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Reference to which String object is returned from the pool when calling intern()

The Java Language Specification states that
When the intern method is invoked, if the pool already contains a
string equal to this String object as determined by the
equals(java.lang.Object) method, then the string from the pool is
returned
In the following code snippet:
class StringPoolTest {
public static void main(String[] args) {
String first = "string";
String second = new String("string");
String third = "string".intern();
System.out.println(System.identityHashCode(first));
System.out.println(System.identityHashCode(second));
System.out.println(System.identityHashCode(third));
}
}
Output:
989184670
268130470
989184670
I added a String object first to the pool (assigning a String literal to the reference) and explicitly created second using a String constructor.
There are now two identical character sequences in the pool.
I wanted to see which one would be returned when calling intern. Since the method hashCode is overridden for the class String, I used System.identityHashCode to see exactly which two String references were the same.
Clearly, intern returned the reference to the object created using a String literal. Why is this so? Are there any rules regarding which reference is returned in the case of multiple identical String objects in the pool?

When a string literal(without using new Operator) is created, the JVM checks whether that String exists in the internal list or not. If it already exists in the list, then it does not create a new String and it uses reference to the existing String Object.
JVM does this type of checking internally for String literal but not for String object, which it creates through 'new' keyword i.e. using new always a new string is returned.

Why does String not require a constructor in Java? [duplicate]

I'm a C++ guy learning Java. I'm reading Effective Java and something confused me. It says never to write code like this:
String s = new String("silly");
Because it creates unnecessary String objects. But instead it should be written like this:
String s = "No longer silly";
Ok fine so far...However, given this class:
public final class CaseInsensitiveString {
private String s;
public CaseInsensitiveString(String s) {
if (s == null) {
throw new NullPointerException();
}
this.s = s;
}
:
:
}
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
String s = "polish";
Why is the first statement ok? Shouldn't it be
CaseInsensitiveString cis = "Polish";
How do I make CaseInsensitiveString behave like String so the above statement is OK (with and without extending String)? What is it about String that makes it OK to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java?

String is a special built-in class of the language. It is for the String class only in which you should avoid saying
String s = new String("Polish");
Because the literal "Polish" is already of type String, and you're creating an extra unnecessary object. For any other class, saying
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
is the correct (and only, in this case) thing to do.

I believe the main benefit of using the literal form (ie, "foo" rather than new String("foo")) is that all String literals are 'interned' by the VM. In other words it is added to a pool such that any other code that creates the same string will use the pooled String rather than creating a new instance.
To illustrate, the following code will print true for the first line, but false for the second:
System.out.println("foo" == "foo");
System.out.println(new String("bar") == new String("bar"));

Strings are treated a bit specially in java, they're immutable so it's safe for them to be handled by reference counting.
If you write
String s = "Polish";
String t = "Polish";
then s and t actually refer to the same object, and s==t will return true, for "==" for objects read "is the same object" (or can, anyway, I"m not sure if this is part of the actual language spec or simply a detail of the compiler implementation-so maybe it's not safe to rely on this) .
If you write
String s = new String("Polish");
String t = new String("Polish");
then s!=t (because you've explicitly created a new string) although s.equals(t) will return true (because string adds this behavior to equals).
The thing you want to write,
CaseInsensitiveString cis = "Polish";
can't work because you're thinking that the quotations are some sort of short-circuit constructor for your object, when in fact this only works for plain old java.lang.Strings.

String s1="foo";
literal will go in pool and s1 will refer.
String s2="foo";
this time it will check "foo" literal is already available in StringPool or not as now it exist so s2 will refer the same literal.
String s3=new String("foo");
"foo" literal will be created in StringPool first then through string arg constructor String Object will be created i.e "foo" in the heap due to object creation through new operator then s3 will refer it.
String s4=new String("foo");
same as s3
so System.out.println(s1==s2);// **true** due to literal comparison.
and System.out.println(s3==s4);// **false** due to object
comparison(s3 and s4 is created at different places in heap)

Strings are special in Java - they're immutable, and string constants are automatically turned into String objects.
There's no way for your SomeStringClass cis = "value" example to apply to any other class.
Nor can you extend String, because it's declared as final, meaning no sub-classing is allowed.

Java strings are interesting. It looks like the responses have covered some of the interesting points. Here are my two cents.
strings are immutable (you can never change them)
String x = "x";
x = "Y";
The first line will create a variable x which will contain the string value "x". The JVM will look in its pool of string values and see if "x" exists, if it does, it will point the variable x to it, if it does not exist, it will create it and then do the assignment
The second line will remove the reference to "x" and see if "Y" exists in the pool of string values. If it does exist, it will assign it, if it does not, it will create it first then assignment. As the string values are used or not, the memory space in the pool of string values will be reclaimed.
string comparisons are contingent on what you are comparing
String a1 = new String("A");
String a2 = new String("A");
a1 does not equal a2
a1 and a2 are object references
When string is explicitly declared, new instances are created and their references will not be the same.
I think you're on the wrong path with trying to use the caseinsensitive class. Leave the strings alone. What you really care about is how you display or compare the values. Use another class to format the string or to make comparisons.
i.e.
TextUtility.compare(string 1, string 2)
TextUtility.compareIgnoreCase(string 1, string 2)
TextUtility.camelHump(string 1)
Since you are making up the class, you can make the compares do what you want - compare the text values.

The best way to answer your question would be to make you familiar with the "String constant pool". In java string objects are immutable (i.e their values cannot be changed once they are initialized), so when editing a string object you end up creating a new edited string object wherease the old object just floats around in a special memory ares called the "string constant pool". creating a new string object by
String s = "Hello";
will only create a string object in the pool and the reference s will refer to it, but by using
String s = new String("Hello");
you create two string objects: one in the pool and the other in the heap. the reference will refer to the object in the heap.

You can't. Things in double-quotes in Java are specially recognised by the compiler as Strings, and unfortunately you can't override this (or extend java.lang.String - it's declared final).

- How do i make CaseInsensitiveString behave like String so the above statement is ok (with and w/out extending String)? What is it about String that makes it ok to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java right?
Enough has been said from the first point. "Polish" is an string literal and cannot be assigned to the CaseInsentiviveString class.
Now about the second point
Although you can't create new literals, you can follow the first item of that book for a "similar" approach so the following statements are true:
// Lets test the insensitiveness
CaseInsensitiveString cis5 = CaseInsensitiveString.valueOf("sOmEtHiNg");
CaseInsensitiveString cis6 = CaseInsensitiveString.valueOf("SoMeThInG");
assert cis5 == cis6;
assert cis5.equals(cis6);
Here's the code.
C:\oreyes\samples\java\insensitive>type CaseInsensitiveString.java
import java.util.Map;
import java.util.HashMap;
public final class CaseInsensitiveString {
private static final Map<String,CaseInsensitiveString> innerPool
= new HashMap<String,CaseInsensitiveString>();
private final String s;
// Effective Java Item 1: Consider providing static factory methods instead of constructors
public static CaseInsensitiveString valueOf( String s ) {
if ( s == null ) {
return null;
}
String value = s.toLowerCase();
if ( !CaseInsensitiveString.innerPool.containsKey( value ) ) {
CaseInsensitiveString.innerPool.put( value , new CaseInsensitiveString( value ) );
}
return CaseInsensitiveString.innerPool.get( value );
}
// Class constructor: This creates a new instance each time it is invoked.
public CaseInsensitiveString(String s){
if (s == null) {
throw new NullPointerException();
}
this.s = s.toLowerCase();
}
public boolean equals( Object other ) {
if ( other instanceof CaseInsensitiveString ) {
CaseInsensitiveString otherInstance = ( CaseInsensitiveString ) other;
return this.s.equals( otherInstance.s );
}
return false;
}
public int hashCode(){
return this.s.hashCode();
}
// Test the class using the "assert" keyword
public static void main( String [] args ) {
// Creating two different objects as in new String("Polish") == new String("Polish") is false
CaseInsensitiveString cis1 = new CaseInsensitiveString("Polish");
CaseInsensitiveString cis2 = new CaseInsensitiveString("Polish");
// references cis1 and cis2 points to differents objects.
// so the following is true
assert cis1 != cis2; // Yes they're different
assert cis1.equals(cis2); // Yes they're equals thanks to the equals method
// Now let's try the valueOf idiom
CaseInsensitiveString cis3 = CaseInsensitiveString.valueOf("Polish");
CaseInsensitiveString cis4 = CaseInsensitiveString.valueOf("Polish");
// References cis3 and cis4 points to same object.
// so the following is true
assert cis3 == cis4; // Yes they point to the same object
assert cis3.equals(cis4); // and still equals.
// Lets test the insensitiveness
CaseInsensitiveString cis5 = CaseInsensitiveString.valueOf("sOmEtHiNg");
CaseInsensitiveString cis6 = CaseInsensitiveString.valueOf("SoMeThInG");
assert cis5 == cis6;
assert cis5.equals(cis6);
// Futhermore
CaseInsensitiveString cis7 = CaseInsensitiveString.valueOf("SomethinG");
CaseInsensitiveString cis8 = CaseInsensitiveString.valueOf("someThing");
assert cis8 == cis5 && cis7 == cis6;
assert cis7.equals(cis5) && cis6.equals(cis8);
}
}
C:\oreyes\samples\java\insensitive>javac CaseInsensitiveString.java
C:\oreyes\samples\java\insensitive>java -ea CaseInsensitiveString
C:\oreyes\samples\java\insensitive>
That is, create an internal pool of CaseInsensitiveString objects, and return the corrensponding instance from there.
This way the "==" operator returns true for two objects references representing the same value.
This is useful when similar objects are used very frequently and creating cost is expensive.
The string class documentation states that the class uses an internal pool
The class is not complete, some interesting issues arises when we try to walk the contents of the object at implementing the CharSequence interface, but this code is good enough to show how that item in the Book could be applied.
It is important to notice that by using the internalPool object, the references are not released and thus not garbage collectible, and that may become an issue if a lot of objects are created.
It works for the String class because it is used intensively and the pool is constituted of "interned" object only.
It works well for the Boolean class too, because there are only two possible values.
And finally that's also the reason why valueOf(int) in class Integer is limited to -128 to 127 int values.

In your first example, you are creating a String, "silly" and then passing it as a parameter to another String's copy constructor, which makes a second String which is identical to the first. Since Java Strings are immutable (something that frequently stings people who are used to C strings), this is a needless waste of resources. You should instead use the second example because it skips several needless steps.
However, the String literal is not a CaseInsensitiveString so there you cannot do what you want in your last example. Furthermore, there is no way to overload a casting operator like you can in C++ so there is literally no way to do what you want. You must instead pass it in as a parameter to your class's constructor. Of course, I'd probably just use String.toLowerCase() and be done with it.
Also, your CaseInsensitiveString should implement the CharSequence interface as well as probably the Serializable and Comparable interfaces. Of course, if you implement Comparable, you should override equals() and hashCode() as well.

CaseInsensitiveString is not a String although it contains a String. A String literal e.g "example" can be only assigned to a String.

CaseInsensitiveString and String are different objects. You can't do:
CaseInsensitiveString cis = "Polish";
because "Polish" is a String, not a CaseInsensitiveString. If String extended CaseInsensitiveString String then you'd be OK, but obviously it doesn't.
And don't worry about the construction here, you won't be making unecessary objects. If you look at the code of the constructor, all it's doing is storing a reference to the string you passed in. Nothing extra is being created.
In the String s = new String("foobar") case it's doing something different. You are first creating the literal string "foobar", then creating a copy of it by constructing a new string out of it. There's no need to create that copy.

Just because you have the word String in your class, does not mean you get all the special features of the built-in String class.

when they say to write
String s = "Silly";
instead of
String s = new String("Silly");
they mean it when creating a String object because both of the above statements create a String object but the new String() version creates two String objects: one in heap and the other in string constant pool. Hence using more memory.
But when you write
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
you are not creating a String instead you are creating an object of class CaseInsensitiveString. Hence you need to use the new operator.

If I understood it correctly, your question means why we cannot create an object by directly assigning it a value, lets not restrict it to a Wrapper of String class in java.
To answer that I would just say, purely Object Oriented Programming languages have some constructs and it says, that all the literals when written alone can be directly transformed into an object of the given type.
That precisely means, if the interpreter sees 3 it will be converted into an Integer object because integer is the type defined for such literals.
If the interpreter sees any thing in single quotes like 'a' it will directly create an object of type character, you do not need to specify it as the language defines the default object of type character for it.
Similarly if the interpreter sees something in "" it will be considered as an object of its default type i.e. string. This is some native code working in the background.
Thanks to MIT video lecture course 6.00 where I got the hint for this answer.

In Java the syntax "text" creates an instance of class java.lang.String. The assignment:
String foo = "text";
is a simple assignment, with no copy constructor necessary.
MyString bar = "text";
Is illegal whatever you do because the MyString class isn't either java.lang.String or a superclass of java.lang.String.

First, you can't make a class that extends from String, because String is a final class. And java manage Strings differently from other classes so only with String you can do
String s = "Polish";
But whit your class you have to invoke the constructor. So, that code is fine.

I would just add that Java has Copy constructors...
Well, that's an ordinary constructor with an object of same type as argument.

String is one of the special classes in which you can create them without the new Sring part
it's the same as
int x = y;
or
char c;

String str1 = "foo";
String str2 = "foo";
Both str1 and str2 belongs to tha same String object, "foo", b'coz Java manages Strings in StringPool, so if a new variable refers to the same String, it doesn't create another one rather assign the same alerady present in StringPool.
String str1 = new String("foo");
String str2 = new String("foo");
Here both str1 and str2 belongs to different Objects, b'coz new String() forcefully create a new String Object.

It is a basic law that Strings in java are immutable and case sensitive.

Java creates a String object for each string literal you use in your code. Any time "" is used, it is the same as calling new String().
Strings are complex data that just "act" like primitive data. String literals are actually objects even though we pretend they're primitive literals, like 6, 6.0, 'c', etc. So the String "literal" "text" returns a new String object with value char[] value = {'t','e','x','t}. Therefore, calling
new String("text");
is actually akin to calling
new String(new String(new char[]{'t','e','x','t'}));
Hopefully from here, you can see why your textbook considers this redundant.
For reference, here is the implementation of String: http://www.docjar.com/html/api/java/lang/String.java.html
It's a fun read and might inspire some insight. It's also great for beginners to read and try to understand, as the code demonstrates very professional and convention-compliant code.
Another good reference is the Java tutorial on Strings:
http://docs.oracle.com/javase/tutorial/java/data/strings.html

In most versions of the JDK the two versions will be the same:
String s = new String("silly");
String s = "No longer silly";
Because strings are immutable the compiler maintains a list of string constants and if you try to make a new one will first check to see if the string is already defined. If it is then a reference to the existing immutable string is returned.
To clarify - when you say "String s = " you are defining a new variable which takes up space on the stack - then whether you say "No longer silly" or new String("silly") exactly the same thing happens - a new constant string is compiled into your application and the reference points to that.
I dont see the distinction here. However for your own class, which is not immutable, this behaviour is irrelevant and you must call your constructor.
UPDATE: I was wrong!
I tested this and realise that my understanding is wrong - new String("Silly") does indeed create a new string rather than reuse the existing one. I am unclear why this would be (what is the benefit?) but code speaks louder than words!

ArrayList modifying value returned by "get" method [duplicate]

This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 6 years ago.
I have below two situations related to ArrayList get method, one with custom class and one with String class:
1. Below is the example of modifying Custom class ArrayList element:
ArrayList<MyClass> mTmpArray1 = new ArrayList<MyClass>();
MyClass myObj1 = new MyClass(10);
mTmpArray1.add(myObj1);
MyClass myObj2 = mTmpArray1.get(0);
myObj2.myInt = 20;
MyClass myObj3 = mTmpArray1.get(0);
Log.d(TAG, "Int Value:"+myObj3.myInt); // Prints "20"
2. And below is the example of modifying String ArrayList element:
ArrayList<String> mTmpArray2 = new ArrayList<String>();
mTmpArray2.add("Test_10");
String myStr1 = mTmpArray2.get(0);
myStr1 = "Test_20";
String myStr2 = mTmpArray2.get(0);
Log.d(TAG, "Str Value:"+myStr2); // Prints "Test_10"
So in case of MyClass ArrayList, when I call get and modify the value, then I see change is reflecting when I do get again.
But same way when I modify String ArrayList, then changes are not reflecting.
What is the different in of the get method in both the scenarios?
Is it that in case of String, String class creating deep copy and returns new object, and in case of Custom class shallow copy is created?
In the first scenario applicable to "LinkedHashMap", "HashMap" and "List"?

Your are not doing the same thing in the two cases.
Here you update the state of an object, so the change affects the object stored in the list :
myObj2.myInt = 20;
Here you are assigning a new object to a local variable, so the list is not affected :
myStr1 = "Test_20";
If String was mutable, you could have modified the String by calling some method, and the change would have been reflected in the object stored in the list :
myStr1.setSomething(...);
On the other hand, if in the first case you would have changed the value of the local variable, the object stored in the list wouldn't have been affected :
myObj2 = new MyClass (...);

Strings are immutable. You're not inserting the new string into the array list.
When you do String myStr2 = mTmpArray2.get(0);, even tho you are pointing to a reference in the ArrayList, any attempt to change the value, will (because of String immutability) create a new String (myStr2) that will not reference the ArrayList anymore.
When you do myStr1 = "xxx", you're not actually changing the ArrayList reference, you're changing a new (copy) (now called myStr1) that was grabbed from the ArrayList and it has a local scope.
Read some more about Strings: Immutability of Strings in Java
Now in the first example, you are pointing to a mutable object (your custom class) so you're literally changing the direct value, through the reference. Welcome to Java. ;)
Unrelated: This code: MyClass myObj1 = new MyClass(10); is (arguably) considered bad. It's better to use a factory pattern that is a lot easier to read. In other words, public constructors with parameters are hard to read (for example, I have no idea what I am constructing when I read your code).
A (perhaps) better approach would be: MyClass myObj = MyClass.createInstanceWithCapacity(10); // i've invented the name because I don't know what your 10 is, but look at both, which one do you think is easier to understand upon first glance?
Disclaimer: The above unrelated comment is my personal opinion and not all developers will agree. ;)

Strings have very nice property called "Immutablity"
This means that String cannot be mutable (changed), when we create/
try to refer to old string, a new instance string is created. And any
changes we do are saved in new instance and it do not affect the old
string
For example,
String s = "Old String";
System.out.println("Old String : "+s); // output : Old String
String s2 = s;
s2 = s2.concat(" made New");
System.out.println("New String : "+s2); // output : Old String made New
System.out.println("Old String is not changed : "+s); // output : Old String

These is no difference between the two "get" calls. The difference is between the types that the ArrayList is holding, and what you're doing the references the "get" method returns.
In your first example, you do this:
MyClass myObj2 = mTmpArray1.get(0);
myObj2.myInt = 20;
Here, you're getting a reference to the MyClass instance in the ArrayList in position 0, and you are modifying a field within this instance.
In your second example, you do this:
String myStr1 = mTmpArray2.get(0);
myStr1 = "Test_20";
Here, you're getting a reference to the String instance in the array list, and then you're giving myStr1 a reference to a different string which you create ("Test_20"). It's as if you did wrote myObj2 = new MyClass(20); in the 2nd line in the 1st example.
So, in short, in the 1st example, you access a field within the object by altering it from the reference you grabbed. In the 2nd example, you simply altered your reference to point at a different String.
I should also mention that in Java, Strings are immutable, meaning once they have been created, they cannot be changed.

String is an immutable class. A line like
myStr1 = "Test_20";
does not change the value of the String object myStr1 is pointing to. Instead a new String is created and myStr1 is modified to point to the new String. The original String is unchanged and can be retrieved from the ArrayList.
Your MyClass object is clearly mutable. Only one instance is created and its state is changed by the assignment
myObj2.myInt = 20;
Hence when this object is retrieved from the ArrayList, its new state is seen.

You simply do NOT change the list in your 2nd example.
In the first example, you are doing this:
Get the first object from the list and store it in the variable called 'myObj2'
Modify the object stored in variable 'myObj2' by setting the int value of this object
But your second example is completely different:
String myStr1 = mTmpArray2.get(0);
myStr1 = "Test_20";
Let me translate that:
Get the first element from the list and store it in the variable called 'myStr1'.
Set the value of the variable 'myStr1' to "Test_20"
So, in case one you are modifying a variable of the object stored in the list. In case two you are retrieving the object stored in the list - and then re-use the variable you stored that retrieved object in and use it for something new - but that does not change the original list, of course.
To modify your list for a type like string, you would need to use set(x, "Test_20").

Does this function create a new String everytime this is called?

public String toString()
{
return "NotSubscribed";
}
We are trying to limit String's creation in our application after the objects are instantiated. So, just wondering if this creates a new String every time this toString is called. It might be a bad question, but I don't know the answer. Please suggest.
--
I just remembered that if a String is created, it is stored in the String Pool and after that it is always retrieved from the pool without being re-created. Am I right?

No, that won't create another string each time. String constants are interned - so actually the constant "NotSubscribed" will refer to the same string object throughout your code, not even just every time this method is executed.
From section 3.10.5 of the Java Language Specification:
Each string literal is a reference (§4.3) to an instance (§4.3.1, §12.5) of class String (§4.3.3). String objects have a constant value. String literals-or, more generally, strings that are the values of constant expressions (§15.28)-are "interned" so as to share unique instances, using the method String.intern.
However, in your edit:
I just remembered that if a String is created, it is stored in the String Pool and after that it is always retrieved from the pool without being re-created. Am I right?
No, that's not right. Only string constants are interned by default. For example, every time you run this code:
public List<String> createStrings() {
List<String> list = new ArrayList<String>();
for (int i = 0; i < 10; i++) {
list.add("foo" + i);
}
return list;
}
... that will create another 10 strings.

Java Strings: "String s = new String("silly");"

I'm a C++ guy learning Java. I'm reading Effective Java and something confused me. It says never to write code like this:
String s = new String("silly");
Because it creates unnecessary String objects. But instead it should be written like this:
String s = "No longer silly";
Ok fine so far...However, given this class:
public final class CaseInsensitiveString {
private String s;
public CaseInsensitiveString(String s) {
if (s == null) {
throw new NullPointerException();
}
this.s = s;
}
:
:
}
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
String s = "polish";
Why is the first statement ok? Shouldn't it be
CaseInsensitiveString cis = "Polish";
How do I make CaseInsensitiveString behave like String so the above statement is OK (with and without extending String)? What is it about String that makes it OK to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java?

String is a special built-in class of the language. It is for the String class only in which you should avoid saying
String s = new String("Polish");
Because the literal "Polish" is already of type String, and you're creating an extra unnecessary object. For any other class, saying
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
is the correct (and only, in this case) thing to do.

I believe the main benefit of using the literal form (ie, "foo" rather than new String("foo")) is that all String literals are 'interned' by the VM. In other words it is added to a pool such that any other code that creates the same string will use the pooled String rather than creating a new instance.
To illustrate, the following code will print true for the first line, but false for the second:
System.out.println("foo" == "foo");
System.out.println(new String("bar") == new String("bar"));

Strings are treated a bit specially in java, they're immutable so it's safe for them to be handled by reference counting.
If you write
String s = "Polish";
String t = "Polish";
then s and t actually refer to the same object, and s==t will return true, for "==" for objects read "is the same object" (or can, anyway, I"m not sure if this is part of the actual language spec or simply a detail of the compiler implementation-so maybe it's not safe to rely on this) .
If you write
String s = new String("Polish");
String t = new String("Polish");
then s!=t (because you've explicitly created a new string) although s.equals(t) will return true (because string adds this behavior to equals).
The thing you want to write,
CaseInsensitiveString cis = "Polish";
can't work because you're thinking that the quotations are some sort of short-circuit constructor for your object, when in fact this only works for plain old java.lang.Strings.

String s1="foo";
literal will go in pool and s1 will refer.
String s2="foo";
this time it will check "foo" literal is already available in StringPool or not as now it exist so s2 will refer the same literal.
String s3=new String("foo");
"foo" literal will be created in StringPool first then through string arg constructor String Object will be created i.e "foo" in the heap due to object creation through new operator then s3 will refer it.
String s4=new String("foo");
same as s3
so System.out.println(s1==s2);// **true** due to literal comparison.
and System.out.println(s3==s4);// **false** due to object
comparison(s3 and s4 is created at different places in heap)

Strings are special in Java - they're immutable, and string constants are automatically turned into String objects.
There's no way for your SomeStringClass cis = "value" example to apply to any other class.
Nor can you extend String, because it's declared as final, meaning no sub-classing is allowed.

Java strings are interesting. It looks like the responses have covered some of the interesting points. Here are my two cents.
strings are immutable (you can never change them)
String x = "x";
x = "Y";
The first line will create a variable x which will contain the string value "x". The JVM will look in its pool of string values and see if "x" exists, if it does, it will point the variable x to it, if it does not exist, it will create it and then do the assignment
The second line will remove the reference to "x" and see if "Y" exists in the pool of string values. If it does exist, it will assign it, if it does not, it will create it first then assignment. As the string values are used or not, the memory space in the pool of string values will be reclaimed.
string comparisons are contingent on what you are comparing
String a1 = new String("A");
String a2 = new String("A");
a1 does not equal a2
a1 and a2 are object references
When string is explicitly declared, new instances are created and their references will not be the same.
I think you're on the wrong path with trying to use the caseinsensitive class. Leave the strings alone. What you really care about is how you display or compare the values. Use another class to format the string or to make comparisons.
i.e.
TextUtility.compare(string 1, string 2)
TextUtility.compareIgnoreCase(string 1, string 2)
TextUtility.camelHump(string 1)
Since you are making up the class, you can make the compares do what you want - compare the text values.

The best way to answer your question would be to make you familiar with the "String constant pool". In java string objects are immutable (i.e their values cannot be changed once they are initialized), so when editing a string object you end up creating a new edited string object wherease the old object just floats around in a special memory ares called the "string constant pool". creating a new string object by
String s = "Hello";
will only create a string object in the pool and the reference s will refer to it, but by using
String s = new String("Hello");
you create two string objects: one in the pool and the other in the heap. the reference will refer to the object in the heap.

You can't. Things in double-quotes in Java are specially recognised by the compiler as Strings, and unfortunately you can't override this (or extend java.lang.String - it's declared final).

- How do i make CaseInsensitiveString behave like String so the above statement is ok (with and w/out extending String)? What is it about String that makes it ok to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java right?
Enough has been said from the first point. "Polish" is an string literal and cannot be assigned to the CaseInsentiviveString class.
Now about the second point
Although you can't create new literals, you can follow the first item of that book for a "similar" approach so the following statements are true:
// Lets test the insensitiveness
CaseInsensitiveString cis5 = CaseInsensitiveString.valueOf("sOmEtHiNg");
CaseInsensitiveString cis6 = CaseInsensitiveString.valueOf("SoMeThInG");
assert cis5 == cis6;
assert cis5.equals(cis6);
Here's the code.
C:\oreyes\samples\java\insensitive>type CaseInsensitiveString.java
import java.util.Map;
import java.util.HashMap;
public final class CaseInsensitiveString {
private static final Map<String,CaseInsensitiveString> innerPool
= new HashMap<String,CaseInsensitiveString>();
private final String s;
// Effective Java Item 1: Consider providing static factory methods instead of constructors
public static CaseInsensitiveString valueOf( String s ) {
if ( s == null ) {
return null;
}
String value = s.toLowerCase();
if ( !CaseInsensitiveString.innerPool.containsKey( value ) ) {
CaseInsensitiveString.innerPool.put( value , new CaseInsensitiveString( value ) );
}
return CaseInsensitiveString.innerPool.get( value );
}
// Class constructor: This creates a new instance each time it is invoked.
public CaseInsensitiveString(String s){
if (s == null) {
throw new NullPointerException();
}
this.s = s.toLowerCase();
}
public boolean equals( Object other ) {
if ( other instanceof CaseInsensitiveString ) {
CaseInsensitiveString otherInstance = ( CaseInsensitiveString ) other;
return this.s.equals( otherInstance.s );
}
return false;
}
public int hashCode(){
return this.s.hashCode();
}
// Test the class using the "assert" keyword
public static void main( String [] args ) {
// Creating two different objects as in new String("Polish") == new String("Polish") is false
CaseInsensitiveString cis1 = new CaseInsensitiveString("Polish");
CaseInsensitiveString cis2 = new CaseInsensitiveString("Polish");
// references cis1 and cis2 points to differents objects.
// so the following is true
assert cis1 != cis2; // Yes they're different
assert cis1.equals(cis2); // Yes they're equals thanks to the equals method
// Now let's try the valueOf idiom
CaseInsensitiveString cis3 = CaseInsensitiveString.valueOf("Polish");
CaseInsensitiveString cis4 = CaseInsensitiveString.valueOf("Polish");
// References cis3 and cis4 points to same object.
// so the following is true
assert cis3 == cis4; // Yes they point to the same object
assert cis3.equals(cis4); // and still equals.
// Lets test the insensitiveness
CaseInsensitiveString cis5 = CaseInsensitiveString.valueOf("sOmEtHiNg");
CaseInsensitiveString cis6 = CaseInsensitiveString.valueOf("SoMeThInG");
assert cis5 == cis6;
assert cis5.equals(cis6);
// Futhermore
CaseInsensitiveString cis7 = CaseInsensitiveString.valueOf("SomethinG");
CaseInsensitiveString cis8 = CaseInsensitiveString.valueOf("someThing");
assert cis8 == cis5 && cis7 == cis6;
assert cis7.equals(cis5) && cis6.equals(cis8);
}
}
C:\oreyes\samples\java\insensitive>javac CaseInsensitiveString.java
C:\oreyes\samples\java\insensitive>java -ea CaseInsensitiveString
C:\oreyes\samples\java\insensitive>
That is, create an internal pool of CaseInsensitiveString objects, and return the corrensponding instance from there.
This way the "==" operator returns true for two objects references representing the same value.
This is useful when similar objects are used very frequently and creating cost is expensive.
The string class documentation states that the class uses an internal pool
The class is not complete, some interesting issues arises when we try to walk the contents of the object at implementing the CharSequence interface, but this code is good enough to show how that item in the Book could be applied.
It is important to notice that by using the internalPool object, the references are not released and thus not garbage collectible, and that may become an issue if a lot of objects are created.
It works for the String class because it is used intensively and the pool is constituted of "interned" object only.
It works well for the Boolean class too, because there are only two possible values.
And finally that's also the reason why valueOf(int) in class Integer is limited to -128 to 127 int values.

In your first example, you are creating a String, "silly" and then passing it as a parameter to another String's copy constructor, which makes a second String which is identical to the first. Since Java Strings are immutable (something that frequently stings people who are used to C strings), this is a needless waste of resources. You should instead use the second example because it skips several needless steps.
However, the String literal is not a CaseInsensitiveString so there you cannot do what you want in your last example. Furthermore, there is no way to overload a casting operator like you can in C++ so there is literally no way to do what you want. You must instead pass it in as a parameter to your class's constructor. Of course, I'd probably just use String.toLowerCase() and be done with it.
Also, your CaseInsensitiveString should implement the CharSequence interface as well as probably the Serializable and Comparable interfaces. Of course, if you implement Comparable, you should override equals() and hashCode() as well.

CaseInsensitiveString is not a String although it contains a String. A String literal e.g "example" can be only assigned to a String.

CaseInsensitiveString and String are different objects. You can't do:
CaseInsensitiveString cis = "Polish";
because "Polish" is a String, not a CaseInsensitiveString. If String extended CaseInsensitiveString String then you'd be OK, but obviously it doesn't.
And don't worry about the construction here, you won't be making unecessary objects. If you look at the code of the constructor, all it's doing is storing a reference to the string you passed in. Nothing extra is being created.
In the String s = new String("foobar") case it's doing something different. You are first creating the literal string "foobar", then creating a copy of it by constructing a new string out of it. There's no need to create that copy.

Just because you have the word String in your class, does not mean you get all the special features of the built-in String class.

when they say to write
String s = "Silly";
instead of
String s = new String("Silly");
they mean it when creating a String object because both of the above statements create a String object but the new String() version creates two String objects: one in heap and the other in string constant pool. Hence using more memory.
But when you write
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
you are not creating a String instead you are creating an object of class CaseInsensitiveString. Hence you need to use the new operator.

If I understood it correctly, your question means why we cannot create an object by directly assigning it a value, lets not restrict it to a Wrapper of String class in java.
To answer that I would just say, purely Object Oriented Programming languages have some constructs and it says, that all the literals when written alone can be directly transformed into an object of the given type.
That precisely means, if the interpreter sees 3 it will be converted into an Integer object because integer is the type defined for such literals.
If the interpreter sees any thing in single quotes like 'a' it will directly create an object of type character, you do not need to specify it as the language defines the default object of type character for it.
Similarly if the interpreter sees something in "" it will be considered as an object of its default type i.e. string. This is some native code working in the background.
Thanks to MIT video lecture course 6.00 where I got the hint for this answer.

In Java the syntax "text" creates an instance of class java.lang.String. The assignment:
String foo = "text";
is a simple assignment, with no copy constructor necessary.
MyString bar = "text";
Is illegal whatever you do because the MyString class isn't either java.lang.String or a superclass of java.lang.String.

First, you can't make a class that extends from String, because String is a final class. And java manage Strings differently from other classes so only with String you can do
String s = "Polish";
But whit your class you have to invoke the constructor. So, that code is fine.

I would just add that Java has Copy constructors...
Well, that's an ordinary constructor with an object of same type as argument.

String is one of the special classes in which you can create them without the new Sring part
it's the same as
int x = y;
or
char c;

String str1 = "foo";
String str2 = "foo";
Both str1 and str2 belongs to tha same String object, "foo", b'coz Java manages Strings in StringPool, so if a new variable refers to the same String, it doesn't create another one rather assign the same alerady present in StringPool.
String str1 = new String("foo");
String str2 = new String("foo");
Here both str1 and str2 belongs to different Objects, b'coz new String() forcefully create a new String Object.

It is a basic law that Strings in java are immutable and case sensitive.

Java creates a String object for each string literal you use in your code. Any time "" is used, it is the same as calling new String().
Strings are complex data that just "act" like primitive data. String literals are actually objects even though we pretend they're primitive literals, like 6, 6.0, 'c', etc. So the String "literal" "text" returns a new String object with value char[] value = {'t','e','x','t}. Therefore, calling
new String("text");
is actually akin to calling
new String(new String(new char[]{'t','e','x','t'}));
Hopefully from here, you can see why your textbook considers this redundant.
For reference, here is the implementation of String: http://www.docjar.com/html/api/java/lang/String.java.html
It's a fun read and might inspire some insight. It's also great for beginners to read and try to understand, as the code demonstrates very professional and convention-compliant code.
Another good reference is the Java tutorial on Strings:
http://docs.oracle.com/javase/tutorial/java/data/strings.html

In most versions of the JDK the two versions will be the same:
String s = new String("silly");
String s = "No longer silly";
Because strings are immutable the compiler maintains a list of string constants and if you try to make a new one will first check to see if the string is already defined. If it is then a reference to the existing immutable string is returned.
To clarify - when you say "String s = " you are defining a new variable which takes up space on the stack - then whether you say "No longer silly" or new String("silly") exactly the same thing happens - a new constant string is compiled into your application and the reference points to that.
I dont see the distinction here. However for your own class, which is not immutable, this behaviour is irrelevant and you must call your constructor.
UPDATE: I was wrong!
I tested this and realise that my understanding is wrong - new String("Silly") does indeed create a new string rather than reuse the existing one. I am unclear why this would be (what is the benefit?) but code speaks louder than words!