Comments inline below:
On Tue, Aug 30, 2016 at 6:38 AM, Jun Zhuang <jun.zhu...@hobsons.com> wrote:
> Hi Kris,
>
>
>
> Thanks for the response. Regarding the 2nd question, I want to make sure
> I understand your answer. The handling of *new String("Cat")* can be
> broken into two steps:
>
>
>
> *Step #1*: Handling the “Cat” expression
>
> If ( Does not already exist )
>
> Create an instance on the heap
>
> Add a reference to the StringTable for the above instance and return the
> reference
>
>
>
> else
>
> Return the reference from the StringTable
>
>
>
Yes, that it correct.
> Question: I assume the reference returned is consumed by
> the String () constructor, is it using the existing object to create a new
> one?
>
In this specific example, yes, the reference is directly passed to the
constructor call as the argument. It's the same as passing any reference
type argument to any method.
>
>
> *Step #2*: Handling the new String()
>
> Action: Create an instance and return reference
>
>
>
Yes.
> Either way, TWO instance of String object representing Cat will end up on
> the heap, right?
>
That is correct. The one created by the "new" expression is required to
have a different identity (meaning being a different object) than the one
from the "Cat" constant expression. Thus:
"Cat" == "Cat" will always be true, and
"Cat" == new String("Cat") will always be false.
> If I have another statement String str2 = new String("Cat"); a third such
> instance will be created?
>
>
> That is correct. The "Cat" part will resolve to a reference to the same
object as any earlier mentions of the interned "Cat" String instance.
- Kris
> Thanks much,
>
> Jun
>
>
>
> ------------------------------------------------------------
> ------------------------------------------------------------
> --------------------
>
> #2: Is the following statement correct?
>
> String str = new String("Cat");
>
> In above statement, either 1 or 2 string will be created. If there is
> already a string literal “Cat” in the pool, then only one string “str” will
> be created in the pool. If there is no string literal “Cat” in the pool,
> then it will be first created in the pool and then in the heap space, so
> total 2 string objects will be created.
>
> From http://www.journaldev.com/797/what-is-java-string-pool#comment-36152
>
> For this statement, yes, either 1 or 2 java.lang.String instances are
> going to be created. But I wouldn't frame it that way, since it's
> potentially confusing.
>
> The correct way to frame this is to make a clear distinction between
> one-time actions and normal runtime actions.
>
>
>
> The "Cat" expression is a reference to a compile-time constant of type
> java.lang.String. At runtime, there will be a one-time resolution for such
> a reference. Indeed, such resolution will probe the StringTable to see if a
> matching String instance is already referenced by the StringTable, if so
> return that reference; otherwise create a java.lang.String instance from
> the Symbol representing "Cat", intern the reference into the StringTable,
> and return that reference.
>
>
>
> The "new String(...)" expression, on the other hand, is a "new"
> expression. Semantically it should always create a new String instance
> every time.
>
>
>
> - Kris
>
>
>
> [1]: http://hg.openjdk.java.net/jdk8u/jdk8u/hotspot/file/
> 312e113bc3ed/src/share/vm/classfile/symbolTable.hpp#l255
>
>
>
>
>
> Apprciate your answers,
>
> Jun
>
>
>
> *Jun Zhuang*
>
> *Sr. Performance QA Engineer* | Hobsons
> <https://www.hobsons.com/?utm_source=outlook&utm_medium=email&utm_campaign=banner_02.12.16_general>
>
> T: +1 513 746 2288 | jun.zhu...@hobsons.com
>
> 50 E-Business Way, Suite 300 | Cincinnati, OH 45241 | USA
>
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