SortedSet<TestClass> comparing on equality one field and sorting by another

Question

Please review code:

/* Run1.java */
package test;

import java.util.Iterator;
import java.util.SortedSet;
import java.util.TreeSet;

public class Run1 
{
    static public void main(String[] args)
    {
        SortedSet<TestClass> s = new TreeSet<TestClass>(); 

        s.add( new TestClass("name1", 100) );
        s.add( new TestClass("name2", 10) );
        s.add( new TestClass("name3", 1) );
        s.add( new TestClass("name4", 10) );
        s.add( new TestClass("name5", 100) );

        Iterator<TestClass> it = s.iterator();

        while(it.hasNext())
        {
            TestClass t = it.next();
            System.out.println( t.name+' '+t.value );
        }
    }
}

/* TestClass.java */
package test;

public class TestClass implements Comparable<TestClass> 
{
    public String name;
    public int value;

    public TestClass(String name, int value) {
        this.name = name;
        this.value = value;
    }

    public int compareTo(TestClass o) 
    {
        return this.value - o.value;
    }

    public boolean equals(Object o) 
    {
        if (!(o instanceof TestClass))
            return false;
        TestClass n = (TestClass)o;

        return this.name.equals(n.name);
    }

    public int hashCode() 
    {
        return 31*name.hashCode();
    }

    public String toString() 
    {
        return name;
    }
}

Print out

name3 1
name2 10
name1 100

as i see because compareTo used for checking to equality (when returned 0). But i need check for unique by field TestClass.name and only sort by TestClass.value

Answer 1

The result of compareTo() and equals() need to be compatible in this case, which means that you need to take into account equality in the comparison. For example:

public int compareTo(TestClass o) 
{
    return (this.value == o.value) ? this.name.compareTo(o.name) : this.value - o.value;
}

which introduces a sub-order on name for objects with the same value, making the result compatible with your equals() implementation.