Immutability of Strings in Java

Question

Consider the following example.

String str = new String();

str  = \"Hello\";
System.out.println(str);  //Prints Hello

str = \"Help!\";
System.out.println(s         


        

Answer 1

Though java tries to ignore it, str is nothing more than a pointer. This means that when you first write str = "Hello";, you create an object that str points to. When you reassign str by writing str = "Help!";, a new object is created and the old "Hello" object gets garbage collected whenever java feels like it.

Answer 2

Light_handle I recommend you take a read of Cup Size -- a story about variables and Pass-by-Value Please (Cup Size continued). This will help a lot when reading the posts above.

Have you read them? Yes. Good.

String str = new String();

This creates a new "remote control" called "str" and sets that to the value new String() (or "").

e.g. in memory this creates:

str --- > ""

---

str  = "Hello";

This then changes the remote control "str" but does not modify the original string "".

e.g. in memory this creates:

str -+   ""
     +-> "Hello"

---

str = "Help!";

This then changes the remote control "str" but does not modify the original string "" or the object that the remote control currently points to.

e.g. in memory this creates:

str -+   ""
     |   "Hello"
     +-> "Help!"

Answer 3

Or you can try:

public class Tester
{
public static void main(String[] args)
{
 String str = "Mississippi"; 
 System.out.println(str); // prints Mississippi 
 System.out.println(str.hashCode());

 str = str.replace("i", "!"); 
 System.out.println(str); // prints M!ss!ss!pp! 
 System.out.println(str.hashCode());
 }
 }

This will show how the hashcode changes.

Answer 4

String in Java in Immutable and Final just mean it can't be changed or modified:

Case 1:

class TestClass{  
 public static void main(String args[]){  
   String str = "ABC";  
   str.concat("DEF");  
   System.out.println(str);  
 }  
} 

Output: ABC

Reason: The object reference str is not changed in fact a new object "DEF" is created which is in the pool and have no reference at all (i.e lost).

Case 2:

class TestClass{  
 public static void main(String args[]){  
   String str="ABC";  
   str=str.concat("DEF");  
   System.out.println(str);  
 }  
}  

Output: ABCDEF

Reason: In this case str is now referring to a new object "ABCDEF" hence it prints ABCDEF i.e. previous str object "ABC" is lost in pool with no reference.

Answer 5

Because String is immutable so changes will not occur if you will not assign the returned value of function to the string.so in your question assign value of swap function  returned value to s.

s=swap(s, n1, n2) ;then the value of string s will change.

I was also getting the unchanged value when i was writing the program to get some permutations string(Although it is not giving all the permutations but this is for example to answer your question)

Here is a example.

> import java.io.*;  public class MyString { public static void
> main(String []args)throws IOException {  BufferedReader br=new
> BufferedReader(new InputStreamReader(System.in));  String
> s=br.readLine().trim(); int n=0;int k=0;  while(n!=s.length()) {
> while(k<n){  swap(s,k,n); System.out.println(s); swap(s,k,n); k++; }
> n++; } }  public static void swap(String s,int n1,int n2) { char temp;
> temp=s.charAt(n1); StringBuilder sb=new StringBuilder(s);
> sb.setCharAt(n1,s.charAt(n2)); sb.setCharAt(n2,temp); s=sb.toString();
> } }

but i was not getting the permuted values of the string from above code.So I assigned the returned value of the swap function to the string and got changed values of string. after assigning the returned value i got the permuted values of string.

/import java.util.*; import java.io.*; public class MyString { public static void main(String []args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); 
String s=br.readLine().trim(); int n=0;int k=0; 
while(n!=s.length()){ while(k<n){ s=swap(s,k,n); 
System.out.println(s); s=swap(s,k,n); k++; } n++; } } 
public static String swap(String s,int n1,int n2){
char temp; temp=s.charAt(n1); StringBuilder sb=new StringBuilder(s); sb.setCharAt(n1,s.charAt(n2)); sb.setCharAt(n2,temp); s=sb.toString(); return s; } }

Answer 6

Here immutability means that instance can point to other reference but the original content of the string would not be modified at the original reference. Let me explain by first example given by you. First str is pointing to "Hello" ,its Ok upto this. Second time its pointing to "Help!". Here str started pointing to "Help!" and the reference of "Hello" string is lost and we can not get that back.

In fact when str would try to modify the existing content,then another new string will be generated and str will start to point at that reference. So we see that string at original reference is not modified but that is safe at its reference and instance of object started pointing at different reference so immutability is conserve.