For no particular reason, I am trying to add a #reverse method to the Integer class:
class Integer
def reverse
self.to_s.reverse.to_i
end
end
puts 1337.reverse # => 7331
puts 1000.reverse # => 1
This works fine except for numbers ending in a 0, as shown when 1000.reverse returns 1 rather than 0001. Is there any way to keep leading zeroes when converting a string into an integer?
Short answer: no, you cant.
2.1.5 :001 > 0001
=> 1
0001 doesn't make sense at all as Integer. In the Integer world, 0001 is exactly as 1.
Moreover, the number of leading integer is generally irrelevant, unless you need to pad some integer for displaying, but in this case you are probably converting it into another kind of object (e.g a String).
If you want to keep the integer as Fixnum you will not be able to add leading zeros.
The real question is: why do you want/need leading zeros? You didn't provide such information in the question. There are probably better ways to achieve your result (such as wrapping the value into a decorator object if the goal is to properly format a result for display).
Does rjust work for you?
1000.to_s.reverse.to_i.to_s.rjust(1000.to_s.size,'0') #=> "0001"
self.to_s.to_i does convert the integer to a string and this string "0001" to an integer value. Since leading zeros are not required for regular numbers they are dropped. In other words: Keeping leading zeros does not make sense for calculations, so they are dropped. Just ask yourself how the integer 1 would look like if leading zeros would be preserved, since it represents a 32 bit number. If you need the leading zeros, there is no way around a string.
BUT 10 + "0001".to_i returns 11, so you probably need to override the + method of the String class.
Related
The following code always outputs "not":
print "input a number please. "
TestNumber = gets
if TestNumber % 2 == 0
print "The number is even"
else
print "The number is not even"
end
What is going wrong with my code?
The gets() method returns an object of type String.
When you call %() on a String object, the return value is a new String object (usually it changes the text. You can read more about string formatting here).
Since there are no String objects that == 0, the if/else will always take the same path.
If you want to use the return value of gets() like a number, you will need to transform it into one first. The simplest approach is probably to use the to_i() method on String objects, which returns a new 'Integer' object. If you're doing something where the user input will not always be an integer (e.g. 3.14 or 1.5), you might need to use a different approach.
One last thing: in your example the result of gets() is saved into a constant called TestNumber. Constants are different to normal variables, and they will probably cause problems if you're not using them intentionally. Normal variables don't start with capital letters. (You can read more about ruby variables here). In ruby you need to write you variable names like this: test_number.
I suspect your Testnumber variable might be interpreted as a string during the operation. make sure the testnum is converted to an integer first even if you put in say 100 it could be its being interpreted as the stirng "100" and not the integer 100.
A similar issue can be found here: Ruby Modulo Division
You have to convert TestNumber from string to integer, as your input has linefeed and/or other unwanted characters that do not match an integer.
Use TestNumber = gets.to_i to convert to integer before testing.
I want to create a 32 bit string that I can use as encryption key. This string/key should be derived from a plain text string, e.g.:
'I am a string'
My approach would first be to hash it:
hashed_string = Digest::SHA1.hexdigest('I am a string') # => 'bd82fb0e81ee9f15f5929e0564093bc9f8015f1d'
And then to use just the first 32 characters:
hashed_string[0..31] # => 'bd82fb0e81ee9f15f5929e0564093bc9'
However, I feel there must be a better approach, and I'm not sure if I risk the possibility of 2 input strings yielding similar keys.
What would be a better approach? I have seen this post that touches on truncation, but can't find an answer that appeals to me there.
If you want a string with 32 bits out of your (weak) password :
Digest::SHA1.digest('I am a string').unpack('B32').first
#=> "10111101100000101111101100001110"
The same amount of information can also be displayed with 8 hexadecimal digits :
Digest::SHA1.hexdigest('I am a string')[0,8]
#=> "bd82fb0e"
or 4 ascii chars :
Digest::SHA1.digest('I am a string')[0,4]
#=> "\xBD\x82\xFB\x0E"
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