Autoboxing wth ++,-- operator in java

Question

I am confuse about autoboxing unboxing in java. Please see my following two progarm.

Integer x = 400;
Integer y = x;
x++; x--;
System.out.println((x==y));

The output is false. 
I known why the output is false. Because of autoboxing x.

Integer x = 100;
Integer y = x;
x++; x--;
System.out.println((x==y));

The output is true.
But the program is same as the upper. Why the output is true? 
Please explain me detail.

Thank you very much.

@TBZ: Use @ when addressing someone in a comment, this will give them a notice. — Keppil, Mar 14 '14 at 05:11
@Keppil Just an FYI, the notification system now pings you regardless :) — Brian Roach, Mar 14 '14 at 05:13
@Brian: What does `regardless` mean? Do you have a link describing when you need to use @ and when you don't? — Keppil, Mar 14 '14 at 05:17

Evgeniy Dorofeev · Accepted Answer · 2014-03-14T05:34:31.763

1

This is because Integers -128 to 127 are cached, so in second example x and y refer to the same Integer instance.

Integer x = 100;          // x refers to cached 100
x++;

is equivalent to

int var = x.intValue();
var++;
x = Integer.valueOf(var);  // returns cached 100

See Integer.valueOf(int) API.

edited Mar 14 '14 at 05:34

answered Mar 14 '14 at 05:06

Evgeniy Dorofeev

133,369
30
199
275

+1, but *may* be cached (but almost always are) – Bohemian Mar 14 '14 at 05:09
I think Jon Skeet's answer to the Q this is a Dup of is far more comprehensive, especially in that regard. – Brian Roach Mar 14 '14 at 05:11
@Evgeniy Dorofeev: Thank you very much +1 :). What is the same Integer instance? – T8Z Mar 14 '14 at 05:25
@T8Z added more details – Evgeniy Dorofeev Mar 14 '14 at 05:34

Autoboxing wth ++,-- operator in java

1 Answers1