Groovy syntax for regular expression matching

m[0] is the first match object.
m[0][0] is everything that matched in this match.
m[0][1] is the first capture in this match.
m[0][2] is the second capture in this match.

Based on what I have read (I don't program in Groovy or have a copy handy), given

def m = "barbaz" =~ /(ba)([rz])/;

m[0][0] will be "bar"
m[0][1] will be "ba"
m[0][2] will be "r"
m[1][0] will be "baz"
m[1][1] will be "ba"
m[1][2] will be "z"

I could not stand not knowing if I was correct or not, so I downloaded groovy and wrote an example:

def m = "barbaz" =~ /(ba)([rz])/;

println "m[0][0] " + m[0][0]
println "m[0][1] " + m[0][1]
println "m[0][2] " + m[0][2]
println "m[1][0] " + m[1][0]
println "m[1][1] " + m[1][1]
println "m[1][2] " + m[1][2]

This was the closest match to the Perl code that I could achieve:

def txt = "abc : groovy : def"
if ((m = txt =~ / : (.+?) : /)) {
  def match = m.group(1)
  println "MATCH=$match"
}

// Prints:
// MATCH=groovy