Comparing the values of two generic Numbers

A working (but brittle) solution is something like this:

class NumberComparator implements Comparator<Number> {

    public int compare(Number a, Number b){
        return new BigDecimal(a.toString()).compareTo(new BigDecimal(b.toString()));
    }

}

It's still not great, though, since it counts on toString returning a value parsable by BigDecimal (which the standard Java Number classes do, but which the Number contract doesn't demand).

Edit, seven years later: As pointed out in the comments, there are (at least?) three special cases toString can produce that you need to take into regard:

• Infinity, which is greater than everything, except itself to which it is equal
• -Infinity, which is less than everything, except itself to which it is equal
• NaN, which is extremely hairy/impossible to compare since all comparisons with NaN result in false, including checking equality with itself.

• it's not likely but it can cause NumberFormatException if I create a subclass of Number and don't override toString in a proper way. - Roman
• Hopefully the programmer will know if any descendants of Number other than the standard ones might be encountered. - DJClayworth
• Zeeks! This is a really bad solution. Though I can't off hand think of a better one, so I won't downvote yet. :-( - user949300
• @user949300 As I said, it's a working (but brittle) solution which is not great, though - so, I agree it's a bad solution! But the whole premise is flawed, generally one shouldn't work with arbitrary Numbers. Best thing here probably would've been constraining the type parameter further, to T extends Number & Comparable<T>. - gustafc
• "toString returning a value parsable by BigDecimal (which the standard Java Number classes do" That is not entirely true, given that Double and Float can return 3 special values (NaN, Infinity, and -Infinity) that are not supported by BigDecimal. - Andreas

This should work for all classes that extend Number, and are Comparable to themselves. By adding the & Comparable you allow to remove all the type checks and provides runtime type checks and error throwing for free when compared to Sarmun answer.

class NumberComparator<T extends Number & Comparable> implements Comparator<T> {

    public int compare( T a, T b ) throws ClassCastException {
        return a.compareTo( b );
    }
}

• Is there any drawback with this answer? Otherwise it should probably be chosen as best answer. - Campa
• It requires the source collection to be a specific type, so it won't work for abstract Number class. - RoninCoder

After having asked a similar question and studying the answers here, I came up with the following. I think it is more efficient and more robust than the solution given by gustafc:

public int compare(Number x, Number y) {
    if(isSpecial(x) || isSpecial(y))
        return Double.compare(x.doubleValue(), y.doubleValue());
    else
        return toBigDecimal(x).compareTo(toBigDecimal(y));
}

private static boolean isSpecial(Number x) {
    boolean specialDouble = x instanceof Double
            && (Double.isNaN((Double) x) || Double.isInfinite((Double) x));
    boolean specialFloat = x instanceof Float
            && (Float.isNaN((Float) x) || Float.isInfinite((Float) x));
    return specialDouble || specialFloat;
}

private static BigDecimal toBigDecimal(Number number) {
    if(number instanceof BigDecimal)
        return (BigDecimal) number;
    if(number instanceof BigInteger)
        return new BigDecimal((BigInteger) number);
    if(number instanceof Byte || number instanceof Short
            || number instanceof Integer || number instanceof Long)
        return new BigDecimal(number.longValue());
    if(number instanceof Float || number instanceof Double)
        return new BigDecimal(number.doubleValue());

    try {
        return new BigDecimal(number.toString());
    } catch(final NumberFormatException e) {
        throw new RuntimeException("The given number (\"" + number + "\" of class " + number.getClass().getName() + ") does not have a parsable string representation", e);
    }
}

• Note that BigDecimal.valueOf should be used instead of the constructor ESPECIALLY for the Float/Double creation or you'll get base2 floating point errors in the conversion. - Eric
• Thanks for your comment! I'm not sure your point is valid though. The Javadoc of the double constructor says: "[...], note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor." It is true that the exact conversion often yields unexpected results (e.g. for 0.1 literals). But since we don't know anything about the numbers' origins, I think the only sensible way to handle them is to assume nothing about it. - rolve

One solution that might work for you is to work not with T extends Number but with T extends Number & Comparable. This type means: "T can only be set to types that implements both the interfaces."

That allows you to write code that works with all comparable numbers. Statically typed and elegant.

This is the same solution that BennyBoy proposes, but it works with all kinds of methods, not only with comparator classes.

public static <T extends Number & Comparable<T>> void compfunc(T n1, T n2) {
    if (n1.compareTo(n2) > 0) System.out.println("n1 is bigger");
}

public void test() {
    compfunc(2, 1); // Works with Integer.
    compfunc(2.0, 1.0); // And all other types that are subtypes of both Number and Comparable.
    compfunc(2, 1.0); // Compilation error! Different types.
    compfunc(new AtomicInteger(1), new AtomicInteger(2)); // Compilation error! Not subtype of Comparable
}

This should work for all classes that extend Number, and are Comparable to themselves.

class NumberComparator<T extends Number> implements Comparator<T> {

    public int compare(T a, T b){
        if (a instanceof Comparable) 
            if (a.getClass().equals(b.getClass()))
                return ((Comparable<T>)a).compareTo(b);        
        throw new UnsupportedOperationException();
    }
}

The most "generic" Java primitive number is double, so using simply

a.doubleValue() > b.doubleValue()

should be enough in most cases, but... there are subtle issues here when converting numbers to double. For example the following is possible with BigInteger:

    BigInteger a = new BigInteger("9999999999999992");
    BigInteger b = new BigInteger("9999999999999991");
    System.out.println(a.doubleValue() > b.doubleValue());
    System.out.println(a.doubleValue() == b.doubleValue());

results in:

false
true

Although I expect this to be very extreme case this is possible. And no - there is no generic 100% accurate way. Number interface have no method like exactValue() converting to some type able to represent number in perfect way without loosing any information.

Actually having such perfect numbers is impossible in general - for example representing number Pi is impossible using any arithmetic using finite space.

• Note that this also does not work for Long either since they can also not be fully represented using Double. - Christopher Barber

if(yourNumber instanceof Double) {
    boolean greaterThanOtherNumber = yourNumber.doubleValue() > otherNumber.doubleValue();
    // [...]
}

Note: The instanceof check isn't necessarily needed - depends on how exactly you want to compare them. You could of course simply always use .doubleValue(), as every Number should provide the methods listed here.

Edit: As stated in the comments, you will (always) have to check for BigDecimal and friends. But they provide a .compareTo() method:

if(yourNumber instanceof BigDecimal && otherNumber instanceof BigDecimal) { 
    boolean greaterThanOtherNumber = ((BigDecimal)yourNumber).compareTo((BigDecimal)otherNumber) > 0;
} 

• What about BigDecimal or other types out of range for doubles? - b_erb
• simply add a if(yourNumber instanceof BigDecimal && otherNumber instanceof BigDecimal){ boolean greaterThanOtherNumber = yourNumber.compareTo(otherNumber) > 0; } - Tedil

You can simply use Number's doubleValue() method to compare them; however you may find the results are not accurate enough for your needs.

• I think that this is too risky - Yaneeve
• BigInteger and BigDecimal implements Number as well. This wouldn't work. - Roman
• True, it could be risky and it could fail. There are more caveats than the precision/accuracy issue I noted. But it may suit the OP's needs if he doesn't need to handle large BigDecimals. - Steven Mackenzie

What about this one? Definitely not nice, but it deals with all necessary cases mentioned.

public class SimpleNumberComparator implements Comparator<Number>
    {
        @Override
        public int compare(Number o1, Number o2)
        {
            if(o1 instanceof Short && o2 instanceof Short)
            {
                return ((Short) o1).compareTo((Short) o2);
            }
            else if(o1 instanceof Long && o2 instanceof Long)
            {
                return ((Long) o1).compareTo((Long) o2);
            }
            else if(o1 instanceof Integer && o2 instanceof Integer)
            {
                return ((Integer) o1).compareTo((Integer) o2);
            }
            else if(o1 instanceof Float && o2 instanceof Float)
            {
                return ((Float) o1).compareTo((Float) o2);
            }
            else if(o1 instanceof Double && o2 instanceof Double)
            {
                return ((Double) o1).compareTo((Double) o2);
            }
            else if(o1 instanceof Byte && o2 instanceof Byte)
            {
                return ((Byte) o1).compareTo((Byte) o2);
            }
            else if(o1 instanceof BigInteger && o2 instanceof BigInteger)
            {
                return ((BigInteger) o1).compareTo((BigInteger) o2);
            }
            else if(o1 instanceof BigDecimal && o2 instanceof BigDecimal)
            {
                return ((BigDecimal) o1).compareTo((BigDecimal) o2);
            }
            else
            {
                throw new RuntimeException("Ooopps!");
            }

        }

    }

• This won't work on a list of mixed number types - finnw
• Like you said, not nice, shouldn't be here :) - keni

Let's assume that you have some method like:

public <T extends Number> T max (T a, T b) {
   ...
   //return maximum of a and b
}

If you know that there are only integers, longs and doubles can be passed as parameters then you can change method signature to:

public <T extends Number> T max(double a, double b) {
   return (T)Math.max (a, b);
}

This will work for byte, short, integer, long and double.

If you presume that BigInteger's or BigDecimal's or mix of floats and doubles can be passed then you cannot create one common method to compare all these types of parameters.

If your Number instances are never Atomic (ie AtomicInteger) then you can do something like:

private Integer compare(Number n1, Number n2) throws SecurityException, NoSuchMethodException, IllegalArgumentException, IllegalAccessException, InvocationTargetException {

 Class<? extends Number> n1Class = n1.getClass();
 if (n1Class.isInstance(n2)) {
  Method compareTo = n1Class.getMethod("compareTo", n1Class);
  return (Integer) compareTo.invoke(n1, n2);
 }

 return -23;
}

This is since all non-Atomic Numbers implement Comparable

EDIT:

This is costly due to reflection: I know

EDIT 2:

This of course does not take of a case in which you want to compare decimals to ints or some such...

EDIT 3:

This assumes that there are no custom-defined descendants of Number that do not implement Comparable (thanks @DJClayworth)

• This assumes that there are no custom-defined descendants of Number. - DJClayworth
• @DJClayworth: true enough. And those are not the only constraints as I had mentioned above. But why would there be 'custom-defined descendants of Number'? - Yaneeve
• An example of a custom-defined descendant of Number is org.apache.commons.math.fraction.BigFraction, which I use in many of my programs. - finnw