Java Why Do Only Certain Integers Share Addresses in Java?

[tmt1]

In java, 2 Integer objects with the same value share the same address if their values fit into the range of an unsigned byte.

My question is, why just Integers that fit into this range?

Wouldn't the jvm be optimized if all Integers shared the same address if they had the same value?

Thanks

 

[I like Serena]

In java, 2 Integer objects with the same value share the same address if their values fit into the range of an unsigned byte.

My question is, why just Integers that fit into this range?

Wouldn't the jvm be optimized if all Integers shared the same address if they had the same value?

Thanks

Hi tmt,

That doesn't sound right.
I don't think two integer objects that happen to have the same value will share the same address, unless they are final.
On the other hand, if we have 2 integer objects, we can only assign one to the other if they have a matching type. Afterwards, they will share their address, and will obviously have the same value.

 

[johng1]

I wasn't aware that the jvm did this. I'll take your word for it; it seems very reasonable since the integer wrapper classes are final and immutable. My only quibble is that Java doesn't have an unsigned byte; all integer types are signed. So I assume you mean 8 bits.

It seems to me that your suggestion is much worse from a storage standpoint. For example:

Byte byteOb=new Byte((byte)10);
Integer intObj=new Integer(10);

If the jvm followed your suggestion, a total of 4 bytes would be allocated; whereas the actual jvm allocates just one byte.

Since all Java methods are "virtual" (C++ terminology), I can't see much problem with your suggestion with regard to methods, but I may be overlooking something.

Final comment. I am very confident that the Java engineers designed a very good language; of course there are bugs, but the basic design is excellent.