校验和方法确定 [英] Checksum method determination
问题描述
嘿，
我正在尝试与串口设备通信，我需要知道用于计算1digit校验和的
方法。
我有一个可以与设备通信的商业程序，以及一个
串口嗅探器，所以我可以获得任意数量的样本
校验和数字。下面是一些列表。
我一直在寻找模式，它似乎围绕着魔法
的数量为7.
如果有人可以帮忙找到这个，或者有人知道确定校验和的b
暴力方法吗？
谢谢，
Ryan Johnson。
1255 | 4
1265 | 7
1256 | 7
1355 | 5
3455 | 0
2455 | 1
1455 | 2
455 | 3
3355 | 7
2355 | 6
1355 | 5
355 | 4
255 | 5
1255 | 4
2255 | 7
3255 | 6
4255 | 1
425 | 4
1 | 6
2 | 5
3 | 4
4 | 3
5 | 2
6 | 1
7 | 0
8 | ？
9 | >
10 | 6
11 | 7
12 | 4
13 | 5
14 | 2
15 | 3
16 | 0
17 | 1
18 | >
19 | ？
21 | 4
23 | 6
2 | 5
22 | 7
222 | 5
3333 | 7
3 | 4
33 | 7
333 | 4
3333 | 7
123456789123456789 | 9
45 | 6
56 | 4
3567 | 0
3566 | 1
356 | 7
1234 | 3
123 | 7
Hey,
I am trying to communicate with serial device and I need to know the
method used to calculate the 1digit checksum.
I have a commercial program that can communicate with the device, and a
serial port sniffer, and so I am able to get any number of sample
checksum digits. Below is a list of some.
I keep finding patterns, and it seems to revolve around the magic
number of 7.
If anyone could please help find this out, or does anyone know of a
brute-force method of determining a checksum?
Thanks,
Ryan Johnson.
1255 | 4
1265 | 7
1256 | 7
1355 | 5
3455 | 0
2455 | 1
1455 | 2
455 | 3
3355 | 7
2355 | 6
1355 | 5
355 | 4
255 | 5
1255 | 4
2255 | 7
3255 | 6
4255 | 1
425 | 4
1 | 6
2 | 5
3 | 4
4 | 3
5 | 2
6 | 1
7 | 0
8 | ?
9 | >
10 | 6
11 | 7
12 | 4
13 | 5
14 | 2
15 | 3
16 | 0
17 | 1
18 | >
19 | ?
21 | 4
23 | 6
2 | 5
22 | 7
222 | 5
3333 | 7
3 | 4
33 | 7
333 | 4
3333 | 7
123456789123456789 | 9
45 | 6
56 | 4
3567 | 0
3566 | 1
356 | 7
1234 | 3
123 | 7
推荐答案
rj * *******@hotmail.com 写道：
我正在尝试与串口设备通信，我需要知道用于计算1digit校验和的
方法。
I am trying to communicate with serial device and I need to know the
method used to calculate the 1digit checksum.
[snip]
这与C编程语言无关。请限制
跟进sci.crypt，comp.programming和sci.math，或者限制为某些
适当的子集。
-
Keith Thompson（The_Other_Keith） ks***@mib.org < http：/ /www.ghoti.net/~kst>
圣地亚哥超级计算机中心< *> < http://users.sdsc.edu/~kst>
我们必须做点什么。这是事情。因此，我们必须这样做。
[snip]
This has nothing to do with the C programming language. Please limit
followups to sci.crypt, comp.programming, and sci.math, or to some
appropriate subset.
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
在消息< 11 ********************** @ g43g2000cwa.googlegroups .com>，
rj********@hotmail.com 写道
In message <11**********************@g43g2000cwa.googlegroups .com>,
rj********@hotmail.com writes
嘿，
我正在尝试与串口设备通信，我需要知道用于计算1digit校验和的
方法。
我有一个可以与设备通信的商业程序，以及一个串口嗅探器，所以我可以得到任意数量的样本校验和数字。下面列出了一些。
我一直在寻找模式，而且它似乎围绕着神奇的数量7个。
如果有人可以请帮助找到这个，还是有人知道确定校验和的蛮力方法？
谢谢，瑞安约翰逊。
Hey,
I am trying to communicate with serial device and I need to know the
method used to calculate the 1digit checksum.
I have a commercial program that can communicate with the device, and a
serial port sniffer, and so I am able to get any number of sample
checksum digits. Below is a list of some.
I keep finding patterns, and it seems to revolve around the magic
number of 7.
If anyone could please help find this out, or does anyone know of a
brute-force method of determining a checksum?
Thanks,
Ryan Johnson.
....< data snipped>
0到7的值似乎表明你只能得到3位。
注意结果输入8或9.
也许你有某种移位寄存器???
-
Jeremy Boden
.... <data snipped>
The 0 - 7 values would seem to indicate that you only get 3 bits back.
Note the results of inputting 8 or 9.
Perhaps you have some kind of shift register???
--
Jeremy Boden
< rj ******** @ hotmail.com>在消息中写道
news：11 ********************** @ g43g2000cwa.googlegr oups.com ...
<rj********@hotmail.com> wrote in message
news:11**********************@g43g2000cwa.googlegr oups.com...
如果有人可以帮忙找到这个，或者有人知道确定校验和的蛮力方法吗？
If anyone could please help find this out, or does anyone know of a
brute-force method of determining a checksum?
???你的问题甚至没有意义。你确定你知道你b $ b要求的是什么吗？
是的我熟悉一种强制校验和但校验和的方法
碰撞可能会让你失望 - 当然，就像你暴力强迫
一样。
-
LTP
:)
??? Your question doesn''t even make sense. Are you sure you know what you
are asking?
Yes I am familiar with one way of bruteforcing a checksum but checksum
collisions could throw you off - of course it is the same way you bruteforce
just about anything.
--
LTP
:)
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