Is there any way to check if 2 linked lists have the same elements regardless of order.
edit question: I have fixed the code and given some more details:
this is the method that compares 2 lists
compare: object2
^ ((mylist asBag) = ((objetc2 getList) asBag)).
the method belongs to the class myClass that has a field : myLList. myList is a linkedList of type element.
I have compiled it in the workspace:
a: = element new id:1.
b:= element new id:2.
c:=element new id:3.
d: = element new id:1.
e:= element new id:2.
f:=element new id:3.
elements1 := myClass new.
elements addFirst:a.
elements addFirst:b.
elements addFirst:c.
elements2 := myClass new.
elements addFirst:d.
elements addFirst:e.
elements addFirst:f.
Transcript show: (elements1 compare:elements2).
so I am getting false.. seems like it checks for equality by reference rather than equality by value..
So I think the correct question to ask would be: how can I compare 2 Bags by value? I have tried the '=='..but it also returned false.
EDIT: The question changed too much - I think it deserves a new question for itself.
The whole problem here is that (element new id: 1) = (element new id: 1) is giving you false. Unless it's particular class (or superclasses) redefine it, the = message is resolved comparing by identity (==) by default. That's why your code only works with a collection being compared with itself.
Test it with, for example, lists of numbers (which have the = method redefined to reflect what humans understand by numeric equality), and it will work.
You should redefine your element's class' = (and hashCode) methods for this to work.
Smalltalk handles everything by reference: all there exist is an object, which know (reference) other objects.
It would be wrong to say that two lists are equivalent if they are in different order, as the order is part of what a list means. A list without an order is what we call a bag.
The asBag message (as all of the other as<anotherCollectionType> messages) return a new collection of the named type with all the elements of the receiver. So, #(1 2 3 2) is an Array of four elements, and #(1 2 3 2) asBag is a bag containing those four elements. As it's a Bag, it doesn't have any particular order.
When you do bagA := Bag new. you are creating a new Bag instance, and reference it with bagA variable. But then you do bagA := myList asBag, so you lose the reference to the previous bag - the first assignment doesn't do anything useful in your code, as you don't use that bag.
Saying aBool ifTrue: [^true] ifFalse: [^false] has exactly the same meaning as saying ^aBool - so we prefer just to say that. And, as you only create those two new bags to compare them, you could simplify your whole method like this:
compareTo: anotherList
^ myList asBag = anotherList asBag
Read it out loud: this object (whatever it is) compares to another list if it's list without considering order is the same than the other list without order.
The name compareTo: is kind of weird for returning a boolean (containsSameElements: would be more descriptive), but you get the point much faster with this code.
Just to be precise about your questions:
1) It doesn't work because you're comparing bag1 and bag2, but just defined bagA and bagB.
2) It's not efficient to create those two extra bags just because, and to send the senseless ifTrue: message, but other way it's OK. You may implement a better way to compare the lists, but it's way better to rely on the implementation of asBag and the Bag's = message being performant.
3) I think you could see the asBag source code, but, yes, you can assume it to be something like:
Collection>>asBag
|instance|
instance := Bag new.
instance addAll: self.
^instance
And, of course, the addAll: method could be:
Collection>>addAll: anotherCollection
anotherCollection do: [ :element | self add: element ]
So, yes - it creates a new Bag with all the receiver's elements.