Escape problem with hex

Question

I need to print escaped characters to a binary file using Ruby. The main problem is that slashes need the whole byte to escape correctly, and I don't know/can't create the byte in such a way.

I am creating the hex value with, basically:

'\x' + char

Where char is some 'hex' value, such as 65. In hex, \x65 is the ASCII character 'e'.

Unfortunately, when I puts this sequence to the file, I end up with this:

\\x65

How do I create a hex string with the properly escaped value? I have tried a lot of things, involving single or double quotes, pack, unpack, multiple slashes, etc. I have tried so many different combinations that I feel as though I understand the problem less now then I did when I started.

How?

Answer 1

If you have the hex value and you want to create a string containing the character corresponding to that hex value, you can do:

irb(main):002:0> '65'.hex.chr
=> "e"

Another option is to use Array#pack; this can be used if you need to convert a list of numbers to a single string:

irb(main):003:0> ['65'.hex].pack("C")
=> "e"
irb(main):004:0> ['66', '6f', '6f'].map {|x| x.hex}.pack("C*")
=> "foo"

Answer 2

You may need to set binary mode on your file, and/or use putc.

File.open("foo.tmp", "w") do |f|
  f.set_encoding(Encoding::BINARY) # set_encoding is Ruby 1.9  
  f.binmode                        # only useful on Windows
  f.putc "e".hex
end

Hopefully this can give you some ideas even if you have Ruby <1.9.

Answer 3

Okay, if you want to create a string whose first byte has the integer value 0x65, use Array#pack

irb> [0x65].pack('U')
#=> "e"
irb> "e"[0]
#=> 101

101₁₀ = 65₁₆, so this works.

If you want to create a literal string whose first byte is '\', second is 'x', third is '6', and fourth is '5', then just use interpolation:

irb> "\\x#{65}"
#=> "\\x65"
irb> "\\x65".split('')
#=> ["\\", "x", "6", "5"]