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1350

answers:

3
+6  Q: 

$1 and \1 in Ruby

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

+13  A: 

\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")

Avdi
+6  A: 

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'
Brian Carper
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
rampion
Right, I'll edit to clear it up.
Brian Carper
You saved my day!
Adrian
A: 

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

David Madison