Java: compareTo not sorting correctly

Question

I have this compareTo code for my list:

public int compareTo(className a) 
{               
    return (this.long1 > a.long1) ? 1 : -1;
}

When I use Collections.sort(list), I get the following error: Comparison method violates its general contract!

When I change it to if (this.long1 >= a.long2), it works, but it does not sort properly. The longs are sorted in order, then out of order, then in order. Using >=, the output looks like this:

Now, duplicates do happen, and they need to be sorted correctly. It doesn't matter if the duplicates appear first or last, as long as they're properly grouped in order, like so:

How would I do this properly? Thank you.

UPDATE

The list is still being sorted out of order with all of the below suggestions. Is this because it's a List<Class> list = new ArrayList<Class>();? I can't use what I'm used to from C#: List<Class> list = new List<Class>().

You do not handle equality, in that case, you need to return `0`. — meskobalazs, Oct 27 '15 at 11:49
Google your error, there're many posts that solve your problem. — Maroun, Oct 27 '15 at 11:50
@MarounMaroun, not really. None of these suggestions are working. — Mark Buffalo, Oct 27 '15 at 11:59

score 9 · Accepted Answer · answered Oct 27 '15 at 11:49

9

You should return 0 when the two numbers are equal, or just use Long.compare :

public int compareTo(className a) 
{               
    return Long.compare(this.long1,a.long1);
}

answered Oct 27 '15 at 11:49

Eran

387,369
54
702
768

1

I like this one. `Long.compare()`. – Mark Buffalo Oct 27 '15 at 11:51
I'd like to state that none of these suggestions are working, including all the information in the `duplicate` post, but since the post was marked as duplicate, there isn't much I can do. Still getting the same defect: list is sorted out of order. – Mark Buffalo Oct 27 '15 at 12:09
1

@EduardoScissaro You can still edit your question. Add relevant code, inputs and outputs. – Eran Oct 27 '15 at 12:10
Output is the same. Code is the same. I'm going to keep at it for a while. – Mark Buffalo Oct 27 '15 at 12:12
1

@EduardoScissaro You can show how you initialize your list, you can show the code of your class that implements `compareTo`. Perhaps there's something there that would help solve your problem. – Eran Oct 27 '15 at 12:14
Agreed. I thought it was an issue with `compareTo`, but it's much deeper than that. I'm going to try and solve it myself, though. – Mark Buffalo Oct 27 '15 at 12:18
1

Found the problem, which was totally my fault, somewhere else deep in the code, and it was messing with `compareTo`. While this doesn't answer the problem I was having, it is indeed correct when removing the code that shouldn't even be interacting with this in the first place. Technically, this answer is still correct... therefore, +1 and check. – Mark Buffalo Oct 27 '15 at 14:10

score 4 · Answer 2 · answered Oct 27 '15 at 11:49

4

return (this.long1 > a.long1) ? 1 : -1;

If both numbers are equal, this returns -1, not 0.

return (this.long1 >= a.long1) ? 1 : -1;

Now 1 is equal, still not 0.
Correct:

if(this.long1 > a.long1) return 1;
if(this.long1 < a.long1) return -1;
return 0;

answered Oct 27 '15 at 11:49

deviantfan

11,268
3
32
49

1

Although this is not incorrect but i would use something stackoverflow.com/a/33367246/1085186 as it uses library code and more readable. – StackFlowed Oct 27 '15 at 11:51

score 3 · Answer 3 · answered Oct 27 '15 at 11:50

3

A correct solution would do this:

public int compareTo(className a) {               
    return this.long1 > a.long1 ? 1 : this.long1 < a.long1 ? -1 : 0;
}

answered Oct 27 '15 at 11:50

meskobalazs

15,741
2
40
63

I just added the same solution, but in a different order. :p – Mark Buffalo Oct 27 '15 at 11:50
1

Although this is not incorrect but i would use something http://stackoverflow.com/a/33367246/1085186 as it uses library code and more readable. – StackFlowed Oct 27 '15 at 11:51

Java: compareTo not sorting correctly

3 Answers3