3.64 -> 1.2.3.64". Earlier versions of the library behaved differently, but that is how all the latest releases for the last few years parse it.
Disclaimer: I am the project manager of that library.
It's interesting because I too can't seem to find any formal specification on why setting the un-fixed bits to 0 is the convention. Looking at RFC 4632 says:
[Classless prefixes] make explicit which bits in a 32-bit IPv4
address are interpreted as the network number (or prefix) associated
with a site and which are the used to number individual end systems
within the site.
Even with the convention of setting the un-fixed bits to 0, you may have CIDR notation that has an IP ending in a non-zero number. For example: 192.168.1.254/31 which represents the range of IPs from 192.168.1.254 to 192.168.1.255.
CIDR notation only specifies the number of bits that are fixed, so even if you put .64 it would appear that the /24 still represents only 24 fixed bits of the IP address.
You can see that the CIDR Calculator from ARIN shows that 1.2.3.64/24 still represents all IP addresses from 1.2.3.0 to 1.2.3.255.
Screenshot here.
Now it may be that different systems are implemented to handle this situation differently, so I would personally follow the convention, but from the perspective of CIDR notation (and what I can seem to find in the RFC) it should still represent the entire range.
Assume that the 160.5.132.224/27 network is split into 2 subnets, with equal number of IP addresses. What are the correct addresses for these subnets
Basically, I am trying to find a fool proof method to calculate the network address for any subnet. The answer options are :
160.5.132.224/28
160.5.132.192/26
160.5.132.240/28
160.5.132.192/27
160.5.132.208/28
160.5.132.192/28
Can someone tell me which one is the correct one and demonstrate their working out.
The specified address is a 32 bit number, and the subnet mask specifies the number of bits to hold constant in the subnet. The notation 160.5.132.224/27 means to use the first 27 bits from the speicified address as the subnet and vary the last 5 bits as addresses in the subnet.
To split the subnet you would add a bit to the subnet, i.e. /28, and specify additional bits in the addresses by adding a 1 to one address and a 0 to the other.
In this case, the last eight bits in the specified address are provided by 224, which is 11100000. You should change the last eight bits of one subnet to 11110000 (an extra one) which is 240. The other subnet will specify an extra zero, but it will remain 1110000 (which is still 224). Then you end up with 160.5.132.224/28 and 160.5.132.240/28.
I am creating an application which uses semi-random numbers to roll on a table. The random numbers are generated from wireless network MAC Addressess, so they should be consistent across multiple devices (i.e. if I go to a wireless network with 2 different devices, I should generate the same random number).
My question is if there are any patterns that wireless network MACs are predisposed towards, that will affect my number generation, or if they are entirely random. It's important because it will affect the probabilities I assign to table values.
This is the image i pulled from Wikipedia that I am having some trouble understanding.
So based on this diagram, does this mean that the first 3 bytes of a MAC Address will be unique across router brands? If I run into two D-Link routers will that be the same? Or if it's two D-Link routers of the same model will it be the same?
And then are the last 3 bytes just randomly generated numbers? Thanks to anyone who takes the time to help me figure this out.
An OUI part with the 0x020000 bit cleared ("globally unique") is an assigned number from a registry. Every manufacturer of a wireless device applies for at least one of them, and uses it in the products they build. There is no requirement that a manufacturer have only one, or that they use different ones depending on who the end-producer of the equipment is. Thus, for example, many Dell laptops have MAC addresses that begin with Intel OUIs, because the laptops are built with Intel wireless chips.
For these MAC addresses, the NIC part has no specified value, but is usually assigned in monotonically-increasing values from pre-divided blocks during manufacture. Thus two Intel wireless chips rolling off the same assembly line will likely have numbers separated by 1.
An OUI part with the 0x020000 bit set ("locally administered") can have any other bits set, depending entirely on what the system it is attached to sets the address to.
Yes, the first 3 bytes of MAC addresses are usually not random at all, they are assigned to a particular corporation. (Some companies have more than 1 assignment). The last 3 bytes are also not "random" because sometimes they are assigned sequentially. Unless it was a really popular router, the high bits could be constant.
Also, a MAC address can be changed on most cards. And there are different types of MAC addresses, such as "link local" and "multicast".
If you want randomness, use /dev/random or whatever your OS supplies.
I'm trying to make a list of all MAC addresses that are reserved, do not exist, should not be used, should only be used locally etc. (Just like the list of reserved IP-addresses on Wikipedia, but for MAC.) Basically I want to loop over all MAC-addresses from a switch and filter out the "real" ones.
This page suggests all addresses starting with 00-00-5E or 01-00-5E are reserved, but when I look them up it seems like 00-00-5E is also assigned to the Information Sciences Institute (part of a university in California).
So 2 questions:
1) Is there any place I can find a list of reserved MAC-adresses?
2) What's up with 00-00-5E? Is only part of that range reserved, or is there some reason they assigned it to ISI?
I was just looking into this myself recently. I believe that the IANA (which you refer to in one of your links) will give the most authoritative answer: IANA Ethernet Number Assignments
I don't think that this means that these addresses can never be used though. According to RFC5342, Section 2.1
"The 2**8 unicast identifiers from 00-00-5E-00-00-00 through 00-00-5E-00-00-FF are reserved and require IESG Ratification for allocation (see Section 5.1)."
So basically, it appears you need special permission from IESG (Internet Engineering Steering Group) to get an address in that range, which I suppose the ISI has obtained somehow.
Section 2.1 of RFC5342 deals with 48-Bit MAC Identifiers and OUIs, and it doesn't make any mention of any address ranges that are strictly forbidden or permanently reserved from what I've understood.
The following OUI are reserved as per RFC 5342:
OUI 01:00:5E:(00:00:00-7f:ff:ff) - Used for IPV4 Multicast and MLPS Multicast.
OUI 00:00:5E:(00:01:00 – 00:01:FF) - Used for Virtual Router Redundancy Protocol (VRRP) IPV4
OUI 00:00:5E:(00:02:00 – 00:02:FF) - Used for Virtual Router Redundancy Protocol (VRRP) IPV6
OUI 33:33:00 – 33:33:FF - Reserved for IPV6 Multicast
OUI CF:00:00 – CF:FF:FF - Reserved by IANA for PPP(Point to Point Protocol)
OUI 00:00:5E (00:00:00 - 00:00:FF) - Requires IESG Ratification for allocation.
Was looking into this myself.. I know it's been a while since the post was active.. but I found these to be ok to use locally:
x2-xx-xx-xx-xx-xx
x6-xx-xx-xx-xx-xx
xA-xx-xx-xx-xx-xx
xE-xx-xx-xx-xx-xx
Source: https://honeywellaidc.force.com/supportppr/s/article/Locally-Administered-MAC-addresses
The registration authority for MAC addresses is the IEEE. It hands out OUIs (Organizationally Unique Identifiers), which give you a three byte prefix, and 2^24 addresses within it, for a fee (currently 2 995USD). You also get the rights to the corresponding multicasts, which have the prefix with the lowest bit of the first byte set. For instance, 00:80:C2 is allocated to the IEEE 802.1 committee, which uses 01:08:C2:00:00:00 for Spanning tree.
So, there isn't really a list of reserved addresses. There is a list of OUIs that have been allocated, unless the buyer has paid (a lot) extra for privacy. You can use any address that has the local bit set freely. A tiny fraction of multicast addresses have a significant meaning because heavyweights like IEEE, Cisco, IANA assign meanings to them. From the IEEE registration point of view, there is no particular significance to these blocks (except possibly to those it has allocated to itself).
Now, how did the 01-00-5E range end up allocated to the Information Sciences Institute? The simple
answer is that they paid for it. So, really the question should be 'how did the Internet get to use part of the range allocated to ISI?'. The answer is that the IANA used to be run from an office in ISI: specifically IANA was the legendary Jon Postel
Bottom line: you are on a bit of a fool's errand. You can distinguish local addresses and multicast addresses, and make some attempt to tie up allocated unicast addresses to vendor blocks. And you can probably do a bit more with well-known multicast addresses but only by tracking down individudal vendor's documentation (IANA is obviously an important one but only definitive for 1 of the 2^22 available blocks). One of the best places to start is probably the Wireshark codebase.
I am told that one can derive the port number if the IP is of the given form.
How does one do that? My searches didnt lead me anywhere
Edit ==
I am leaving the question containing the term IP address as this is something many people are likely to search for. Technically, from the document linked in the answer
If
this command is used, the argument is the concatenation of a
32-bit internet host address and a 16-bit TCP port address.
This address information is broken into 8-bit fields and the
value of each field is transmitted as a decimal number (in
character string representation). The fields are separated
by commas. A port command would be:
PORT h1,h2,h3,h4,p1,p2
where h1 is the high order 8 bits of the internet host
address.
I assume you're talking about the PORT ftp command.
As seen in the RFC 959, page 28, those two numbers represents the two bytes of a 16-bit word, as decimal numbers, with the high one first.
The general formula is:
(b1 << 8) | b2
Applied to your specific case:
(193 << 8) | 148 = 49 556