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POIs String Literal Pool a collection of references to the String Object, Or a collection of Objects
 

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    <p>I am all confused after reading the article on javaranch site by Corey McGlone, The author of The SCJP Tip Line. named Strings, Literally and the SCJP Java 6 Programmer Guide by <code>Kathy Sierra (co-founder of javaranch) and Bert Bates</code>.</p> <p>I will try to quote what Mr. Corey and Ms Kathy Sierra have quoted about String Literal Pool.</p> <p><strong>1.</strong> According to Mr Corey McGlone :</p> <ul> <li><p>String Literal Pool is a Collection of references that points to the String Objects.</p></li> <li><p><code>String s = "Hello";</code> (Assume there is No object on the Heap named "Hello"), will create an String object <code>"Hello"</code> on the heap, and will place an reference to this object in the String Literal Pool (Constant Table)</p></li> <li><p><code>String a = new String("Bye");</code> (Assume there is No object on the Heap named "Bye", <code>new</code> operator will oblige the JVM to create an object on the Heap.</p></li> </ul> <p>Now the explanation of <code>"new"</code> operator for the creation of a String and its reference is bit confusing in this article, so i am putting the code and explanation from the article itself as it-is below.</p> <pre><code>public class ImmutableStrings { public static void main(String[] args) { String one = "someString"; String two = new String("someString"); System.out.println(one.equals(two)); System.out.println(one == two); } } </code></pre> <p>In this case, we actually end up with a slightly different behavior because of the keyword <code>"new."</code> In such a case, references to the two String literals are still put into the constant table (the String Literal Pool), but, when you come to the keyword <code>"new,"</code> the JVM is obliged to create a new String object at run-time, rather than using the one from the constant table.</p> <p>Here is the diagram explaining it..</p> <p><img src="https://i.stack.imgur.com/8FAKX.jpg" alt="enter image description here"></p> <p><strong>So does it mean, that String Literal Pool too have a reference to this Object ?</strong></p> <p>Here is the link to the Article by Corey McGlone</p> <p><a href="http://www.javaranch.com/journal/200409/Journal200409.jsp#a1">http://www.javaranch.com/journal/200409/Journal200409.jsp#a1</a></p> <p><strong>2.</strong> According to Kathy Sierra and Bert Bates in SCJP book:</p> <ul> <li><p>To make Java more memory efficient, the JVM set aside a special area of memory called the "String constant pool", when the compiler encounters a String Literal, it checks the pool to see if an identical String already exits or not. If not then it creates a new String Literal Object. </p></li> <li><p><code>String s = "abc";</code> // Creates one String object and one reference variable.... </p> <p>thats fine, but Now the below statement got me confused.</p></li> <li><p><code>String s = new String("abc")</code> // Creates two objects, and one reference variable.</p> <p>It says in the book that.... a new String object in normal(non-pool) memory , and "s" will refer to it... whereas an additional the literal "abc" will be placed in the pool.</p> <p>The above lines in the book collides with the one in the article by Corey McGlone.</p> <ul> <li><p>If String Literal Pool is a collection of references to the String object as mentioned by Corey McGlone, then how come literal object "abc" will be placed in the pool, as mentioned in the book.</p></li> <li><p>And where do this String Literal Pool resides.</p></li> </ul></li> </ul> <p>Please clear this doubt, though it won't matter too much while writing a code, but is very important from the aspect of memory management, and thats the reason i want to clear this funda.</p>
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    1. COHow the pool is managed might to some degree depend on the JVM implementation. As long as something isn't fixed by the [language specification](http://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.5), they are free to experiment. So whether the pool holds references or objects might well depend, I believe.
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