findWithinHorizon(".", 0).charAt(0); is really bugging me.

[Joseph Alrawi]

Hello readers!

After hours of working on this code, I really can not understand what "findWithinHorizon..." does. Well I understand that it looks for a character based on what you typed in "charAt(number);". So far so good. But the problem is, it acts really, really strange. I have a code, not too long that I've been working with for a while now, it is very simple, but still complicated for me since I'm a beginner.

package Default;

import java.util.Scanner;
import java.util.Random;

public class Horizon {
	public static void main(String args[]) {
		char letter;
		String word = null;
		int length = 0;
		int number = 0;
		String answer;
		
		Scanner Scan = new Scanner(System.in);
		Random Rand = new Random();
		
		
		System.out.println
		("Hello, please type any word.");
		
		word = Scan.nextLine();
		
		length = word.length();
		
		number = Rand.nextInt(length);
		
		letter = Scan.findWithinHorizon(".", 0).charAt(number);
		
		System.out.println(letter); //This is just to make sure I'm working with the correct letter.
		
		String letter2 = Character.toString(letter);
		
		number = (number += 1);

		System.out.println
		("You have answered:");
		
		System.out.println(word);
		
		System.out.println();
		
		System.out.println
		("What letter is at the number " + number + " of the word you typed?");
		
		Scan.nextLine();
		answer = Scan.nextLine();
		
		if (answer.equals(letter2)) {
			System.out.println("You are correct.");
			
		
			} else {
				System.out.println("Wrong answer.");
			
		}
		
		
		
	}

}

First of all, the program needs to scan 2 lines in order for it to move on, which has probably something to do with 2 scanner being called (letter and word one). The point of this program is simple, it scans for a word typed by the user. The program grabs a random letter in that word and then requires the user to type the correct letter in the command box. So the issue is, as mentioned just now, it needs to scan twice, so I have to type two lines. Second of all, it only works with words being 2 or less letters long, if I try something with 3 letters it returns:

Hello, please type any word.
fff
fff
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 1
at java.lang.String.charAt(Unknown Source)
at Default.Horizon.main(Horizon.java:27)
Picked up _JAVA_OPTIONS: -Xmx512M (what is this anyways?)

I hope this wasn't too much bother-some for you to read, any help greatly appreciated, thanks!

[Paweł Baczyński]

Welcome to the Ranch!

Please, UseCodeTags (← click this) when posting. I added them for you this time. Doesn't it look better?

Have you looked at Scanner#findWithinHorizon javadoc? (link here)
It describes what the method does.

It has nothing to do with charAt method. This method is called on a result of findWithinHorizon which happen to be a String.

Print out whatever is returned from findWithinHorizon and you will see what it is.

From the exception message I can see that you want to extract a character at index 1 from a string that is of length 0 or 1 which causes the exception.

The method findWithinHorizon workd with Pattern class (as javadoc says). Pattern class works with regular expressions (or regexes).
A dot character "." in a regex has a special meaning. Are you aware of that?

[Joseph Alrawi]

Paweł Baczyński wrote:Welcome to the Ranch!

Please, UseCodeTags (← click this) when posting. I added them for you this time. Doesn't it look better?

Have you looked at Scanner#findWithinHorizon javadoc? (link here)
It describes what the method does.

It has nothing to do with charAt method. This method is called on a result of findWithinHorizon which happen to be a String.

Print out whatever is returned from findWithinHorizon and you will see what it is.

From the exception message I can see that you want to extract a character at index 1 from a string that is of length 0 or 1 which causes the exception.

The method findWithinHorizon workd with Pattern class (as javadoc says). Pattern class works with regular expressions (or regexes).
A dot character "." in a regex has a special meaning. Are you aware of that?

Hey! Thanks for your help! I tried using findInLine instead since it really made it a whole lot simpler. But what can I do so the program only needs to scan one line and not 2 in order to fetch the word and the character? Meaning this:

Hello, please type any word.
Hello //this bugs me.
Hello
l
You have answered:
Hello

What letter is at the number 4 of the word you typed?
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Thank you for your time and the warm welcome!