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Java
import java.util.*;
public class test {
    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        System.out.println("enter the number");
        String k = sc.next();

        int s = k.length()/2;
       boolean b = true;
        
      while(s>0){
          for (int i = 0; i<=s; i++) {
          if(k.charAt(i)==k.charAt(s)){
             b = true;
          }
          }
        if (b)
             s--;
        else
              System.out.println("exit");
    }  
     if(b)
            System.out.println("palindrome");
                
            
      }
    }


What I have tried:

I know it's very puny thing for experts here but I want to check why my palindrome program is not working as expected, it shows as palindrome to every number or string i enter, can you please look into it where the issue is, please. actually i'm trying to create this program on my own and not checking any ready made method for it so asking here. please help.
Posted
Updated 27-Jun-21 18:58pm
v2

Look at what you do with b :
Java
int s = k.length() / 2;
boolean b = true;

while (s > 0) {
    for (int i = 0; i <= s; i++) {
        // here print k.charAt(i) and k.charAt(s), just to see what you are comparing
        if (k.charAt(i) == k.charAt(s)) {
            b = true;
        }
    }

And that is only the first error.
Algorithm error: with palindromes, you know a string is a palindrome when you have found no difference in the whole check, you know a string is not a palindrome on first difference.
-----
Advice: Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.
Java
import java.util.*;
public class test {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        System.out.println("enter the number");
        String k = sc.next();

        int s = k.length() / 2;
        boolean b = true;

        while (s > 0) {
            for (int i = 0; i <= s; i++) {
                if (k.charAt(i) == k.charAt(s)) {
                    b = true;
                }
            }
            if (b)
                s--;
            else
                System.out.println("exit");
        }
        if (b)
            System.out.println("palindrome");


    }
}

Indentation style - Wikipedia[^]
Best C++ Formatter and Beautifier[^]
Online C/C++ Formatter, Indenter and Beautifier – Techie Delight[^]

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]
Enabling Open Innovation & Collaboration | The Eclipse Foundation[^]
-----
Your code do not behave the way you expect, or you don't understand why !

There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

jdb - The Java Debugger[^]
https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html[^]

The debugger is here to only show you what your code is doing and your task is to compare with what it should do.
   
v3
Comments
Abhishek Sharma Jun2021 29-Jun-21 7:27am
   
Thanks Patrice for adding code as well as important suggestions require to fulfill as mature developer like indentation.
I found the mistake it was unnecessary while loop. fixed it.
your advice is very helpful for beginner programmer. Thanks again👍
Patrice T 29-Jun-21 7:36am
   
Thank you.
Feel free to accept useful solutions, it will reward helpers and close the question as solved.
You can also add your own solution.
All you're doing by dividing "s" by 2, is that you're comparing the "first" half of the word with itself (instead of the second half).

It would be easier if you reversed all the letters, and compared the new word to the original: if they're equal, it's a palindrome.

StringBuffer reverse() Method in Java with Examples - GeeksforGeeks[^]
   
Comments
Abhishek Sharma Jun2021 29-Jun-21 7:23am
   
Thanks Gerry, I found the issue Yes that while loop had no use there, it got fixed with your help and yes reversing then comparing is much easier.
thanks for reply :)

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