Comparison method violates its general contract Exception

Question

Below is a block of code that results in exception as indicated,

Code :

Collections.sort( arrayList, new Comparator() 
{
    public int compare( Object o1, Object o2 )
    {
        TypeAdapterSort tas1 = ( TypeAdapterSort ) o1;
        TypeAdapterSort tas2 = ( TypeAdapterSort ) o2;
        if ( tas1.order < tas2.order )
            return -1;
        else
            return 1;
    }
} );

Exception :

java.lang.IllegalArgumentException: Comparison method violates its general contract!
                at java.util.TimSort.mergeLo(TimSort.java:747)
                at java.util.TimSort.mergeAt(TimSort.java:483)
                at java.util.TimSort.mergeForceCollapse(TimSort.java:426)
                at java.util.TimSort.sort(TimSort.java:223)
                at java.util.TimSort.sort(TimSort.java:173)
                at java.util.Arrays.sort(Arrays.java:659)
                at java.util.Collections.sort(Collections.java:217)

When I run the same code as a standalone program, the issue never occurs. What is the issue with the comparator here? Is there a way to reproduce the issue in a standalone code?

This issue occurs only on Java 1.7 as there has been change in the implementation on Arrays.sort & Collections.sort. How to change the above code to avoid the issue?. Also, how to reproduce this issue in a standalone code?

Arrangement

if order is of type Integer you could return tas1.order.compareTo(tas2.order) – Analyzer

what do you mean by 'standalone program'? Or rather, what other option of running the code you are referring to? – Wb

possible duplicate of "Comparison method violates its general contract!" – Interfile

You already asked this question: compare method violates its general contract exception java 7 – Benjie

Answer 1

You need to return 0 on equal objects.

        if ( tas1.order < tas2.order ){
            return -1;
        } else if ( tas1.order == tas2.order ){
            return 0;
        } else {
            return 1;
        }

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Deluxe

But, why is the issue not reproduced when run as a standalone program where the value of intance variable order is maintained the same? – Arrangement

compare must be transitive. Your code isn't. So instead of working incorrectly, it states to you that your code is buggy. I think it is a very good behavior. – Deluxe

@baraky You might expect to write tas1.order == tas2.order, you surely missed the double "=" instead of ... assignment ;) – Palinode

Answer 2

In your comparator, for A == B, compare(A,B) returns 1 (A>B), but compare(B,A) still returns 1 (meaning B>A). If the sorting algorithm detects this contradiction, it throws an exception instead of maybe endlessly looping or recursing.

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