I am working on an NTP Client. A few other threads indicate that the a message containing "\x1b' + 47 * '\0" is sent to the NTP server, but none of these threads give an explanation about what this message actually means or why it is sent. I've tried looking at the NTP RFC but I was unable to find any information about it in there either.
"\x1b' + 47 * '\0" represents a data field of 48 bytes. 0x1B followed by 47 times
0. 48 bytes is the size of an NTP UDP packet. The first byte (0x1B) specifies LI, VN, and Mode.
RFC 5905 NTP Specification (7.3. Packet Header Variables) specifies the message header as follows:
0 1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|LI | VN |Mode | Stratum | Poll | Precision |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Setting the first byte of the data to 0x1B or 00 011 011 means
LI = 0 (Leap indicator)
VN = 3 (Version number)
Mode = 3 (Mode, mode 3 is client mode)
You may also use the more recent version (VN = 4). This would require the first header byte to be set to
0x23 (00 100 011).
The modes are defined as
+-------+--------------------------+
| Value | Meaning |
+-------+--------------------------+
| 0 | reserved |
| 1 | symmetric active |
| 2 | symmetric passive |
| 3 | client |
| 4 | server |
| 5 | broadcast |
| 6 | NTP control message |
| 7 | reserved for private use |
+-------+--------------------------+
Specifying Mode = 3 indicates the message as a client request message.
Sending such a packet to port 123 of an NTP server will force the server to send a reply package.
Related
I am trying to understand that Why 255.255.255.0 regarded as class B subnet mask,while many students in my class argue that the mask address is of class C.
Consider reading one of the relevant RFCs. RFC 870, for instance.
Here is an excerpt from that memo:
The third type of address, class C, has a 21-bit network number
and a 8-bit local address. The three highest-order bits are set
to 1-1-0. This allows 2,097,152 class C networks.
1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|1 1 0| NETWORK | Local Address |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Class C Address
In it, class C networks are defined to have 8-bit local address, hence the mask 255.255.255.0.
I am wondering if the stem function in R is producing the stem and leaf plot correctly for this example. The code
X <- c(rep(1,1000),2:15)
stem(X,width = 20)
produces the output
The decimal point is at the |
1 | 00000000+980
2 | 0
3 | 0
4 | 0
5 | 0
6 | 0
7 | 0
8 | 0
9 | 0
10 | 0
11 | 0
12 | 0
13 | 0
14 | 0
15 | 0
There are 1000 ones in the data, but the output of the stem function seems to indicate that there are 988 ones (if you count the zeros in the first row and add 980). Instead of +980, I think it should display +992 at the end of the first row.
Is there an error in the stem function or am I not reading the output correctly?
I have gone through the RFC 4443 and 8201, perhaps I did not understand as I am new to some terminology described in these RFC but I want to understand the implication of using payload in ICMP packet to big message?
As per the RFC 4443
Linkt to 4443 RFC
The payload will contain as much of invoking packet not exceeding the minimum path MTU in packet to big message.
I don't understand the use case of such payload, even in the RFC 8201 there is no mention about the usages of payload.
only one comment present was
Added clarification in Section 4, "Protocol Requirements", that
nodes should validate the payload of ICMP PTB messages per RFC
4443, and that nodes should detect decreases in PMTU as fast as
possible.
What can be the implication validating payload in the PTB message and how should we validate the payload and based on what conditions?
Packet Too Big Message as per RFC 4443.
0 1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Type | Code | Checksum |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| MTU |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| As much of invoking packet |
+ as possible without the ICMPv6 packet +
| exceeding the minimum IPv6 MTU [IPv6] |
I'm reading about raft, but I got a bit confused when it comes to consensus after a network partition.
So, considering a cluster of 2 nodes, 1 Leader, 1 Follower.
Before partitioning, there where X messages written, successfully replicated, and then imagine that a network problem caused partitioning, so there are 2 partitions, A(ex-leader) and B(ex-follower), which are now both leaders (receiving writes):
before partition | Messages |x| Partition | Messages
Leader | 0 1 2 3 4 |x| Partition A | 5 6 7 8 9
Follower | 0 1 2 3 4 |x| Partition B | 5' 6' 7' 8' 9'
After the partition event, we've figured it out, what happens?
a) We elect 1 new leader and consider its log? (dropping messages of the new follower?
e.g:
0 1 2 3 4 5 6 7 8 9 (total of 10 messages, 5 dropped)
or even:
0 1 2 3 4 5' 6' 7' 8' 9' (total of 10 messages, 5 dropped)
(depending on which node got to be leader)
b) We elect a new leader and find a way to make consensus of all the messages?
0 1 2 3 4 5 5' 6 6' 7 7' 8 8' 9 9' (total of 15 messages, 0 dropped)
if b, is there any specific way of doing that? or it depends on client implementation? (e.g.: message timestamp...)
The leaders log is taken to be "the log" when the leader is elected and has successfully written its initial log entry for the term. However in your case the starting premise is not correct. In a cluster of 2 nodes, a node needs 2 votes to be leader, not 1. So given a network partition neither node will be leader.
I'm trying to encode a TCP header myself, but can't understand what is the right order of bits/octets in it. This is what RFC 793 says:
0 1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Source Port | Destination Port |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Sequence Number |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
...
This means that Source Port should take first two octets and the lowest bit should be in the first octet. This means to me that in order to encode source port 180 I should start my TCP header with these two bytes:
B4 00 ...
However, all examples I can find tell me to do it the other way around:
00 B4 ...
Why?
This means that Source Port should take first two octets
Correct.
and the lowest bit should be in the first octet.
Incorrect. It doesn't mean that. It doesn't say anything about it.
All multi-byte integers in all IP headers are represented in network byte order, which is big-endian. This is specified in RFC 1700.