20

Java Wrapper classes are supposed to be immutable. This means that once an object is being created, e.g., 

Integer i = new Integer(5);

its value cannot be changed. However, doing 

i = 6;

is perfectly valid.

So, what does immutability in this context mean? Does this have to do with auto-boxing/unboxing? If so, is there any way to prevent the compiler from doing it?

Thank you

PetrosB
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6 Answers6

26

i is a reference. Your code change the reference i to point to a different, equally immutable, Integer.

final Integer i = Integer.valueOf(5);

might be more useful.

bmargulies
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19

Immutable means that the object state cannot be changed. In your case you haven't changed the object new Integer(5), but you have changed the reference i to point to another object. Hope it is clear:)

Petar Minchev
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15

The compiler autoboxes primitive values, this means that 

Integer value = 6;

will be compiled as

Integer value = Integer.valueOf(6);

Integer.valueOf will return an Integer instance with the given value. In your case i will now reference the Integer(6) instead of the Integer(5), the Integer(5) object itself will not change.

To see this you can do following

Integer i = new Integer(5);//assign new integer to i
Integer b = i;//b refences same integer as i
i = 6;//modify i
System.out.println(i +"!="+b);

This will print 6!=5, if the integer instance had been modified it would print 6!=6 instead.

To clarify this is only meant to show how an assignment to Integer only modifies the reference and does not alter the Integer instance itself. As user @KNU points out it does not prove or show the immutability of Integer, as far as I can tell the immutability is only indirectly given by the lack of modifying methods in its API and the requirement that instances returned by Integer.valueOf have to be cached for a certain range.

josefx
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  • This is quite an interesting blog entry on primitive/object equality: http://marxsoftware.blogspot.com/2010/08/subtle-nuance-of-new-keyword-with.html – Richard Fearn Nov 07 '10 at 13:42
  • this example doesn't proves immutability of `Integer` in any way. "....if the integer instance had been modified it would print 6!=6 instead." OK how about repeating above test on know mutable object `AtomicInteger`??? it would still print `6!=5` ,throwing water on above argument. – KNU Oct 02 '14 at 15:30
  • Note: for above suggested test one needs to use `i = new AtomicInteger(6);` instead of `i = 6;` – KNU Oct 02 '14 at 15:32
  • @KNU I added a clarification. – josefx Oct 03 '14 at 16:05
  • the `will be compiled as...` thing was really enlightening for me. Until now I never cared to now how it actually works.Great – KNU Oct 03 '14 at 20:53
6

The reason i = 6 works is that auto-boxing is intercepting and turning it into i = new Integer(6). Thus as @Peter said, you are now pointing at a new object.

drekka
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  • Almost - it effectively turns it into `Integer.valueOf(6)`. See this for more info: http://marxsoftware.blogspot.com/2010/08/subtle-nuance-of-new-keyword-with.html – Richard Fearn Nov 07 '10 at 13:43
  • Oops. My Bad :-) I had not actually bothered to look up exactly what it translated to. – drekka Nov 07 '10 at 22:54
0
Integer i = new Integer(5);
i++;                         // i  will become 6

where i++ is the same with i = new Integer( i.intValue() + 1);

Java bee
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  • Interesting does the Java compiler still recommend using ++? Swift got rid of it strictly += now – Greg Price Jul 21 '20 at 00:46
  • @Greg Pricei I'm lazy, so i++ is shorter than i+=1 and cooler than i=i+1 – Java bee Jul 21 '20 at 19:05
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    Duracell De Monaco This is a nice read even if you haven't touched swift. :) https://stackoverflow.com/questions/35158422/the-and-operators-have-been-deprecated-xcode-7-3/35158570 – Greg Price Jul 23 '20 at 20:07
-1

All wrapper classes in java are immutable. We can't change the value of a wrapper class object once created, i.e., can't change the value wrapped inside the object. Because wrapper classes are used as object form of primitive data types and if they are mutable, data inconsistencies will occur in runtime. However, you can change the wrapper class reference variable to hold another object.

Patchikori Rajeswari
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