# Sets

Neil Muya

Ranch Hand

Posts: 30

posted 7 years ago

Hi ranchers this is a code from ExamLab just edited a part to make it work.

I know that :

negative if objOne < objTwo

zero if objOne == objTwo

positive if objOne > objTwo

Here's my question:

how does the set arrange the elements?

Because as of now what I just know is:

if you compare the "1st argument to the 2nd argument" in compare method you get an natural order, or you get a duplicate

if you compare the "2nd argument to the 1st argument" in compare method you get an reverse order , or you get a duplicate

In this method does the set insert the 1st element I added comparing to the second element?

and if it gets a "1"?it puts "ob1" to cs[0]?

sorry I am kinda confused.

Thanks in advance!

I know that :

negative if objOne < objTwo

zero if objOne == objTwo

positive if objOne > objTwo

Here's my question:

how does the set arrange the elements?

Because as of now what I just know is:

if you compare the "1st argument to the 2nd argument" in compare method you get an natural order, or you get a duplicate

if you compare the "2nd argument to the 1st argument" in compare method you get an reverse order , or you get a duplicate

In this method does the set insert the 1st element I added comparing to the second element?

and if it gets a "1"?it puts "ob1" to cs[0]?

sorry I am kinda confused.

Thanks in advance!

Ralph Jaus

Ranch Hand

Posts: 342

posted 7 years ago

Hi Neil,

to simplify writing I set a1 = new A(1) and respectively a2, a3.

Now, when cs.add(a1) is performed, the objects a1 and a3 are compared. Actually it doesn't matter if compare(a1,a3) or

compare(a3,a1) is inoked, since in any case the comparison yields a1 < a3. For:

compare(a1,a3) yields -1, meaning a1 < a3.

compare(a3,a1) yields +1, meaning a3 > a1, which is equivalent to a1 < a3.

So after adding a1, the order of cs is {a1,a3}.

When adding a2, the three elements a1, a2, a3 are compared, yielding the order {a1,a2,a3}.

Hope it helps.

to simplify writing I set a1 = new A(1) and respectively a2, a3.

Now, when cs.add(a1) is performed, the objects a1 and a3 are compared. Actually it doesn't matter if compare(a1,a3) or

compare(a3,a1) is inoked, since in any case the comparison yields a1 < a3. For:

compare(a1,a3) yields -1, meaning a1 < a3.

compare(a3,a1) yields +1, meaning a3 > a1, which is equivalent to a1 < a3.

So after adding a1, the order of cs is {a1,a3}.

When adding a2, the three elements a1, a2, a3 are compared, yielding the order {a1,a2,a3}.

Hope it helps.

SCJP 5 (98%) - SCBCD 5 (98%)