Java: String is Immutable. What exactly is the meaning?

Before proceeding further with the fuss of immutability, let’s just take a look into the String class and its functionality a little before coming to any conclusion.

This is how String works:

String str = "knowledge";

This, as usual, creates a string containing "knowledge" and assigns it a reference str. Simple enough? Lets perform some more functions:



// assigns a new reference to the 
// same string "knowledge"
String s = str;     

Let’s see how the below statement works:

  str = str.concat(" base");

This appends a string " base" to str. But wait, how is this possible, since String objects are immutable? Well to your surprise, it is.

When the above statement is executed, the VM takes the value of String str, i.e. "knowledge" and appends " base", giving us the value "knowledge base". Now, since Strings are immutable, the VM can’t assign this value to str, so it creates a new String object, gives it a value "knowledge base", and gives it a reference str.

An important point to note here is that, while the String object is immutable, its reference variable is not. So that’s why, in the above example, the reference was made to refer to a newly formed String object.



At this point in the example above, we have two String objects: the first one we created with value "knowledge", pointed to by s, and the second one "knowledge base", pointed to by str. But, technically, we have three String objects, the third one being the literal "base" in the concat statement.

Important Facts about String and Memory usage

What if we didn’t have another reference s to "knowledge"? We would have lost that String. However, it still would have existed, but would be considered lost due to having no references.
Look at one more example below

filter_none

edit
close

play_arrow

link
brightness_4
code

/*package whatever // do not write package name here */
  
import java.io.*;
  
class GFG {
    public static void main(String[] args)
    {
        String s1 = "java";
        s1.concat(" rules");
  
        // Yes, s1 still refers to "java"
        System.out.println("s1 refers to " + s1);
    }
}

chevron_right


Output:

s1 refers to java

What’s happening:

  1. The first line is pretty straightforward: create a new String "java" and refer s1 to it.
  2. Next, the VM creates another new String "java rules", but nothing refers to it. So, the second String is instantly lost. We can’t reach it.

The reference variable s1 still refers to the original String "java".

Almost every method, applied to a String object in order to modify it, creates new String object. So, where do these String objects go? Well, these exist in memory, and one of the key goals of any programming language is to make efficient use of memory.

As applications grow, it’s very common for String literals to occupy large area of memory, which can even cause redundancy. So, in order to make Java more efficient, the JVM sets aside a special area of memory called the “String constant pool“.

When the compiler sees a String literal, it looks for the String in the pool. If a match is found, the reference to the new literal is directed to the existing String and no new String object is created. The existing String simply has one more reference. Here comes the point of making String objects immutable:

In the String constant pool, a String object is likely to have one or many references. If several references point to same String without even knowing it, it would be bad if one of the references modified that String value. That’s why String objects are immutable.

Well, now you could say, what if someone overrides the functionality of String class? That’s the reason that the String class is marked final so that nobody can override the behavior of its methods.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.