Comparing Integers in Java

Question

I got the following code from a Java book that didn't have explanation about the results. I ran it and got the results noted below inline, but I need help understanding the results:

1. Integer i = -10;
2. Integer j = -10;
3. System.out.print(i==j);  //  ==> true
4. System.out.print(i.equals(j));   //  ==> true
5. Integer n = 128;
6. Integer m = 128;
7. System.out.print(n==m);  //  ==> false
8. System.out.print(n.equals(m));   //  ==> true

My questions are:

why does line #3 resolve to true? are they not two separate objects?
if line #3 is true, why does line #7 not true?

Thanks if advance.

Hint, integers in a specific range are cached ;) Have a look into the Integer class. — Thomas Jungblut
– Thomas Jungblut, Commented Dec 2, 2012 at 19:07
@ThomasJungblut Thats the reason why should prefer Integer.valueOf(int) to new Integer(int) — AlexWien
– AlexWien, Commented Dec 2, 2012 at 19:26
But here's an interesting question: Is it possible for #7 to ever be true? The answer is YES, since the upper limit (but not lower limit) of the Integer cache is configurable. — Hot Licks
– Hot Licks, Commented Dec 2, 2012 at 19:31

sehe · Accepted Answer · 2012-12-02 19:17:12Z

9

See Integer.valueOf(int i) source code.

It caches (use integer pool) the Integers between -128 and 127 (those are the default values, which can be customized via java.lang.Integer.IntegerCache.high)

edited Dec 2, 2012 at 19:17

sehe

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answered Dec 2, 2012 at 19:08

Bela Vizer

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Comments

Marko Topolnik · Accepted Answer · 2012-12-02 19:53:02Z

2

To emphasize the level of absurdity resulting from using == on Integers, consider this line:

Integer a = 200, b = 200;
System.out.println(a < b || a == b || a > b);

Apparently, this should print true regardless of values because it looks like a tautology. It prints false, of course. The following, apparently a tautology of the same kind, prints the expected true value:

System.out.println(a <= b || a > b);

edited Dec 2, 2012 at 19:53

answered Dec 2, 2012 at 19:19

Marko Topolnik

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1 Comment

dantuch Over a year ago

great one! I really liked it :)

Hot Licks · Accepted Answer · 2012-12-02 19:25:54Z

0

Integer is an object, whereas int is a "scalar". Something like Integer i = 270;, under the covers, becomes Integer i = Integer.valueOf(270);. So when you create two Integer objects they are distinct (as demonstrated with ==).

However, the JDK, in valueOf, caches the creation of Integer objects with a small value -- -128 to 127, and previous versions of those are reused if available. So two apparently distinct instances of small Integers may actually be the same, although they appear to have been instantiated separately.

If you tried the same thing with int values the == compares would always be true, and the equals() calls would generate compile errors (because int isn't an object).

answered Dec 2, 2012 at 19:25

Hot Licks

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Comments

Kneel-Before-ZOD · Accepted Answer · 2012-12-02 19:27:35Z

0

On question 1

According to the doc, a Integer class is wrapped around a primitive value, which simply means that you can expect the object to behave as a primitive when required (the comparison performed in Step. 3), and equally as an object when required (Step 4; this step isn't possible if int instead of an Integer class was used.).
If you swap the types in Steps 1 and 2 to String instead of an Integer, and perform the tests again, you'll obtained the desired results you were hoping to get.

On question 2

It has already been answered above: Integers in specific ranges (-128 and 127) are cached.

Hope this helps.

answered Dec 2, 2012 at 19:27

Kneel-Before-ZOD

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