How to compare two non-primitive Long in Java?

Question

I'm searching for a method that can compare two non-primitive Long objects in Java. Something like this : Longs.compare(Long x, Long y). As I didn't find anything on the web yet, I am wondering if a such library implement that.

So far, I compared them with Long.compare(x, y) but now I've realized that x and y can be null, so sometimes I can get a NullPointerException.

Do you know any library that can do that?

PS : My need is not the same as this post. I want to know if x > y not if x == y.

Possible duplicate of [Compare non-primitive Long values 127 and 128](http://stackoverflow.com/questions/20541636/compare-non-primitive-long-values-127-and-128) — burglarhobbit, Oct 12 '15 at 15:11
@Mr.Robot Despite the similar summary line, I don't think this is a dupe of that question. That question is looking for equality comparisons only, and is basically about the confusion between `==` and `equals`. This question is asking for a `compareTo`-style comparison that includes nullity. — yshavit, Oct 12 '15 at 15:14
Yes, that's what I'm doing for now. But I think it's too bad there isn't library that implement this... — Clemzd, Oct 12 '15 at 15:19
While it's true that you don't need a whole library just for this, if you already depend on a library that gives you this behavior (say, Guava or Apache Commons), you may as well use it. (In fact, pretty much _all_ of Guava is stuff for which you could argue, "you don't need a whole library just for this." But you add all of them together, and that library starts being worthwhile.) — yshavit, Oct 12 '15 at 15:25

azurefrog · Answer 1 · 2015-10-12T15:28:12.890

I believe what you're looking for is the ObjectUtils class from apache commons. It's specifically designed to handle general null-safe comparison.

Its compare() method will never throw an exception based on null input, and instead just assumes that

null is ... less than a non-null value.

So for instance:

    System.out.println(ObjectUtils.compare(null, null));                   // 0
    System.out.println(ObjectUtils.compare(new Long(1), null));            // 1
    System.out.println(ObjectUtils.compare(null, new Long(1)));            // -1
    System.out.println(ObjectUtils.compare(new Long(1), new Long(2)));     // -1

commons-lang3 3.12 do not have compare for Long. – Mubasher Aug 17 '23 at 07:57 — Mubasher, Aug 17 '23 at 07:57

Puce · Accepted Answer · 2015-10-12T15:36:24.013

2

With Java SE 8 try:

Comparator<Long> nullsFirstLongComparator = Comparator.nullsFirst(Comparator.naturalOrder());

or

Comparator<Long> nullsLastLongComparator = Comparator.nullsLast(Comparator.naturalOrder());

edited Oct 12 '15 at 15:36

answered Oct 12 '15 at 15:25

Puce

37,247
13
80
152

score 0 · Answer 3 · answered Oct 12 '15 at 15:14

0

Documentation is your best guide.

http://docs.oracle.com/javase/7/docs/api/java/lang/Long.html

Please check the public int compareTo(Long anotherLong) method.

answered Oct 12 '15 at 15:14

Pavan Dittakavi

3,013
5
27
47

1

This will still throw NPEs, since all that `compareTo()` does is try compare `this.value` with `anotherLong.value`. – azurefrog Oct 12 '15 at 15:16
1

neither `null.compareTo(1L)` nor `1L.compareTo(null)` works (it will throw a null pointer exception) – Clemzd Oct 12 '15 at 15:17
AFAIK there is no built in API which does this. You could perhaps write your own version and check if either of the inputs is null?. – Pavan Dittakavi Oct 12 '15 at 15:19

yshavit · Answer 4 · 2015-10-12T18:56:42.933

0

The longhand answer is to check for nulls first, but you can also use Guava's Ordering<T>, which is an extension of Comparator<T> that also supports a fluent builder style:

Comparator<Long> nullsFirst = Ordering.natural().nullsFirst();
// there's also a nullsLast()

edited Oct 12 '15 at 18:56

answered Oct 12 '15 at 15:19

yshavit

42,327
7
87
124

How to compare two non-primitive Long in Java?

4 Answers4