compareTo with primitives -> Integer / int

Question

Is it better to write

int primitive1 = 3, primitive2 = 4;
Integer a = new Integer(primitive1);
Integer b = new Integer(primitive2);
int compare = a.compareTo(b);

or

int primitive1 = 3, primitive2 = 4;
int compare = (primitive1 > primitive2) ? 1 : 0;
if(compare == 0){
    compare = (primitive1 == primitive2) ? 0 : -1;
}

I think the second one is better, should be faster and more memory optimized. But aren't they equal?

never call "new Integer" instead using Integer.valueOf(int) (http://docs.oracle.com/javase/6/docs/api/java/lang/Integer.html#valueOf(int)) this method accesses an internal cache and will often avoid an allocation for small integers. — luke, Feb 05 '12 at 15:35
FYI: On modern JVMs (like HotSpot), the two programs will probably become the same machine code after optimization. The best way to know for sure is (as always) benchmark both solutions. — Adam Paynter, Feb 05 '12 at 15:35
In addition to @luke 's comment: new Integer() will be deprecated in Java9 http://download.java.net/java/jdk9/docs/api/java/lang/Integer.html#Integer-int- — Naxos84, Oct 27 '16 at 07:11
https://docs.oracle.com/javase/9/docs/api/java/lang/Integer.html#Integer-int- is a working link — MartyIX, Jan 18 '19 at 14:18

score 147 · Accepted Answer · edited Jan 18 '19 at 14:21

147

For performance, it usually best to make the code as simple and clear as possible and this will often perform well (as the JIT will optimise this code best). In your case, the simplest examples are also likely to be the fastest.

I would do either

int cmp = a > b ? +1 : a < b ? -1 : 0;

or a longer version

int cmp;
if (a > b)
   cmp = +1;
else if (a < b)
   cmp = -1;
else
   cmp = 0;

or

int cmp = Integer.compare(a, b); // in Java 7
int cmp = Double.compare(a, b); // before Java 7

It's best not to create an object if you don't need to.

Performance wise, the first is best.

If you know for sure that you won't get an overflow you can use

int cmp = a - b; // if you know there wont be an overflow.

you won't get faster than this.

edited Jan 18 '19 at 14:21

MartyIX

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answered Feb 05 '12 at 15:33

Peter Lawrey

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6

Integer doesn't have an equivalent static method before Java 7. :( But Double.compare() has the same effect. – Peter Lawrey Feb 05 '12 at 15:37
Never use `cmp = a - b`, even if you "know" there won't be an overflow. You don't know there won't be an overflow. – Clement Cherlin Jun 09 '20 at 13:35
1

There won't be an overflow if both `a` and `b` are positive integers, for example. – DV82XL Nov 08 '21 at 20:15

score 67 · Answer 2 · answered Feb 05 '12 at 15:34

67

Use Integer.compare(int, int). And don'try to micro-optimize your code unless you can prove that you have a performance issue.

answered Feb 05 '12 at 15:34

MForster

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this method is in JDK since `1.7`, I'm sorry, but for compatibility issues I can't use it so far. – Marek Sebera Feb 05 '12 at 15:37
Any pre-1.7 example like this? – Gaʀʀʏ Jan 16 '14 at 19:50
@Gaʀʀʏ, see Peter's answer above. – MForster Jan 17 '14 at 07:55
1

This is the only correct answer. Additionally, the entire implementation of `Integer.compare(x, y)` is `return (x < y) ? -1 : ((x == y) ? 0 : 1);` so on JDK < 1.7 you can paste that into your own static utility method – Clement Cherlin Jun 09 '20 at 13:38

score 15 · Answer 3 · edited Feb 05 '12 at 15:35

15

May I propose a third

((Integer) a).compareTo(b)

edited Feb 05 '12 at 15:35

Peter Lawrey

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answered Feb 05 '12 at 15:33

Johan Sjöberg

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score 5 · Answer 4 · answered Feb 05 '12 at 15:44

5

Wrapping int primitive into Integer object will cost you some memory, but the difference will be only significant in very rare(memory demand) cases (array with 1000+ elements). I will not recommend using new Integer(int a) constructor this way. This will suffice :

Integer a = 3;

About comparision there is Math.signum(double d).

compare= (int) Math.signum(a-b);

answered Feb 05 '12 at 15:44

This will overflow, since the int to double conversion happens after the subtraction. – Clement Cherlin Jun 09 '20 at 13:41

score 4 · Answer 5 · answered May 27 '15 at 22:57

4

They're already ints. Why not just use subtraction?

compare = a - b;

Note that Integer.compareTo() doesn't necessarily return only -1, 0 or 1 either.

answered May 27 '15 at 22:57

MQDuck

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This will produce erroneous results in the case of overflow. Try `a = Integer.MAX_VALUE` and `b = -3`. The result will be negative, indicating that `a < b`. – Jim Mischel Oct 03 '16 at 12:43

score 4 · Answer 6 · answered Nov 17 '15 at 15:59

4

For pre 1.7 i would say an equivalent to Integer.compare(x, y) is:

Integer.valueOf(x).compareTo(y);

answered Nov 17 '15 at 15:59

Dave

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score 3 · Answer 7 · answered May 20 '17 at 06:53

3

If you are using java 8, you can create Comparator by this method:

Comparator.comparingInt(i -> i);

if you would like to compare with reversed order:

Comparator.comparingInt(i -> -i);

answered May 20 '17 at 06:53

tanghao

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WebComer · Answer 8 · 2017-12-15T19:17:52.913

-1

If you need just logical value (as it almost always is), the following one-liner will help you:

boolean ifIntsEqual = !((Math.max(a,b) - Math.min(a, b)) > 0);

And it works even in Java 1.5+, maybe even in 1.1 (i don't have one). Please tell us, if you can test it in 1.5-.

This one will do too:

boolean ifIntsEqual = !((Math.abs(a-b)) > 0);

edited Dec 15 '17 at 19:17

answered Dec 15 '17 at 16:35

WebComer

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These expressions perform a lot of wasted computation to (in the first case, incorrectly) compute a == b, which is not what the question is about. – Clement Cherlin Jun 09 '20 at 13:50

compareTo with primitives -> Integer / int

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