Ruby character .ord and .chr [duplicate] - ruby

I've been working with the Ruby chr and ord methods recently and there are a few things I don't understand.
My current project involves converting individual characters to and from ordinal values. As I understand it, if I have a string with an individual character like "A" and I call ord on it I get its position on the ASCII table which is 65. Calling the inverse, 65.chr gives me the character value "A", so this tells me that Ruby has a collection somewhere of ordered character values, and it can use this collection to give me the position of a specific character, or the character at a specific position. I may be wrong on this, please correct me if I am.
Now I also understand that Ruby's default character encoding uses UTF-8 so it can work with thousands of possible characters. Thus if I ask it for something like this:
'好'.ord
I get the position of that character which is 22909. However, if I call chr on that value:
22909.chr
I get "RangeError: 22909 out of char range." I'm only able to get char to work on values up to 255 which is extended ASCII. So my questions are:
Why does Ruby seem to be getting values for chr from the extended ASCII character set but ord from UTF-8?
Is there any way to tell Ruby to use different encodings when it uses these methods? For instance, tell it to use ASCII-8BIT encoding instead of whatever it's defaulting to?
If it is possible to change the default encoding, is there any way of getting the total number of characters available in the set being used?

According to Integer#chr you can use the following to force the encoding to be UTF_8.
22909.chr(Encoding::UTF_8)
#=> "好"
To list all available encoding names
Encoding.name_list
#=> ["ASCII-8BIT", "UTF-8", "US-ASCII", "UTF-16BE", "UTF-16LE", "UTF-32BE", "UTF-32LE", "UTF-16", "UTF-32", ...]
A hacky way to get the maximum number of characters
2000000.times.reduce(0) do |x, i|
begin
i.chr(Encoding::UTF_8)
x += 1
rescue
end
x
end
#=> 1112064

After tooling around with this for a while, I realized that I could get the max number of characters for each encoding by running a binary search to find the highest value that doesn't throw a RangeError.
def get_highest_value(set)
max = 10000000000
min = 0
guess = 5000000000
while true
begin guess.chr(set)
if (min > max)
return max
else
min = guess + 1
guess = (max + min) / 2
end
rescue
if min > max
return max
else
max = guess - 1
guess = (max + min) / 2
end
end
end
end
The value input to the method is the name of the encoding being checked.

Related

How to convert bytes in number into a string of characters? (character representation of a number)

How do I easily convert a number, e.g. 0x616263, equivalently 6382179 in base 10, into a string by dividing the number up into sequential bytes? So the example above should convert into 'abc'.
I've experimented with Array.pack but cant figure out how to get it to convert more than one byte in the number, e.g. [0x616263].pack("C*") returns 'c'.
I've also tried 0x616263.to_s(256), but that throws an ArgumentError: invalid radix. I guess it needs some sort of encoding information?
(Note: Other datatypes in pack like N work with the example I've given above, but only because it fits within 4 bytes, so e.g. [0x616263646566].pack("N") gives cdef, not abcdef)
This question is vaguely similar to this one, but not really. Also, I sort of figured out how to get the hex representation string from a character string using "abcde".unpack("c*").map{|c| c.to_s(16)}.join(""), which gives '6162636465'. I basically want to go backwards.
I don't think this is an X-Y problem, but in case it is - I'm trying to convert a number I've decoded with RSA into a character string.
Thanks for any help. I'm not too experienced with Ruby. I'd also be interested in a Python solution (for fun), but I don't know if its right to add tags for two separate programming languages to this question.
To convert a single number 0x00616263 into 3 characters, what you really need to do first is separate them into three numbers: 0x00000061, 0x00000062, and 0x00000063.
For the last number, the hex digits you want are already in the correct place. But for the other two, you have to do a bitshift using >> 16 and >> 8 respectively.
Afterwards, use a bitwise and to get rid of the other digits:
num1 = (0x616263 >> 16) & 0xFF
num2 = (0x616263 >> 8) & 0xFF
num3 = 0x616263 & 0xFF
For the characters, you could then do:
char1 = ((0x616263 >> 16) & 0xFF).chr
char2 = ((0x616263 >> 8) & 0xFF).chr
char3 = (0x616263 & 0xFF).chr
Of course, bitwise operations aren't very Ruby-esque. There are probably more Ruby-like answers that someone else might provide.
64 bit integers
If your number is smaller than 2**64 (8 bytes), you can :
convert the "big-endian unsigned long long" to 8 bytes
remove the leading zero bytes
Ruby
[0x616263].pack('Q>').sub(/\x00+/,'')
# "abc"
[0x616263646566].pack('Q>').sub(/\x00+/,'')
# "abcdef"
Python 2 & 3
In Python, pack returns bytes, not a string. You can use decode() to convert bytes to a String :
import struct
import re
print(re.sub('\x00', '', struct.pack(">Q", 0x616263646566).decode()))
# abcdef
print(re.sub('\x00', '', struct.pack(">Q", 0x616263).decode()))
# abc
Large numbers
With gsub
If your number doesn't fit in 8 bytes, you could use a modified version of your code. This is shorter and outputs the string correctly if the first byte is smaller than 10 (e.g. for "\t") :
def decode(int)
if int < 2**64
[int].pack('Q>').sub(/\x00+/, '')
else
nhex = int.to_s(16)
nhex = '0' + nhex if nhex.size.odd?
nhex.gsub(/../) { |hh| hh.to_i(16).chr }
end
end
puts decode(0x616263) == 'abc'
# true
puts decode(0x616263646566) == 'abcdef'
# true
puts decode(0x0961) == "\ta"
# true
puts decode(0x546869732073656e74656e63652069732077617920746f6f206c6f6e6720666f7220616e20496e743634)
# This sentence is way too long for an Int64
By the way, here's the reverse method :
def encode(str)
str.reverse.each_byte.with_index.map { |b, i| b * 256**i }.inject(:+)
end
You should still check if your RSA code really outputs arbitrary large numbers or just an array of integers.
With shifts
Here's another way to get the result. It's similar to #Nathan's answer, but it works for any integer size :
def decode(int)
a = []
while int>0
a << (int & 0xFF)
int >>= 8
end
a.reverse.pack('C*')
end
According to fruity, it's twice as fast as the gsub solution.
I'm currently rolling with this:
n = 0x616263
nhex = n.to_s(16)
nhexarr = nhex.scan(/.{1,2}/)
nhexarr = nhexarr.map {|e| e.to_i(16)}
out = nhexarr.pack("C*")
But was hoping for a concise/built-in way to do this, so I'll leave this answer unaccepted for now.

How can I increase the number of decimal digits when converting BigDecimal to String?

I am facing a problem with BigDecimal.
This code:
x = BigDecimal.new('1.0') / 7
puts x.to_s
outputs:
0.142857142857142857E0
I want to increase the number of digits.
In JAVA, I could do:
BigDecimal n = new BigDecimal("1");
BigDecimal d = new BigDecimal("7");
n = n.divide(d,200, RoundingMode.HALF_UP);
System.out.println(n);
The output is:
0.1428571428571428571428571428571428571428571428571428571428... (200 digits)
I looked at BigDecimal documentation, and tried to set the digits when instantiating the number, then tried to set the limit with the BigDecimal.limit, but I couldn't print more than 18 digits.
What am I missing?
I am running ruby 1.9.3p0 (2011-10-30) [i386-mingw32] on Windows 7 64bits
The div method allows you to specify the digits:
x = BigDecimal.new('1.0').div( 7, 50 )
puts x
With a result of:
0.14285714285714285714285714285714285714285714285714E0
Despite the internal representation of a big decimal, the to_s method is responsible for converting it to a string. I see to_s supports a format string:
Converts the value to a string.
The default format looks like 0.xxxxEnn.
The optional parameter s consists of either an integer; or an optional ‘+’ or ‘ ’, followed by an optional number, followed by an optional ‘E’ or ‘F’.
If there is a ‘+’ at the start of s, positive values are returned with a leading ‘+’.
A space at the start of s returns positive values with a leading space.
If s contains a number, a space is inserted after each group of that many fractional digits.
If s ends with an ‘E’, engineering notation (0.xxxxEnn) is used.
If s ends with an ‘F’, conventional floating point notation is used.
Examples:
BigDecimal.new('-123.45678901234567890').to_s('5F') -> '-123.45678 90123 45678 9'
BigDecimal.new('123.45678901234567890').to_s('+8F') -> '+123.45678901 23456789'
BigDecimal.new('123.45678901234567890').to_s(' F') -> ' 123.4567890123456789'

UTF-8 Encoding in Ruby using a variable

I am using Ruby 1.8.7 (and upgrading isn't an option). I would like to create a string of all UTF-8 code points from 0 to 127, written as "\uXXXX".
My problem is that this is being interpreted as (for example): 'u0008'. If I try to use '\u0008', the string becomes "\u0008" which IS NOT what I want.
I have tried many different ways, but it seems impossible to create a string that is exactly just "\uXXXX" ie. "\u000B". it always is either "\u000B" or "u000B"
Escaping the '\' isn't an option. I need to send a string to a server, such that the server will receive '\u000B' for example. It is so that other server can test its parsing of the \uXXXX syntax. This seems impossible to do in Ruby however.
Happy if someone can prove me wrong :)
Use Integer #chr to get the character. Here's a clean version:
(1..127).each do |i|
value << "U+#{i} = #{i.chr}, hex = \\x#{"%02x" % i}; "
end
The "%02x" % i is the equal to sprintf("%02x", i). It returns the integer as a 2-digit hexadecimal number.
Escaped output (see comments):
(1..127).each do |i|
value << "U+#{i} = \\u#{"%04x" % i}, hex = \\x#{"%02x" % i}; "
end

Converting a hexadecimal number to binary in ruby

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

Ruby: Fuzzing through all unicode characters ‎(UTF8/Encoding/String Manipulation)

I can't iterate over the entire range of unicode characters.
I searched everywhere...
I am building a fuzzer and want to embed into a url, all unicode characters (one at a time).
For example:
http://www.example.com?a=\uff1c
I know that there are some built tools but I need more flexibility.
If i could do someting like the following: "\u" + "ff1c" it would be great.
This is the closest I got:
char = "\u0000"
...
#within iteration
char.succ!
...
but after the character "\u0039", which is the number 9, I will get "10" instead of ":"
You could use pack to convert numbers to UTF8 characters but I'm not sure if this solves your problem.
You can either create an array with numeric values of all the characters and use pack to get an UTF8 string or you can just loop from 0 to whatever you need and use pack within the loop.
I've written a small example to explain myself. The code below prints out the hex value of each character followed by the character itself.
0.upto(100) do |i|
puts "%04x" % i + ": " + [i].pack("U*")
end
Here's some simpler code, albeit slightly obfuscated, that takes advantage of the fact that Ruby will convert an integer on the right hand side of the << operator to a codepoint. This only works with Ruby 1.8 up for integer values <= 255. It will work for values greater than 255 in 1.9.
0.upto(100) do |i|
puts "" << i
end

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