Strings in Java
In the given example only one object will be created. Firstly JVM will not find any string object with the value “Welcome” in string constant pool, so it will create a new object. After that it will find the string with the value “Welcome” in the pool, it will not create new object but will return the reference to the same instance.
Note: String objects are stored in a special memory area known as string constant pool.
Why java uses the concept of string literal?
To make Java more memory efficient (because no new objects are created if it exists already in string constant pool).
2. By new keyword
- String s=new String(“Welcome”);
- In such a case, JVM will create a new string object in normal (non pool) heap memory and the literal “Welcome” will be placed in the string constant pool. The variable s will refer to the object in heap (non pool).
Example:
Java
public class StringExample{ public static void main(String args[]) { String s1=new String("example");//creating java string by new keyword //this statement create two object i.e first object is created in heap //memory area and second object is create in String constant pool. System.out.println(s1); }} |
example
Syntax:
<String_Type> <string_variable> = "<sequence_of_string>";
Example:
String str = "Geeks";
Memory allotment of String: Whenever a String Object is created as a literal, the object will be created in the String constant pool. This allows JVM to optimize the initialization of String literal.
For example:
String str = "Geeks";
The string can also be declared using new operator i.e. dynamically allocated. In case of String are dynamically allocated they are assigned a new memory location in heap. This string will not be added to String constant pool.
For example:
String str = new String("Geeks");If you want to store this string in the constant pool then you will need to “intern” it.
For example:
String internedString = str.intern(); // this will add the string to string constant pool.
It is preferred to use String literals as it allows JVM to optimize memory allocation.
An example that shows how to declare a String
Java
// Java code to illustrate Stringimport java.io.*;import java.lang.*;class Test { public static void main(String[] args) { // Declare String without using new operator String s = "GeeksforGeeks"; // Prints the String. System.out.println("String s = " + s); // Declare String using new operator String s1 = new String("GeeksforGeeks"); // Prints the String. System.out.println("String s1 = " + s1); }} |
String s = GeeksforGeeks String s1 = GeeksforGeeks
Interfaces and Classes in Strings in Java
CharBuffer: This class implements the CharSequence interface. This class is used to allow character buffers to be used in place of CharSequences. An example of such usage is the regular-expression package java.util.regex.
String: It is a sequence of characters. In java, objects of String are immutable which means a constant and cannot be changed once created.
Ways of Creating a String
There are two ways to create a string in Java:
String literal
String s = “GeeksforGeeks”;
Using new keyword
String s = new String (“GeeksforGeeks”);
StringBuffer is a peer class of String that provides much of the functionality of strings. The string represents fixed-length, immutable character sequences while StringBuffer represents growable and writable character sequences.
Syntax:
StringBuffer s = new StringBuffer("GeeksforGeeks");StringBuilder in Java represents a mutable sequence of characters. Since the String Class in Java creates an immutable sequence of characters, the StringBuilder class provides an alternate to String Class, as it creates a mutable sequence of characters.
Syntax:
StringBuilder str = new StringBuilder();
str.append("GFG");StringTokenizer class in Java is used to break a string into tokens.
Example:
A StringTokenizer object internally maintains a current position within the string to be tokenized. Some operations advance this current position past the characters processed. A token is returned by taking a substring of the string that was used to create the StringTokenizer object.
StringJoiner is a class in java.util package which is used to construct a sequence of characters(strings) separated by a delimiter and optionally starting with a supplied prefix and ending with a supplied suffix. Though this can also be with the help of StringBuilder class to append delimiter after each string, StringJoiner provides an easy way to do that without much code to write.
Syntax:
public StringJoiner(CharSequence delimiter)
Above we saw we can create string by String Literal.
For ex- // String s=”Welcome”;
Here the JVM checks the String Constant Pool. If the string does not exist, then a new string instance is created and placed in a pool. If the string exists, then it will not create a new object. Rather, it will return the reference to the same instance. The cache that stores these string instances is known as the String Constant pool or String Pool. In earlier versions of Java up to JDK 6 String pool was located inside PermGen(Permanent Generation) space. But in JDK 7 it is moved to the main heap area.
Why did the String pool move from PermGen to the normal heap area?
PermGen space is limited, the default size is just 64 MB. it was a problem with creating and storing too many string objects in PermGen space. That’s why the String pool was moved to a larger heap area. To make Java more memory efficient, the concept of string literal is used. By the use of the ‘new’ keyword, The JVM will create a new string object in the normal heap area even if the same string object is present in the string pool.
For example:
String a=new String(“Bhubaneswar”)
Let us have a look at the concept with a java program and visualize the actual JVM memory structure:
Program:
Java
class StringStorage { public static void main(String args[]) { String s1 = "TAT"; String s2 = "TAT"; String s3 = new String("TAT"); String s4 = new String("TAT"); System.out.println(s1); System.out.println(s2); System.out.println(s3); System.out.println(s4); }} |
TAT TAT TAT TAT
Note: All objects in Java are stored in a heap. The reference variable is to the object stored in the stack area or they can be contained in other objects which puts them in the heap area also.
Example 1:
Java
//Construct String from subset of char arrayclass GFG{public static void main(String args[]){byte ascii[]={71,70,71}; String s1= new String(ascii); System.out.println(s1); String s2= new String(ascii,1,2); System.out.println(s2);}} |
GFG FG
Example 2:
Java
// Construct one string from anotherclass GFG{public static void main(String args[]){ char c[]={'G','f','g'}; String s1=new String (c); String s2=new String (s1); System.out.println(s1); System.out.println(s2);}} |
Gfg Gfg
Immutable String in Java
- In java, string objects are immutable. Immutable simply means unmodifiable or unchangeable.
- Once string object is created its data or state can’t be changed but a new string object is created.
Java
class Demo{ public static void main(String args[]) { String s="Sachin"; s.concat(" Tendulkar"); //concat() method appends the string at the end System.out.println(s); //will print Sachin because strings are immutable objects } } |
Sachin
Now it can be understood by the diagram given below. Here Sachin is not changed but a new object is created with “Sachin Tendulkar”. That is why string is known as immutable.
As you can see in the given figure that two objects are created but s reference variable still refers to “Sachin” not to “Sachin Tendulkar”.
But if we explicitly assign it to the reference variable, it will refer to “Sachin Tendulkar” object. For example:
Java
class Demo{ public static void main(String args[]) { String s="Sachin"; s=s.concat(" Tendulkar"); System.out.println(s); } } |
Sachin Tendulkar
Why string objects are immutable in java?
Because java uses the concept of string literal. Suppose there are 5 reference variables, all referes to one object “sachin”. If one reference variable changes the value of the object, it will be affected to all the reference variables. That is why string objects are immutable in java.
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