$1 and \1 in Ruby

When using regular expressions in Ruby, what is the difference between $1 and \1?

From stackoverflow

• \1 is a backreference which will only work in the same sub or gsub method call, e.g.:

  "foobar".sub(/foo(.*)/, "\1\1") # => "barbar"

  $1 is a global variable which can be used in later code:

  if "foobar" =~ /foo(.*)/ then 
    puts "The matching word was #{$1}"
  end
  

  (prints "The matching word was bar")

• Keep in mind there's a third option, the block form of sub. Sometimes you need it. Say you want to replace some text with the reverse of that text. You can't use $1 because it's not bound quickly enough:

  "foobar".sub(/(.*)/, $1.reverse)  # WRONG: either uses a PREVIOUS value of $1, 
                                    # or gives an error if $1 is unbound
  

  You also can't use \1, because the sub method just does a simple text-substitution of \1 with the appropriate captured text, there's no magic taking place here:

  "foobar".sub(/(.*)/, '\1'.reverse) # WRONG: returns '1\'
  

  So if you want to do anything fancy, you should use the block form of sub ($1, $2, $`, $' etc. will be available):

  "foobar".sub(/.*/){|m| m.reverse} # => returns 'raboof'
  "foobar".sub(/(...)(...)/){$1.reverse + $2.reverse} # => returns 'oofrab'
  

  rampion : Your example could be misleading - the match is what's passed to the block, not the matchgroups. So, if you wanted to change "foobar" to "foorab", you'd have to do `str.sub(/(foo)(\w+)/) { $1 + $2.reverse }`

  rampion : See ri String#sub: In the block form, the current match string is passed in as a parameter, and variables such as $1, $2, $`, $&, and $' will be set appropriately. The value returned by the block will be substituted for the match on each call.

  Brian Carper : Right, I'll edit to clear it up.

  Adrian : You saved my day!

• That "\1\1" in the main answer should be '\1\1'