How to avoid accidental mutation of a mutable collection? - scala

Question

So, I wrote a code to compare sorting algorithms. I implemented a few algs as separate classes. Here is a couple for explanation sake.
class BubbleSort(input:ListBuffer[Int]){
...
}
class InsertionSort(input:ListBuffer[Int]){
...
}
From my driver object, I first create a list, called input.
Next, I pass this to new instances of BubbleSort and InsertionSort.
Then I call a function sort, which sorts the ListBuffer I passed during object creation.
val input = new ListBuffer[Int]
//Populated input here
val bs = new BubbleSort(input)
val is = new InsertionSort(input)
bs.sort
is.sort
The issue I'm facing is, when is.sort is called, the list I passed during object creation is already sorted. I guess this is because I called bs.sort first, but why does it affect a list in another class?
Any help is appreciated. If you want more code snippets, I'll be glad to share. Thanks in advance!
EDIT
As a temporary fix, instead of creating only one list input, I resorted to boilerplate code and created two lists,bsinput and isinput. Each is then passed to separate classes.
val bsinput = new ListBuffer[Int]
val isinput = new ListBuffer[Int]
//Populated both arrays with the same data
val bs = new BubbleSort(bsinput)
val is = new InsertionSort(isinput)
bs.sort
is.sort
This solves the problem, but is this the only way? Is there an idiomatic way of doing this?
Edit 2
As #LuisMiguelMejíaSuárez said it was a reference issue. Used input.clone and it works now.

Answer 1

You're passing twice a reference to the same mutable collection ListBuffer.
It means both of your sorting algorithms will be working on the same collection and therefore on the second time, your list buffer will be already sorted.
To avoid accidental mutations you just need to create a defensive copy each time you pass it:
val input1 = new ListBuffer[Int]
//Populated input here
val input2 = input1.clone
val bs = new BubbleSort(input1)
val is = new InsertionSort(input2)