Java code snippet - playing with variable final - explanation required

Question

final Integer a = 1;
Integer b = a;

System.out.println("a: " + a); // prints 1
System.out.println("b: " + b); // prints 1

b++;
System.out.println("a: " + a); // prints 1
System.out.println("b: " + b); // prints 2

b = b + 1;
System.out.println("a: " + a); // prints 1
System.out.println("b: " + b); // prints 3

b = 10;
System.out.println("a: " + a); // prints 1
System.out.println("b: " + b); // prints 10

It would be great if somebody could explain the code output, expecially with relation to variable B.

Your output is wrong. After b++, b prints 2. Then, it prints 3, and finally 10. I just ran your code verbatim. — Jonathon Faust
– Jonathon Faust, Commented Jul 23, 2010 at 18:31
I modified your question to reflect actual output. Given that nothing strikes me as odd about the output, could you clarify what is confusing you? — Jonathon Faust
– Jonathon Faust, Commented Jul 23, 2010 at 18:33
I was using eclipses jpage/scrappage facility, hence the confusion that b++ didn't work like i would have wanted it to. Running it as a java class gave the result you state above and exactly the one I was expecting. I guess eclipse scrappage implementation has a bug. — AM01
– AM01, Commented Jul 23, 2010 at 18:35
In this case please try to add another print out after that b++ This may it be cause of that the value b was not evluated. — Damian Leszczyński - Vash
– Damian Leszczyński - Vash, Commented Jul 23, 2010 at 18:47

Kylar · Accepted Answer · 2010-07-23 18:35:50Z

7

Ok, let's start with this:

final Integer a = 1;

You've created a final reference to an Integer object, which was autoboxed from a primitive int.

This reference can be assigned exactly once, and never again.

Integer b = a;

here you've created a second reference to the same object, but this reference is not final, so you can reassign it at your leisure.

b++;

This is a shorthand for the following statement:

b = new Integer(b.intValue() + 1);

And, coincidentally, the same for

b = b + 1;

The last statement:

b = 10

Is using autoboxing to shorthand this statement:

b = new Integer(10);

Hope this helps.

answered Jul 23, 2010 at 18:35

Kylar

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9 Comments

Jonathon Faust Over a year ago

I believe autoboxing uses Integer.valueOf, not new Integer.

Kylar Over a year ago

You're probably right - I was more interested in the concepts than the specifics. In that it creates a new one (or returns from the internal cache) for the assignment.

j flemm Over a year ago

Probably doesn't matter for this example, but I believe it uses Integer.valueOf(...) and not new Integer(...) so == still works.

Damian Leszczyński - Vash Over a year ago

But why if b++ is b = new Integer(b.intValue() +1) then print out print 1 ? This is standard post incremenation so the b have the value 2 in case after that sysout

Mark Peters Over a year ago

@j flemm: Never, ever, ever count on that in real code. Just way too easy to make mistakes, even if the JLS specifies a minimum integer cache size. == only works if the value the Integer represents is within the pool of cached integers.

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Mark Peters · Accepted Answer · 2010-07-23 18:39:19Z

2

First of all, you must know that auto-boxing is happening here, you can read about that here.

Now only the b++ strikes me as non-straightforward. It is functionally equivalent to this code:

int bTemp = b.intValue();
bTemp++;
b = Integer.valueOf(bTemp);

Though the bytecode may be slightly different.

answered Jul 23, 2010 at 18:39

Mark Peters

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