Question

I want to perform a copy and get two different objects so that I can work on the copy without impacting the original.

I have this code (groovy 2.0.5):

def a = [[1,5,2,1,1], ["one", "five", "two", "one", "one"]]
def b = a
b.add([6,6,6,6,6,6])
println a
println b

that produces:

[[1, 5, 2, 1, 1], [one, five, two, one, one], [6, 6, 6, 6, 6, 6]]
[[1, 5, 2, 1, 1], [one, five, two, one, one], [6, 6, 6, 6, 6, 6]]

seems like b and a are actually the same object

I can fix it in this way:

def a = [[1,5,2,1,1], ["one", "five", "two", "one", "one"]]
def b = []

a.each {
    b.add(it)
}

b.add([6,6,6,6,6])
println a
println b

that produces my wanted result:

[[1, 5, 2, 1, 1], [one, five, two, one, one]]
[[1, 5, 2, 1, 1], [one, five, two, one, one], [6, 6, 6, 6, 6]]

But now look at this, where I want the original object and a copy with unique and sorted elements:

def a = [[1,5,2,1,1], ["one", "five", "two", "one", "one"]]
def b = a

b.each {
    it.unique().sort()
}

println a
println b

that produces:

[[1, 2, 5], [five, one, two]]
[[1, 2, 5], [five, one, two]]

If I try the same fix this time it doesn’t work:

def a = [[1,5,2,1,1], ["one", "five", "two", "one", "one"]]
def b = []

a.each { 
    b.add(it)
}

b.each {
    it.unique().sort()
}

println a
println b

that still produces:

[[1, 2, 5], [five, one, two]]
[[1, 2, 5], [five, one, two]]

What’s going on ?

Answer 1

Just calling b = a sets b to being the same instance of the list (containing the same list instances) as a

Calling the second method with the a.each { b.add(it) } means b points to a different instance of List, but the contents of b are the same instances of lists as in a

What you need is something like:

def b = a*.collect()

b.each {
  it.unique().sort()
}

So the a*.collect() makes a new instance of a list for each list in a

You can also do this in one line:

def b = a*.unique( false )*.sort( false ) 

Passing false to unique and sort stop those method changing the original lists, and force them to return new list instances.

_{(Indeed the observant amongst you will notice that the sort does not need false passed to it, as we already have a new instance thanks to unique)}