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String object in java

As we know String is immutable, which means a new instance is created every time.
My question is that if I write:
System.out.println("Java"+"is"+"programming");
then how many objects are created in pool?

Your example will create a single string object in the string pool.
after that, if you do:
String x = "Javaisprogramming";
it will still point to the same object in the string pool. You can read more here

Your example will create a single string object. This is noted throughout the Java documentation.
The way I understand it, Java only adds new strings to the string pool when they're initially created. So,
String str1 = "hello";
Would be a single string in the pool.
So would
String str2 = "Java" + "is" + "cool";

Your Example will create 'single' string only which is
String x = "Javaisprogramming";
because '+' is a string concatenation operator in java(it internally call append()) so when you call System.out.println("Java"+"is"+"programming"); then compiler create one string object from concatenation and send it with println() method.
for more about '+' operator see this answer

In Java, does `String s = "lol";` create two objects? [duplicate]

This question already has answers here:
Java Strings and StringPool
(2 answers)
Closed 9 years ago.
I have heard that two objects are created when you execute String s = new String("lol");. One object is created for the string constant pool and one for s on the heap.
So, are 2 objects created when we execute the following? String s = "lol"; Is the object creation the same?
Edit:
how many objects are created by :
String s1 = new String("lol1");
and how many by :
String s2 = "lol2";

No, with String s = "lol";, only one object is created. With every string literal, a String object is created and placed in the string pool. Here, s just refers to that pooled string. When you say s = new String("lol"), the string literal is created and pooled, and another string is allocated and assigned to s, which is a different, yet equal, string.
UPDATE
I had forgotten about the char[] that is used internally by a String object.
String s1 = "lol";
2 objects are created, the char[] that holds {'l', 'o', 'l'} and the String object that references it internally. It's interned in the string pool.
String s2 = new String("lol");
3 objects are created. First, the string literal: 2 objects are created, the char[] that holds {'l', 'o', 'l'} and the String object that references it. It's interned in the string pool as before. Then, the new String object that gets assigned to s2: A new String is created, but it references the same char array as the original string. Two String objects, and one char[] object. (The String(String) constructor may make a copy of the char[] in the circumstance that the original string's array's length is somehow greater than its count, but that doesn't appear to be the case here.)
Grepcode for java.lang.String(String)

"lol" is a String literal - when you reference it in your code, you force Java to create this object. The second object you're seeing is when you explicitly call String's constructor with the new operator. Assigning these values to other variables does not create additional objects.

Strings [= new String vs = ""]

So my question is in relation to declaring and assigning strings.
The way I usually declare strings is by doing the following:
String s1 = "Stackoverflow";
And then if I ever need to change the value of s1 I would do the following:
s1 = "new value";
Today I found another way of doing it and declaring a string would go like:
String s2 = new String("Stackoverflow");
And then changing the value would be:
s2 = new String("new value");
My question is what is the difference between the two or is it simply preferential. From looking at the code the fourth line
s2 = new String ("new value");
I'm assuming that doing that would create a new memory location and s2 would then point to it so I doubt it would be used to change the value but I can see it being used when declaring a string.

From the javadoc :
Initializes a newly created String object so that it represents the
same sequence of characters as the argument; in other words, the newly
created string is a copy of the argument string. Unless an explicit
copy of original is needed, use of this constructor is unnecessary
since Strings are immutable.
So no, you have no reason not to use the simple literal.
Simply do
String s1 = "Stackoverflow";
Historically, this constructor was mainly used to get a lighter copy of a string obtained by splitting a bigger one (see this question). Now, There's no normal reason to use it.

String s1 = "Stackoverflow"; // declaring & initializing s1
String s2 = new String("Stackoverflow"); // declaring & initializing s2
in above cases you are declaring & initializing String object.
The difference between
//first case
String s2 = new String("new String"); // declaring & initializing s2 with new memory
// second case
s2 = "new String" // overwriting previous value of s2
is in the first case you are creating a new object, i-e; allocating memory for a new object which will be refrenced by s2. The previous address to which s2 was pointing/referencing hasn't released the memory yet which will be gc when the program ends or when system needs to.
The good programming practice (second case) is to initialize an object once, and if you want to change its value either assign null to it and then allocate new memory to it or in case of string you can do like this
s2= "new String";

String s1 = "Stackoverflow";
This line will create a new object in the String pool if it does not exist already. That means first it will first try searching for "Stackoverflow" in the String pool, if found then s1 will start pointing to it, if not found then it will create a new object and s1 will refer to it.
String s2 = new String("Stackoverflow");
Will always create a new String object, no matter if the value already exists in the String pool. So the object in the heap would then have the value "Stackovrflow" and s2 will start pointing to it.
s2 = new String("new value");
This will again create a new Object and s2 will start pointing to it. The earlier object that s2 was pointing to is not open for garbage collection (gc).
Let me know if this helps.

The main difference is that the constructor always creates a totally new instance of String containing the same characters than the original String.
String s1 = "Stackoverflow";
String s2 = "Stackoverflow";
Then s1 == s2 will return true
String s1 = "Stackoverflow";
String s2 = new String("Stackoverflow");
Then s1 == s2 will return false
Just using the double quotes option is often better:
With a constructors you might create two instance of String.
Easier to read and less confusing

#user2612619 here i want to say is .. when ever ur creating object with "new" operator it always falls in heap memory . so wenever u have same content but different objects than also it will create new objects on heap by this u cant save memory ....
but in order to save memory java people brought concept of immutable where we can save memory .. if u r creating a different object with same content .. string will recognize dat difference and creates only one object with same content and points the two references to only one object ..
i can solve ur doubts from this figure ..
case 1:
String s = new String("stackoverflow");
String s1 = new String("stackoverflow");
as they are two different objects on heap memory with two different values of hashcode .. so s==s1 (is reference comparison) it is false .. but s.equals(s1) is content comparison ..so it is true
case 2:
String s = "stackoverflow";
String s1 = "stackoverflow";
here objects fall in scp(string constant pool memory)
same object for two different refernces .. so hashcode also same .. same reference value .. so s==s1 is true here [u can observe from figure clearly]
s.equals(s1) is true as content comparison.. this is very beautiful concept .. u will love it wen u solve some problems ... all d best

Creating String object Directly:
String s1 = "Hip Hop"
will create a string object but first JVM checks the String constant or literal pool and if the string does not exist, it creates a new String object “Hip jop” and a reference is maintained in the pool. The variable s1 also refers the same object. Now, if we put a statement after this:
String s2 = "Hip Hop"
JVM checks the String constant pool first and since the string already exists, a reference to the pooled instance is returned to s2.
System.out.println(s1==s2) // comparing reference and it will print true
java can make this optimization since strings are immutable and can be shared without fear of data corruption.
Creating String Object using new
String s3 = new String("Hip Hop")
For new keyword, a String object is created in the heap memory wither or not an equal string object already exists in the pool and s3 will refer to the newly created one.
System.out.println(s3==s2) // two reference is different, will print false
String objects created with the new operator do not refer to objects in the string pool but can be made to using String’s intern() method. The java.lang.String.intern() returns an interned String, that is, one that has an entry in the global String literal pool. If the String is not already in the global String literal pool, then it will be added to the pool.
String s4 = s3.intern();
Systen.out.println(s4 == s2); // will print `true` because s4 is interned,
//it now have the same reference to "Hip Hop" as s2 or s1
But try:
Systen.out.println(s4 == s3) // it will be false,
As the reference of s4, s2 and s1 is the reference to the pooled instance while s3 is referring to the created object in heap memory.
use of new for creting string:
prior to OpenJDK 7, Update 6, Java String.susbtring method had potential memory leak. substring method would build a new String object keeping a reference to the whole char array, to avoid copying it. You could thus inadvertently keep a reference to a very big character array with just a one character string. If we want to have minimal strings after substring, we use the constructor taking another string :
String s2 = new String(s1.substring(0,1));
The issue is resolved in JDK 7 update 6. So No need to create string with new any more for taking advantage provided by String literal pool mechanism.
Referance:
String literal pool
using new to prevent memory leak for using substring

Checking For Equal Instances of 2 Different (Included Example)

I use the == in the code below and prints out "Equals!", why? Can someone explain why these two different strings a and b are equal?
public class test
{
public static void main()
{
String a = "boy";
String b = "boy";
if(a == b)
{
System.out.println("Equals!");
}
else
{
System.out.println("Does not equal!");
}
}
}

This is due to String interning.
Java (The JVM) keeps a collection of String literals that is uses to save memory. So, whenever you create a String like so:
String s = "String";
Java 'interns' the string. However, if you create the String like so:
String s = new String("String");
Java will not automatically intern the String. If you created your strings this way, your code would produce different results.
A quick Google search reveals lots of good resources regarding String interning.

This article will explain it in details:
What is the difference between == and equals() in Java?
After the execution of String a =
“boy”; the JVM adds the
string “boy” to the string
pool and on the next line of the code, it
encounters String b = ”boy” again; in this case the JVM already
knows that this string is already
there in the pool, so it does not create a
new string. So both strings a and b point to the same string what means they
point to the same reference.

String a = "boy"; will create a new string object with value ("boy"), place it in the string pool and make a refer to it.
When the interpreter sees String b = "boy";, it first checks to see if string "boy" is present in the string pool, since it is present, no new object is created and b is made to refer to the same object that a is referring to.
Since both references contain the same content they pass the equality test.

Because the run time will have a string pool and when you need to assign a new constant string, the run time look inside the pool, if the pool contains it, then they set the variable point to the same String object inside the pool.
But you should never depends on this to check for content string equals. You should use the method: equals

Whenever we create a string like below :
String str1 = "abc";
String str2 = "abc";
JVM will check the str2 = "abc" in the string constant pool, if it is present then it wont create a new String instead it point to the string one in the string constant pool.
But in case of this String str = new String("abc"); it will always create a new String Object but we can use intern() function to force JVM to look into the string constant pool.

As rightly explained above, in the case of '==' comparison, the runtime will look into the String pool for the existence of the string. However, it very much possible that during garbage collection, or during memory issues, the virtual machine might destroy the string pool. The "==" operator therefor might or might not return the correct value.
Lesson - Always use equals() for comparison.

Strings are objects in Java, so why don't we use 'new' to create them?

We normally create objects using the new keyword, like:
Object obj = new Object();
Strings are objects, yet we do not use new to create them:
String str = "Hello World";
Why is this? Can I make a String with new?

In addition to what was already said, String literals [ie, Strings like "abcd" but not like new String("abcd")] in Java are interned - this means that every time you refer to "abcd", you get a reference to a single String instance, rather than a new one each time. So you will have:
String a = "abcd";
String b = "abcd";
a == b; //True
but if you had
String a = new String("abcd");
String b = new String("abcd");
then it's possible to have
a == b; // False
(and in case anyone needs reminding, always use .equals() to compare Strings; == tests for physical equality).
Interning String literals is good because they are often used more than once. For example, consider the (contrived) code:
for (int i = 0; i < 10; i++) {
System.out.println("Next iteration");
}
If we didn't have interning of Strings, "Next iteration" would need to be instantiated 10 times, whereas now it will only be instantiated once.

Strings are "special" objects in Java. The Java designers wisely decided that Strings are used so often that they needed their own syntax as well as a caching strategy. When you declare a string by saying:
String myString = "something";
myString is a reference to String object with a value of "something". If you later declare:
String myOtherString = "something";
Java is smart enough to work out that myString and myOtherString are the same and will store them in a global String table as the same object. It relies on the fact that you can't modify Strings to do this. This lowers the amount of memory required and can make comparisons faster.
If, instead, you write
String myOtherString = new String("something");
Java will create a brand new object for you, distinct from the myString object.

String a = "abc"; // 1 Object: "abc" added to pool
String b = "abc"; // 0 Object: because it is already in the pool
String c = new String("abc"); // 1 Object
String d = new String("def"); // 1 Object + "def" is added to the Pool
String e = d.intern(); // (e==d) is "false" because e refers to the String in pool
String f = e.intern(); // (f==e) is "true"
//Total Objects: 4 ("abc", c, d, "def").
Hope this clears a few doubts. :)

We usually use String literals to avoid creating unnecessary objects. If we use new operator to create String object , then it will create new object everytime .
Example:
String s1=“Hello“;
String s2=“Hello“;
String s3= new String(“Hello“);
String s4= new String(“Hello“);
For the above code in memory :

It's a shortcut. It wasn't originally like that, but Java changed it.
This FAQ talks about it briefly. The Java Specification guide talks about it also. But I can't find it online.

String is subject to a couple of optimisations (for want of a better phrase). Note that String also has operator overloading (for the + operator) - unlike other objects. So it's very much a special case.

In Java, Strings are a special case, with many rules that apply only to Strings. The double quotes causes the compiler to create a String object. Since String objects are immutable, this allows the compiler to intern multiple strings, and build a larger string pool. Two identical String constants will always have the same object reference. If you don't want this to be the case, then you can use new String(""), and that will create a String object at runtime. The intern() method used to be common, to cause dynamically created strings to be checked against the string lookup table. Once a string in interned, the object reference will point to the canonical String instance.
String a = "foo";
String b = "foo";
System.out.println(a == b); // true
String c = new String(a);
System.out.println(a == c); // false
c = c.intern();
System.out.println(a == c); // true
When the classloader loads a class, all String constants are added to the String pool.

Well the StringPool is implemented using The Hashmap in java. If we are creating always with a new keyword its not searching in String Pool and creating a new memory for it which might be needed later if we have a memory intensive operation running and if we are creating all the strings with new keyword that would affect performance of our application. So its advisable to not to use new keywords for creating string because then only it will go to String pool which in turn is a Hashmap ,(memory saved , imagine if we have lots of strings created with new keyword ) here it will be stored and if the string already exists the reference of it(which would usually reside in Stack memory) would be returned to the newly created string.
So its done to improve performance .

Syntactic sugar. The
String s = new String("ABC");
syntax is still available.

You can still use new String("string"), but it would be harder to create new strings without string literals ... you would have to use character arrays or bytes :-) String literals have one additional property: all same string literals from any class point to same string instance (they are interned).

There's almost no need to new a string as the literal (the characters in quotes) is already a String object created when the host class is loaded. It is perfectly legal to invoke methods on a literal and don, the main distinction is the convenience provided by literals. It would be a major pain and waste of tine if we had to create an array of chars and fill it char by char and them doing a new String(char array).

Feel free to create a new String with
String s = new String("I'm a new String");
The usual notation s = "new String"; is more or less a convenient shortcut - which should be used for performance reasons except for those pretty rare cases, where you really need Strings that qualify for the equation
(string1.equals(string2)) && !(string1 == string2)
EDIT
In response to the comment: this was not intended to be an advise but just an only a direct response to the questioners thesis, that we do not use the 'new' keyword for Strings, which simply isn't true. Hope this edit (including the above) clarifies this a bit. BTW - there's a couple of good and much better answers to the above question on SO.

The literal pool contains any Strings that were created without using the keyword new.
There is a difference : String without new reference is stored in String literal pool and String with new says that they are in heap memory.
String with new are elsewhere in memory just like any other object.

Because String is an immutable class in java.
Now why it is immutable?
As String is immutable so it can be shared between multiple threads and we dont need to synchronize String operation externally.
As String is also used in class loading mechanism. So if String was mutable then java.io.writer could have been changed to abc.xyz.mywriter

TString obj1 = new TString("Jan Peter");
TString obj2 = new TString("Jan Peter");
if (obj1.Name == obj2.Name)
System.out.println("True");
else
System.out.println("False");
Output:
True
I created two separate objects, both have a field(ref) 'Name'. So even in this case "Jan Peter" is shared, if I understand the way java deals..