How does the sort() method in Java work?

[Biniman Idugboe]

Guys,
Bear with me! A newbie is only trying to get a visual cue as to what is going on behind the scene.

public static <T extends Comparable<? super T>> void sort(List<T> list)

Sorts the specified list into ascending order, according to the natural ordering of its elements. All elements in the list must implement the Comparable interface.

My sources of confusion:

Collections.sort(list);

1.  The sort() method has not been overridden. So, at what point is the compareTo() method invoked?
2.  On which element in the list is the compareTo() method invoked; that is (e1.name).compareTo(e2.name).  How are e1 and e2 chosen from the list?
3.  How does the sort() method iterate through the list in order to sort it?

[Piet Souris]

hi Biniman,

1) The Collections class is a subclass of Object. Since Object does not have a 'sort' method, the 'sort' method of Collections can't be an override. The 'compareTo' is invoked any time the sort method wants to compare two elements of the list

2) As said: the compareTo method is invoked any time when two elements must be compared. So, if the elements e1 and e2 are involved, 'sort' invokes 'e1.compareTo(e2)' (and since a decent 'compareTo' method requires that 'e1.compareTo(e2)' is equal to 'e2.compareTo(e1), it doesn't matter if e1. or e2. is being used).

3) The API of 'Collections.sort' does not specify which particular sort algorithm is used. What elements and how many are involved in the sorting, is depending on that algorithm used.

Look at Wiki for some well-known sorting algorithms, and try to implement one of the simpler sort algorithms, using a List of, say, Integers or Strings (these implement the Comparable interface). See what elements you use, and when you invoke the 'compareTo' method.

[Biniman Idugboe]

Thank you. I grab hold of this part of your reply

'sort' invokes 'e1.compareTo(e2)'

Did I hear somebody still saying Java is easy to learn?

[Philip Freeman]

I may be wrong by I remember reading somewhere that for small lists (ones where you do not need to think of BigO) Insertion sort is used while for larger ones quicksort or mergesort. 

[Campbell Ritchie]

Philip Freeman wrote:. . .  Insertion sort is used while for larger ones quicksort or mergesort. 

You shou‍ld look at the documentation, for example for List#sort(), because it tells you which algorithm it uses. Arrays#sort() may be different.

[Pete Letkeman]

Biniman Idugboe wrote:Did I hear somebody still saying Java is easy to learn?

Easy can be a relative term, which can be hard to define as it can change from person to person.

That being said, there are some things in Java which can be done rather quickly, like writing "Hello World" to the screen.
And then there are other things which can be rather difficult to learn like using some of the classes in the collections library.
Then there is also the fact that not everyone has the same level of experience when they start learning something new like Java and not everyone learns the same way.

The big thing here is, 'Do not give up'.
Learning something like Java can be a daunting task to undertake, and even if you do not end up using this skill you will still be the better for learning it.
Learning is the best way to keep your mind active and delay some of the illnesses associated with old age.

[Biniman Idugboe]

I hear you, loud and clear!