Hadean Fall's Profile User Rating: -----

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User is offline Mar 19 2014 03:39 PM
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Icon   Hadean Fall do ur wurst

Posts I've Made

  1. In Topic: Converting letters to numbers easily

    Posted 27 Feb 2014

    so im getting a loss of precision error with my code
    import java.util.*;
    public class caesarcipher {
    
    
            public static void main (String args[]) {
                    Scanner in = new Scanner(System.in);
                String message;
                    char key = 3;
                    char num = 26;
    
                    System.out.println("please input your message to be encoded");
                    message = in.next();
                    System.out.println("please input the key");
    
                    message = message.toLowerCase();
                    char[] encrypt = message.toCharArray();
                    char[] ans = new char[message.length()];
    
                    for (int count = 0; count <= message.length(); count++)
                    {
                    ans[count] = (encrypt[count] + key) % (char) num;
                            System.out.println(count);
                    }
    
            }
    }
    
    

    i don't understand how, it says its in this line
    ans[count] = (encrypt[count] + key) % (char) num;
    
    
  2. In Topic: Converting letters to numbers easily

    Posted 27 Feb 2014

    so could i use .charAt to read a string into a char array then convert that to numbers?
  3. In Topic: Converting letters to numbers easily

    Posted 27 Feb 2014

    ok thank you, but if i wanted to convert an entire string to numbers such as "helloworld" how would i go about that? Would i have to take substrings for each individual letter, convert it to a char put it in an array then convert the array to numbers? or is there a shortcut i can take
  4. In Topic: if then equalsIgnoreCase()

    Posted 29 Nov 2013

    It worked, thanks so much guys for answering this silly question, it's really appreciated
  5. In Topic: if then equalsIgnoreCase()

    Posted 29 Nov 2013

    im still getting the error "variable in of type Scanner"
    here's my code
    public class rubix {
    	
    	public static void main (String args[]) {
    		Scanner in = new Scanner (System.in);
    		
    		
    		System.out.println("Welcome to my algorithm training program!");
    		System.out.println("Would you like to start? (y/n)");
    		String choice = in.next();
    		if (in.equalsIgnoreCase("y"))
    		{
    		System.out.println("Cool! lets get started!");
    	}
    		else
    	{
    	}
    	}
    }
    
    
    
    

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  1. Photo

    brep Icon

    18 Sep 2011 - 14:53
    Thanks you so much for the positive rep. I greatly appreciate it :)
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