From: Mario Ruiz <lists@...>
Date: 2013-02-22T18:27:01+09:00
Subject: Re: backslash substitution

I found this solution... but i thought that Ruby had a better way:


x="xxxxx P jjjjjj\0\1\2\3.4.1\4\5\6\7\8\9"
m=""
x.each_byte {|a|
  puts a
  if a<32 then
    m+="/" + a.to_s()
  else
    m+=a.chr
  end
}
puts m

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