I am currently working with UDP packets and I need to craft and send custom data. Because it is easier for me to read I work with strings representing hexadecimal values. I have something like this :
a = "12"
b = "15"
header = "c56b4040003300" + a + "800401" + b + "90000000"
Now, what I want to do is converting my header variable into hexadecimal (but not with the hexadecimal value of every character in header). It means that if I write my header variable in a file and I open it with a hexadecimal editor, I want to see
c5 6b 40 40 00 33 00 12 80 04 01 15 90 00 00 00
I don't have a good knowledge of ruby and I couldn't find a way to do it so far. The pack function converts characters in hexa but not hexadecimal string representation as hexadecimal value. And doing something like
header = "\xc5\x6b\x40\x40\x00\x33\x00\x" + a + "\x80\x04\x01\x" + b + "\x90\x00\x00\x00"
will throw me an error saying "invalid hex escape" (which make sense).
So if you have a solution to this problem please tell me (if possible without using any external library)
require 'strscan'
s = StringScanner.new('hexstring here')
s.scan(/../).map { |x| x.hex.chr }.join
String#to_i takes a base argument that will do what you want:
["c56b4040003300", a, "800401", b, "90000000"].join.to_i(16)
But it may not make sense to represent your data as an large integer. If you just want a blob of binary data, you can concatenate everything together and use Array#pack:
[["c56b4040003300", a, "800401", b, "90000000"].join].pack('H*')
Or you can pack the individual components and concatenate the results:
["c56b4040003300", a, "800401", b, "90000000"].map { |s| [s].pack('H*') }.join
Or you can just work with an array of bytes throughout your program:
bytes = "c56b4040003300".scan(/../)
bytes << a
bytes.concat "800401".scan(/../)
bytes << b
bytes.concat "90000000".scan(/../)
bytes.unpack('H*' * bytes.size)
Hope that helps!
Related
I have a two requirements:
I must concatenate some fields from a file in a Cobol program. The way i must concatenate is based on one of the aforementioned field. The concatenated fields must be outputted in a new file.
I must then sort this new file with a sort utility invoked by JCL.
The Issue
I need to sort same file for 2 conditions. I have tried with ifthen outrec build. How can I sort it in one pass?
Here is a source-code example :
ID DIVISION.
PROGRAM-ID. FOO.
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
Select Infil assign to inp001.
Select Outfil assign to out001.
DATA DIVISION.
FILE SECTION.
FD Infil.
01 Main.
03 A.
05 ws-Pc. Pic x(1).
05 filler Pic x(5).
03 B. Pic x(4).
03 C. Pic x(4).
03 D. Pic 9(13)V99.
03 E. Pic x(13).
FD Outfil.
01 Temp Pic x(42).
WORKING-STORAGE SECTION.
01 file-flag Pic x(01).
88 file-end value 'Y'.
88 not-file-end value 'N'.
PROCEDURE DIVISION.
Open input Infil
Open output Outfil
read Infil
at end
set file-end to true
not at end
set not-file-end to true
end-read
Perform until file-end
If ws-Pc = 3
String A B C Delimited by size
into Temp
End-String
Else
String A B C E Delimited by size
into Temp
End-String
End-if
Write Temp
read Infil
at end
Set file-end to true
end-read
end-Perform.
end program foo.
Here is the logic I need for the sort utility :
If ws-Pc=3
Sort(fieldA,fieldB)
Else
Sort(fieldA,fieldB,fieldE)
End-if.
I can propose you three variations :
If you are sure that your last 13 characters are the same (spaces for instance) in the case without a field E (ws-pc equal to three) you can just have this sysin:
SORT FIELDS=(1,6,CH,A,7,4,CH,A),EQUALS
Indeed, thanks to the equals all your inputs will keep their relative order once sorted. For the case ws-pc=3 it will be sorted according to field A and B. Field E plays no importance because it is the same for all.
If you are not sure that the last 13 characters are the same you can do it yourself:
INREC IFTHEN=(WHEN=(1,1,CH,EQ,C'3'),BUILD=(1,14,13X))
SORT FIELDS=(1,6,CH,A,7,4,CH,A),EQUALS
This will force your last characters to be spaces.
If you don't want to use the "EQUALS" you can create your own ordering by appending a line number in the field E when it is unused. You then have to remove it.
INREC IFTHEN=(WHEN=(1,1,CH,EQ,C'3'),BUILD=(1,14,SEQNUM,13,ZD))
SORT FIELDS=(1,6,CH,A,7,4,CH,A)
OUTREC IFTHEN=(WHEN=(1,1,CH,EQ,C'3'),BUILD=(1,14,13X))
I'm using hyperterminal and trying to send strings a to 6 digit scoreboard. I was sent a sample string from the manufacturer to test with and it worked, but to be able change the displayed message I was told to calculate a new Checksum value.
The sample string is: &AHELLO N-12345\71
Charactors A and N are addresses for the scoreboards(allowing two displays be used through one RS232 connection). HELLO and -12345 are the characters to be shown on the display. The "71" is where I am getting stuck.
How can you obtain 71 from "AHELLO N-12345"?
In the literature supplied with the scoreboard, the "71" from the sample string is described as a character by character logical XOR operation on characters "AHELLO N-12345". The manufacturer however called it a checksum. I'm not trained in this type of language and I did try to research but I can't put it together on my own.
The text below is copied from the supplied literature and describes the "71" (ckck) in question...
- ckck = 2 ASCII control characters: corresponds to the two hexadecimal digits obtained by
performing the character by character logical XOR operation on characters
"AxxxxxxByyyyyy". If there is an error in these characters, the string is ignored
Example: if the byte by byte logical XOR operation carried out on the ASCII codes of the
characters of the "AxxxxxxByyyyyy" string returns the hexadecimal value 0x2A,
the control characters ckck are "2" and "A".
You don't specify a language but here's the algorithm in C#. Basically xor the values of the string all together and you'll end up with a value of 113, 71 in hex. Hence 71 is on the end of the input string.
string input = "AHELLO N-12345";
UInt16 chk = 0;
foreach(char ch in input) {
chk ^= ch;
}
MessageBox.Show("value is " + chk);
Outputs "value is 113"
I am trying to split a String in Ruby based on a regex.
The String has the following pattern:
00 0 0 00 00 0
I want to be able to split the string on every second space but I am quite new to Ruby and my experience with Regexes is limited.
I have tried the following:
line.split(/[0-9._%+-]+\s+[0-9._%+-]+/)
But this just returns an array of blank values. I have tried various different combinations of the regex pattern but have not got close to what I want. The result should be an array like this:
Array[0] => '00 0'
Array[1] => '0 00'
Array[2] => '00 0'
Could anyone explain how I could best do this in a Regex? And if possible explain why my attempt doesn't work and why you're working example does work, I want to increase my knowledge of Regexes by solving this problem.
Use String#scan
line = "00 0 0 00 00 0"
line.scan(/[0-9._%+-]+\s+[0-9._%+-]+/)
#=> ["00 0", "0 00", "00 0"]
When you use String#split, you pass a regex to match the things that you don't want in your output. That is, the things that should be in between the strings in the output array.
When using RegEx to split a given string, the matched part is removed from the result set. Therefore you cannot use the RegEx syntax to split those numbers and keep the values at the same time.
Use the following code instead:
"00 0 0 00 00 0".scan(/[0-9]+\s[0-9]+/)
I am having a very difficult time with this:
# contained within:
"MA\u008EEIKIAI"
# should be
"MAŽEIKIAI"
# nature of string
$ p string3
"MA\u008EEIKIAI"
$ puts string3
MAEIKIAI
$ string3.inspect
"\"MA\\u008EEIKIAI\""
$ string3.bytes
#<Enumerator: "MA\u008EEIKIAI":bytes>
Any ideas on where to start?
Note: this is not a duplicate of my previous question.
\u008E means that the unicode character with the codepoint 8e (in hex) appears at that point in the string. This character is the control character “SINGLE SHIFT TWO” (see the code chart (pdf)). The character Ž is at the codepoint u017d. However it is at position 8e in the Windows CP-1252 encoding. Somehow you’ve got your encodings mixed up.
The easiest way to “fix” this is probably just to open the file containing the string (or the database record or whatever) and edit it to be correct. The real solution will depend on where the string in question came from and how many bad strings you have.
Assuming the string is in UTF-8 encoding, \u008E will consist of the two bytes c2 and 8e. Note that the second byte, 8e, is the same as the encoding of Ž in CP-1252. On way to convert the string would be something like this:
string3.force_encoding('BINARY') # treat the string just as bytes for now
string3.gsub!(/\xC2/n, '') # remove the C2 byte
string3.force_encoding('CP1252') # give the string the correct encoding
string3.encode('UTF-8') # convert to the desired encoding
Note that this isn’t a general solution to fix all issues like this. Not all CP-1252 characters, when mangled and expressed in UTF-8 this way will amenable to conversion like this. Some will be two bytes c2 xx where xx the correct byte (like in this case), others will be c3 yy where yy is a different byte.
What about using Regexp & String#pack to convert the Unicode escape?
str = "MA\\u008EEIKIAI"
puts str #=> MA\u008EEIKIAI
str.gsub!(/\\u(.{4})/) do |match|
[$1.to_i(16)].pack('U')
end
puts str #=> MA EIKIAI
I am trying to convert a hex value to a binary value (each bit in the hex string should have an equivalent four bit binary value). I was advised to use this:
num = "0ff" # (say for eg.)
bin = "%0#{num.size*4}b" % num.hex.to_i
This gives me the correct output 000011111111. I am confused with how this works, especially %0#{num.size*4}b. Could someone help me with this?
You can also do:
num = "0ff"
num.hex.to_s(2).rjust(num.size*4, '0')
You may have already figured out, but, num.size*4 is the number of digits that you want to pad the output up to with 0 because one hexadecimal digit is represented by four (log_2 16 = 4) binary digits.
You'll find the answer in the documentation of Kernel#sprintf (as pointed out by the docs for String#%):
http://www.ruby-doc.org/core/classes/Kernel.html#M001433
This is the most straightforward solution I found to convert from hexadecimal to binary:
['DEADBEEF'].pack('H*').unpack('B*').first # => "11011110101011011011111011101111"
And from binary to hexadecimal:
['11011110101011011011111011101111'].pack('B*').unpack1('H*') # => "deadbeef"
Here you can find more information:
Array#pack: https://ruby-doc.org/core-2.7.1/Array.html#method-i-pack
String#unpack1 (similar to unpack): https://ruby-doc.org/core-2.7.1/String.html#method-i-unpack1
This doesn't answer your original question, but I would assume that a lot of people coming here are, instead of looking to turn hexadecimal to actual "0s and 1s" binary output, to decode hexadecimal to a byte string representation (in the spirit of such utilities as hex2bin). As such, here is a good method for doing exactly that:
def hex_to_bin(hex)
# Prepend a '0' for padding if you don't have an even number of chars
hex = '0' << hex unless (hex.length % 2) == 0
hex.scan(/[A-Fa-f0-9]{2}/).inject('') { |encoded, byte| encoded << [byte].pack('H2') }
end
Getting back to hex again is much easier:
def bin_to_hex(bin)
bin.unpack('H*').first
end
Converting the string of hex digits back to binary is just as easy. Take the hex digits two at a time (since each byte can range from 00 to FF), convert the digits to a character, and join them back together.
def hex_to_bin(s) s.scan(/../).map { |x| x.hex.chr }.join end