CODEWITHSUNDEEP
androidarrayssorting
HydroHeiperGen
HydroHeiperGen

Reputation: 121

Arrays.sort(), how to compare to long values?

I have an Array with 100 objects in it, every object has a long value. I can get the long value by object.getlongValue(). How do I use Comparable to compare those two values? What I did:

public class objects implements Comparable<objects>{

private long longValue;

public object(long longValue){
this.longValue = longValue;
}

public long getLongValue(){
return longValue;
}

@Override
    public int compareTo(object obj) {
        return this.compareTo(obj.getLongValue); //now he can't find compareTo
    }

And later in when I need to sort the Array, I do the following:

Arrays.sort(ListOfObj, new Comparator<objects>() {
                @Override
                public int compare(objects obj, objects obj1) {
                    return obj.getLongValue().compareTo(obj1.getLongValue());
                 // Here he dosen't know "compareTo" either
                }
            });

Of cause this is NOT the real Object class, this is just for debugging.

Upvotes: 0

Views: 733

Answers (1)

S-Sh
S-Sh

Reputation: 3893

long is primitive in Java and has no compareTo method. But, you can box it to Long:

    @Override
    public int compareTo(object obj) 
        return new Long(this.getLongValue()).compareTo(new Long(obj.getLongValue()));
    }

or just use signum function:

 Long.signum(this.getLongValue() - obj.getLongValue());

Upvotes: 1

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