Working with strings

[gman1199]

I am making a decoder for a cipher called the Caesar Shift Cipher. I am making a first build of it and it will not be efficent at all so please ignore that. Here is my code so far:

import java.util.Scanner;
public class decoder {

	public static void main(String[] args) {
		Scanner number = new Scanner(System.in);
		System.out.println("Please enter your shift number");
		int shift = number.nextInt();
		int counter = 1;
		
		if (shift == 1)
			while (counter  > 0) {
				System.out.println("Enter a letter");
				String letter = number.next();
				if (letter == "a")
					System.out.println("z");
				else if (letter == "b")
					System.out.println("a");

I thought this would work but whenever I run it it does not print out the new letter. I have no previous work with stings and I would really appriciate any information. Thanks.

[DarrylBurke]

Forum Rules
Guide For New Members
BB Code List - Java Programming Forum

db

[awinston]

I am making a decoder for a cipher called the Caesar Shift Cipher. I am making a first build of it and it will not be efficent at all so please ignore that. Here is my code so far:

       if (shift == 1)
			while (counter  > 0) {
				System.out.println("Enter a letter");
				String letter = number.next();
				if (letter == "a")
					System.out.println("z");
				else if (letter == "b")
					System.out.println("a");

I thought this would work but whenever I run it it does not print out the new letter. I have no previous work with stings and I would really appriciate any information.

1. You don't have a closing bracket for your class.

2. You don't have a closing bracket for your while-loop.

3. A letter will only print out if the user enters a or b.

4. A letter will only print out if the user enters 1 for the shift.

5. Inside your while-loop counter is never decremented, so the while-loop (if it had a closing bracket) would run forever.

6. You compare strings using the == operator when you should use the equals() method instead.

7. No precautions against invalid user input are taken.

8. You don't have a closing bracket for your main method.

Also, I don't understand the connection between the shift number and the actual cipher.


Since you haven't worked with Strings before, I recommend that you read this: Strings (The Java™ Tutorials > Learning the Java Language > Numbers and Strings)

[gman1199]

awinston -

Thank you. The reason for a lot of the errors you pointed out is the program is not finished and I only posted a small section because it is, as I say in the post, very lengthy. The shift number is basically the amount of letters that are moved from the back of the alphabet to the front. For example, a cipher where the shift number is 4 would look like this: a=w, b=x, c=y, d=z, e=a, ... z=v. The cipher was used by Caesar to communicate with his generals. It's very easily solved because there are only 25 possible combinations.

[awinston]

That's interesting!

Why are you choosing to go about this in an inefficient manner (as you pointed out)?

[gman1199]

Sorry for not getting back sooner, out of town for a few weeks. The answer to your question is I have no idea how else to do it. Also, I was working on my code today, trying to make it analise a whole string, and I came up with this (using that web page you gave me) :

import java.util.Scanner;
public class decoder {

	public static void main(String[] args) {
		Scanner number = new Scanner(System.in);
		System.out.println("Please enter your shift number");
		int shift = number.nextInt();
		int counter = 0;
		int i = 0;
		System.out.println("Enter a string");
		String cipher = number.next();
		int len = cipher.length();
		if (shift == 1)
			while (counter < len) {
				Character letter = cipher.charAt(i);
				if (letter.equals("a"))
					System.out.println("z");
				else if (letter.equals("b"))
					System.out.println("a");
				
				counter++;
				i++;	
			}
	}
}

I thought this would work, but when I ran it, nothing was printed out. I originally got this error -

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 3
at java.lang.String.charAt(Unknown Source)
at decoder.main(decoder.java:15)

- but this was fixed when I changed to while loop to while (counter < len). Any thoughts you have on this would be appriciated, I have no idea what is wrong.

Thanks,
gman1199

[awinston]

Can you post the code that was giving you the error?

Also, it looks like you can use either counter or i without using both of them. They are always the same value, so you only need one.

[gman1199]

import java.util.Scanner;
public class decoder {
 
    public static void main(String[] args) {
        Scanner number = new Scanner(System.in);
        System.out.println("Please enter your shift number");
        int shift = number.nextInt();
        int counter = 0;
        int i = 0;
        System.out.println("Enter a string");
        String cipher = number.next();
        int len = cipher.length();
        if (shift == 1)
            while (counter < len) {
                Character letter = cipher.charAt(i);
                if (letter.equals("a"))
                    System.out.println("z");
                else if (letter.equals("b"))
                    System.out.println("a");
                 
                counter++;
                i++;    
            }
    }
}

This is what the error was coming from

[awinston]

I can compile and run that code with no errors. Didn't you say that you were receiving the error before you changed the condition of the while-loop to (counter < len)? What was the condition before?

[awinston]

Also, you should be aware that using the equals() method to compare a Character with a String will return false.

[gman1199]

Sorry, I forgot to change it from while (counter < len) to while (counter <= len). It was (counter <= len) that was causing the error. Also, is there a way to fix that equals() thing?

[quad64bit]

is there a way to fix that equals() thing?

Yeah, just don't use a character.

In java, a literal 'a' is the character a where as "a" is a string. As long as you are not using a literal character, you're fine. The alternative is to use only characters in which case the == will work. You're current solution seems fine though!

[gman1199]

So if I replace "a" with a (or is it 'a'?) it should work?

[gman1199]

Thank you so much! I replaced everything with the '' and it works!

[awinston]

Sorry, I forgot to change it from while (counter < len) to while (counter <= len). It was (counter <= len) that was causing the error.

Yes, (counter <= len) will allow counter to equal length. Since the last character of any String has an index of length-1, this will cause the StringIndexOutOfBoundsException to be thrown.

[Junky]

it works!

Congrats, but here's a tip on making your code more efficient.

System.out.println((char) 'b' - 1);

The only issue is how to wrap 'a' back around to 'z'.

[quad64bit]

So if I replace "a" with a (or is it 'a'?) it should work?

It's the ' vs the " that determines wether a literal is a char or a String in java.
So:

char someCharacter = 'a';
String someString = "a";