smalltalk - check if 2 linked list have the same elements regardless of order -

is there way check if 2 linked lists have same elements regardless of order.

edit question: have fixed code , given more details:

this method compares 2 lists

 compare: object2      ^ ((mylist asbag) = ((objetc2 getlist) asbag)). 

the method belongs class myclass has field : myllist. mylist linkedlist of type element.

i have compiled in 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 getting false.. seems checks equality reference rather equality value..

so think correct question ask be: how can compare 2 bags value? have tried '=='..but returned false.

edit: question changed too - think deserves new question itself.

the whole problem here (element new id: 1) = (element new id: 1) giving false. unless it's particular class (or superclasses) redefine it, = message resolved comparing identity (==) default. that's why code works collection being compared itself.

test with, example, lists of numbers (which have = method redefined reflect humans understand numeric equality), , work.

you should redefine element's class' = (and hashcode) methods work.

smalltalk handles everything reference: there exist object, know (reference) other objects.

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it wrong 2 lists equivalent if in different order, order is part of list means. list without order call bag.

the asbag message (as of other as<anothercollectiontype> messages) return new collection of named type elements of receiver. so, #(1 2 3 2) array of 4 elements, , #(1 2 3 2) asbag bag containing 4 elements. it's bag, doesn't have particular order.

when baga := bag new. creating new bag instance, , reference baga variable. baga := mylist asbag, lose reference previous bag - first assignment doesn't useful in code, don't use bag.

saying abool iftrue: [^true] iffalse: [^false] has same meaning saying ^abool - prefer that. and, create 2 new bags compare them, simplify whole method this:

compareto: anotherlist   ^ mylist asbag = anotherlist asbag 

read out loud: this object (whatever is) compares to list if it's list without considering order same other list without order.

the name compareto: kind of weird returning boolean (containssameelements: more descriptive), point much faster code.

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just precise questions:

1) doesn't work because you're comparing bag1 , bag2, defined baga , bagb.

2) it's not efficient create 2 bags because, , send senseless iftrue: message, other way it's ok. may implement better way compare lists, it's way better rely on implementation of asbag , bag's = message being performant.

3) think see asbag source code, but, yes, can assume like:

collection>>asbag   |instance|   instance := bag new.   instance addall: self.   ^instance 

and, of course, addall: method be:

collection>>addall: anothercollection   anothercollection do: [ :element | self add: element ] 

so, yes - creates new bag receiver's elements.