Ruby正则表达式捕获gsub中的组

Question

Say I want to switch each letter in a message with it's place in the reverse alphabet. Why can't I seem to use the captured group and do it in one gsub?

Perhaps someone could explain in general about using captured groups in gsub, can the back references be bare(no ' ')? Can I use #{\\1}?

def decode(message)
  a = ('a'..'z').to_a
  z = a.reverse
  message.gsub!(/([[:alpha:]])/, z[a.index('\1')]) 
end

decode("the quick brown fox")

Answer 1

Remember that arguments to methods are evaluated immediately and the result of that is passed in to the method. If you want to make the substitution adapt to the match:

message.gsub!(/([[:alpha:]])/) { |m| z[a.index($1)] }

That employs a block that gets evaluated for each match.

Answer 2

Using gsub :

Your code was not working because '\\1' is not yet being evaluated as its regex match, at the point you desire. This can be solved by using a block, so the match variable is defined:

message.gsub(/[[:alpha:]]/) { |char| z[a.index(char)] }

Using tr :

A more efficient way to solve problems like this, where you are simply "replacing one set of characters with another set", is to instead use String#tr . This could be done as follows:

message.tr(a.join(''), z.join(''))

Answer 3

One way is to manipulate ASCII values.

def code(message)
  message.gsub(/[[:alpha:]]/) { |s| (((s < 'a') ? 155 : 219 ) - s.ord).chr }
end

coded = code("the quick brown fox")
  #=> "gsv jfrxp yildm ulc" 
code(coded)
  #=> "the quick brown fox" 

Note:

'A'.ord + 'Z'.ord
  #=> 155 
'a'.ord + 'z'.ord
  #=> 219 

Another is to use a hash:

a = ('a'..'z').to_a
Ch = a.zip(a.reverse).to_h
  #=> {"a"=>"z", "b"=>"y",..., "y"=>"b", "z"=>"a"} 

def code(message)
  message.gsub(/[[:alpha:]]/, Ch)
end

coded = code("the quick brown fox")
  #=> "gsv jfrxp yildm ulc"
code(coded)
  #=> "the quick brown fox"