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.
Example:
String demoString = "Geeks";
The string can also be declared using a new operator i.e. dynamically allocated. In case of String are dynamically allocated they are assigned a new memory location in the heap. This string will not be added to the String constant pool.
Example:
String demoString = new String("Geeks");
If you want to store this string in the constant pool then you will need to “intern” it.
Example:
String internedString = demoString.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 code to illustrate String
import java.io.*;
import java.lang.*;
class Test {
public static void main(String[] args)
{
// Declare String without using new operator
String name = "w3wiki";
// Prints the String.
System.out.println("String name = " + name);
// Declare String using new operator
String newString = new String("w3wiki");
// Prints the String.
System.out.println("String newString = " + newString);
}
}
Output
String name = w3wiki String newString = w3wiki
Note: String Object is created in Heap area and Literals are stored in special memory area known as string constant pool.
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 the 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 a new object but will return the reference to the same instance. In this article, we will learn about Java Strings.