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    <p>Since you say <code>distance</code> is a double, you likely have the same problem as described here:</p> <p><a href="https://stackoverflow.com/questions/9469308/java-error-comparison-method-violates-its-general-contract">Java error: &quot;Comparison method violates its general contract!&quot;</a></p> <p>Perhaps:</p> <pre><code>public int compareTo(Object o) { Match m = (Match)o; int diff = m.matches - matches; if (diff == 0) { return Double.compare(distance, m.distance); } else { return diff; } } </code></pre> <p><strong><em>However</em></strong> ideally you should use the built-in comparison methods as I state below. The above code is an example of the "minimum change needed", illustrating the key problem.</p> <h2><strong>Using existing comparison methods</strong></h2> <p>Also, as @fabian-barney states in his answer, you should avoid taking the direct difference, and instead utilise the built-in comparison methods. So you should have something like:</p> <pre><code>public int compareTo(Object o) { Match m = (Match) o; return m.matches == matches ? Double.compare(m.distance, distance) : Integer.compare(m.matches, matches); } </code></pre> <p>This way, <code>Double.compare</code> will handle the NaN values for you. For any number <code>x</code> (other than NaN) <code>Double.compare(x, Double.NaN) == -1</code> will return true (i.e., NaN is considered greater than any other number).</p> <p>Note here that you are OK using <code>==</code> with <code>int</code>s but it is more complicated with <code>double</code> because <code>Double.NaN != Double.NaN</code>. However, <code>new Double(Double.NaN).equals(Double.NaN)</code> is true. See <a href="https://stackoverflow.com/questions/1726254/why-is-javas-double-comparedouble-double-implemented-the-way-it-is">Why is Java&#39;s Double.compare(double, double) implemented the way it is?</a> for a nice discussion.</p> <h2><strong>Contract breaking:</strong></h2> <p>To see an example of why your original implementation might break the contract if you have NaNs, see <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/Comparable.html#compareTo%28T%29" rel="nofollow noreferrer">Java compareTo documentation</a>. There we have:</p> <blockquote> <p>Finally, the implementer must ensure that x.compareTo(y)==0 implies that sgn(x.compareTo(z)) == sgn(y.compareTo(z)), for all z.</p> </blockquote> <p>So imagine you have <code>x = NaN</code> and <code>y = 5</code> and <code>z = 6</code>, then:</p> <ol> <li><code>x.compareTo(y) == 0</code> (since <code>NaN &gt; 5</code> and <code>NaN &lt; 5</code> are false)</li> <li><code>x.compareTo(z) == 0</code> (same reasoning)</li> <li><code>y.compareTo(z) == -1</code> (y &lt; z).</li> </ol> <p>So 2 and 3 (+<code>sgn</code>) are not equal as required.</p>
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    1. COIndeed, this does work, though I'm not sure why - or where I had gotten any NaNs in my list.
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    2. COI've updated my answer with an example of where the contract is broken if NaNs are present. I can't say where they'd be present in your code or is that is definitely the issue :(
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    3. COYou were right - there are NaNs in the calculated values for the distance: http://ideone.com/R9lTB
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