Regular expression with back reference

regex, ruby

Solution

Here is what this regular expression means:

regex = /(..) [cs]\1/
#        ├──┘ ├──┘├┘
#        │    │   └─ A reference to whatever was in the first matching group.
#        │    └─ A "character class" matching either "c" or "s".
#        └─ A "matching group" referenced by "\1" containing any two characters.

Note that after matching a regular expression with a matching group, the special variables `$1` (`$2`, etc) will contain what matched.

/(..) [cs]\1/.match('The cat sat in the hat') # => #<MatchData...>
$1 # => "at"

Note also that the `Regexp#match` method returns a MatchData object, which contains the string which caused the entire match ("at sat", aka `$&`) and then each matching group ("at", aka `$1`):

/(..) [cs]\1/.match('The cat sat in the hat')
=> #<MatchData "at sat" 1:"at">

Problem

Can anybody explain how exactly the back reference works in ruby regular expression? I particularly want to know exactly how `(..)` grouping works. For example: ``` s = /(..) [cs]\1/.match("The cat sat in the hat") puts s ``` for the code snippet above, the output is: `at sat`. Why/How is it getting this output ?

Original source