Misunderstandings with java string..

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; } }