I'm trying to write a function that will .unpack('Q') (unpack to uint64_t) without access to the unpack method.
When I manually convert from string to binary to uint64, I get a different result than .unpack('Q'):
Integer('abcdefgh'.unpack('B*').first, 2) # => 7017280452245743464
'abcdefgh'.unpack('Q').first # => 7523094288207667809
I don't understand what's happening here.
I also don't understand why the output of .unpack('Q') is fixed regardless of the size of the input. If I add a thousand characters after 'abcdefgh' and then unpack('Q') it, I still just get [7523094288207667809]?
Byte order matters:
Integer('abcdefgh'.
each_char.
flat_map { |c| c.unpack('B*') }.
reverse.
join, 2)
#⇒ 7523094288207667809
'abcdefgh'.unpack('Q*').first
#⇒ 7523094288207667809
Your code produces the wrong result because after converting to binary, bytes should be reversed.
For the last part of your question, the reason the output of .unpack('Q') doesn't change with a longer input string is because the format is specifying a single 64-bit value so any characters after the first 8 are ignored. If you specified a format of Q2 and a 16 character string you'd decode 2 values:
> 'abcdefghihjklmno'.unpack('Q2')
=> [7523094288207667809, 8029475498074204265]
and again you'd find adding additional characters wouldn't change the result:
> 'abcdefghihjklmnofoofoo'.unpack('Q2')
=> [7523094288207667809, 8029475498074204265]
A format of Q* would return as many values as multiples of 64-bits were in the input:
> 'abcdefghihjklmnopqrstuvw'.unpack('Q*')
=> [7523094288207667809, 8029475498074204265, 8608196880778817904]
> 'abcdefghihjklmnopqrstuvwxyz'.unpack('Q*')
=> [7523094288207667809, 8029475498074204265, 8608196880778817904]
Related
I was trying to convert the bytesize to hex string in ruby.
I tried various combinations of pack/unpack/chr/hex to no avail.
Here is what I managed to comeup with,
"1".bytesize.chr.to_s.split.pack('a4') => "\x01\x00\x00\x00"
As you know this wont work for larger strings, for example,
"11111111".bytesize.chr().to_s.split.pack('a4') => "\b\x00\x00\x00"
where first byte is expected to be \x0B, and
"testtttttttttttttttttttttttttttttttttttttttttt".bytesize.chr.to_s.split.pack('a4')
# => ".\x00\x00\x00"
which is wrong.
i.e, just the opposite of this, Unpack 4 bytes to a signed integer, where, if the <string>.bytesize is 4294967295, it should be "\xFF\xFF\xFF\xFF".
I'm trying to decode this file that is in IBM437 into readable UTF I'm at the point where I think I've almost got it but I'm getting an ArgumentError where the string contains nul bytes, I'm aware of how to gsub out nul bytes using:
.gsub("\u0000", '') however I can't figure out where to gsub the bytes out.
Here's the source:
def gather_info
file = './lib/SETI_message.txt'
File.read(file).each_line do |gather|
packed = [gather].pack('b*')
ec = Encoding::Converter.new(packed, 'utf-8')
encoding_forced = packed.encode(ec)
File.open('packed.txt', 'a+'){ |s| s.puts(encoding_forced.gsub("\u0000", '')) }
end
end
gather_info
And here's the file
Can anyone tell me what I'm doing wrong here?
The following works for me :
file = File.read('SETI.txt')
packed = file.scan(/......../).map{|s| s.to_i(2)}.pack('U*')
File.write('packed.txt', packed)
Let's break file.scan(/......../).map{|s| s.to_i(2)}.pack('U*') down :
file.scan(/......../)
Here we break the huge string of 0s and 1s (the file) into an array of strings containing 8 characters each. It looks like that : ['00001111', '11110000', ...].
arr.map{|s| s.to_i(2)}
From step 1 we got an array of strings representing the different characters in binary notation. We can convert one of those strings (called s) by applying s.to_i(2) because the parameter '2' says to the method to_i to use base 2. So '00000011'.to_i(2) returns 3.
We apply this to all the characters by using map.
So we now have an array that looks like [98, 82, 49, 39, ...].
arr.pack('U*')
From step 2 we have an array of integers representing each a character. We can now use the pack method to transform our array of integers into a string. The parameter we use for pack is U to tell him that the integers are in fact UTF-8 characters.
I have a data structure that I'd like to convert back and forth from hex to binary in Ruby. The simplest approach for a binary to hex is '0010'.to_i(2).to_s(16) - unfortunately this does not preserve leading zeroes (due to the to_i call), as one may need with data structures like cryptographic keys (which also vary with the number of leading zeroes).
Is there an easy built in way to do this?
I think you should have a firm idea of how many bits are in your cryptographic key. That should be stored in some constant or variable in your program, not inside individual strings representing the key:
KEY_BITS = 16
The most natural way to represent a key is as an integer, so if you receive a key in a hex format you can convert it like this (leading zeros in the string do not matter):
key = 'a0a0'.to_i(16)
If you receive a key in a (ASCII) binary format, you can convert it like this (leading zeros in the string do not matter):
key = '101011'.to_i(2)
If you need to output a key in hex with the right number of leading zeros:
key.to_s(16).rjust((KEY_BITS+3)/4, '0')
If you need to output a key in binary with the right number of leading zeros:
key.to_s(2).rjust(KEY_BITS, '0')
If you really do want to figure out how many bits might be in a key based on a (ASCII) binary or hex string, you can do:
key_bits = binary_str.length
key_bits = hex_str.length * 4
The truth is, leading zeros are not part of the integer value. I mean, it's a little detail related to representation of this value, not the value itself. So if you want to preserve properties of representation, it may be best not to get to underlying values at all.
Luckily, hex<->binary conversion has one neat property: each hexadecimal digit exactly corresponds to 4 binary digits. So assuming you only get binary numbers that have number of digits divisible by 4 you can just construct two dictionaries for constructing back and forth:
# Hexadecimal part is easy
hex = [*'0'..'9', *'A'..'F']
# Binary... not much longer, but a bit trickier
bin = (0..15).map { |i| '%04b' % i }
Note the use of String#% operator, that formats the given value interpreting the string as printf-style format string.
Okay, so these are lists of "digits", 16 each. Now for the dictionaries:
hex2bin = hex.zip(bin).to_h
bin2hex = bin.zip(hex).to_h
Converting hex to bin with these is straightforward:
"DEADBEEF".each_char.map { |d| hex2bin[d] }.join
Converting back is not that trivial. I assume we have a "good number" that can be split into groups of 4 binary digits each. I haven't found a cleaner way than using String#scan with a "match every 4 characters" regex:
"10111110".scan(/.{4}/).map { |d| bin2hex[d] }.join
The procedure is mostly similar.
Bonus task: implement the same conversion disregarding my assumption of having only "good binary numbers", i. e. "110101".
"I-should-have-read-the-docs" remark: there is Hash#invert that returns a hash with all key-value pairs inverted.
This is the most straightforward solution I found that preserves leading zeros. 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
I have a binary string, say
x = "c1\x98\xCCf3\x1C\x00.\x01\xC7\x00\xC0"
(actually much longer). I need to have it represented as Bignum, for the purposes of further conversion to base-something sequences (something > 36).
x.unpack('H*')[0].to_i
yields an Integer from first bytes of the value, and not a Bignum.
There's no need to use unpack and go through an intermediate hex string representation.
To convert a binary string directly to a number (which will automatically be a Bignum as needed), you can do:
"\xc1\x98\xCC\xf3\x1C\x00".bytes.inject {|a, b| (a << 8) + b }
=> 212862017674240
The default base for String#to_i is, of course, 10 but you're trying to convert hex so you want .to_i(16). If you don't specify the base, to_i will stop when it sees the first non-decimal value and that's where your truncation comes from.
You want to say this:
x.unpack('H*')[0].to_i(16)
For example:
>> "633198cc66331c0001c700c0633198cc66331c0001c700c063312e98cc66331c0001c700c0".to_i
=> 633198
>> "633198cc66331c0001c700c0633198cc66331c0001c700c063312e98cc66331c0001c700c0".to_i(16)
=> 49331350698902676183344474146684368690988113012187221237314170009285390086987127695278272
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