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Thread: Very basic code problem

  1. #1
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    Default Very basic code problem

    Hi guys, I've just started trying to teach myself Java less than two weeks ago. I've been reading the "Beginning Programming Java for Dummies" and I've writing small things to practice as I go that review what I've learned so far. I tried to make a small program that checked your age to see if you could buy cigarettes, asked what kind you'd like, ask how many, and then give you a total price. It compiles fine but after the cigarette selection there's an error and it terminates itself and I don't know why.
    This is the error that shows up

    Exception in thread "main" java.lang.NullPointerException
    at CigarettesOrNo.main(CigarettesOrNo.java:56)


    and this is the code

    import java.util.Scanner;
    import static java.lang.System.out;
     
     
    public class CigarettesOrNo {
     
    	public static void main(String args[]) {
    		Scanner myScanner = new Scanner(System.in);
    		int month, day, year;
    		char reply,sure;
    		double marlboro = 5.95;
    		double newport = 6.49;
    		double generic = 4.39;
    		int whichCigarette;
    		int howMany;
     
    		out.println("What is your birthday? Please type as follows MM DD YYYY");
    		month = myScanner.nextInt();
    		day = myScanner.nextInt();
    		year = myScanner.nextInt();
     
    		if (year < 1993) {
    			reply = 'y';
    		}
     
    		else if (year == 1993 && month < 4) {
    			reply = 'y';
    		}
     
    		else if (year == 1993 && month == 4 && day < 27) {
    			reply = 'y';
     
    		}	else {
    			reply = 'n';
    		}
     
    		if (reply == 'y') {
    			out.println("Which cigarettes would you like? 1. Marlboro, 2. Newport, or 3. Generics? Please select a number.");
    			whichCigarette = myScanner.nextInt();
     
    			if (whichCigarette == 1) {
    				out.println("Marlboros are $5.99 per pack, is that alright? (Y/N)");
    				sure = myScanner.findInLine(".").charAt(0);
     
    				if (sure == 'y') {
    					out.println("How many packs would you like?");
    					howMany = myScanner.nextInt();
    					out.print("Your total will be $" + marlboro * howMany + ". Thank you for your purchase!");
    				} 	else {
    					out.println("Thanks for looking.");
    				}	
    		    }
     
    			if (whichCigarette == 2) {
    				out.println("Newports are $6.49 per pack, is that alright? (Y/N)");
    				sure = myScanner.findInLine(".").charAt(0);
     
    				if (sure == 'y') {
    					out.println("How many packs would you like?");
    					howMany = myScanner.nextInt();
    					out.print("Your total will be $" + newport * howMany + ". Thank you for your purchase!");
    				} 	else {
    					out.println("Thanks for looking.");
    				}	
    		    }
     
    			if (whichCigarette == 3) {
    				out.println("Generics are $4.39 per pack, is that alright? (Y/N)");
    				sure = myScanner.findInLine(".").charAt(0);
     
    				if (sure == 'y') {
    					out.println("How many packs would you like?");
    					howMany = myScanner.nextInt();
    					out.print("Your total will be $" + generic * howMany + ". Thank you for your purchase!");
    				} 	else {
    					out.println("Thanks for looking.");
    				}	
    		    }
     
    		} 	else {
    			out.println("I'm sorry, you are too young to buy cigarettes!");
    		}
     
    	}
     
     
    }

    The error points to this line of code "sure = myScanner.findInLine(".").charAt(0);" but I don't know what's wrong with it. Maybe I'm too much of a beginner to see it haha. Anyway, thanks to anyone that can help me


  2. #2
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    Default Re: Very basic code problem

    I think it's System.out.println.

    Also, I think I found something odd.

    sure = myScanner.findInLine(".").charAt(0);

    What is the findInLine thing for?

    I'm going to check it quickly to be sure.....

    Yep, that's what's causing the problems.

    As you only scan in integers or yes or no, I don't see where you'd be entering a "." anyway.

    I must admit I don't really understand the findInLine method.
    Last edited by javapenguin; April 28th, 2011 at 04:26 AM. Reason: Found something else

  3. #3
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    Default Re: Very basic code problem

    The thing is I only know so much right now and I'm trying to use what I know to get it done. The findInLine was a variable for the question above it which states the cigarette's price and asks (Y/N) if the price is okay with the customer.

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    Default Re: Very basic code problem

    The problem is this line:

    sure = myScanner.findInLine(".").charAt(0);

    The findInLine call is returning null, so you get an exception when you try to call charAt on the result.

    You've just read in the cigarette number, but you haven't skipped to a new line for the next entry. The nextXxx methods do that, but findInLine will only look at the current line. If you insert myScanner.nextLine(); immediately before your findInLine calls, it should work as expected.

  5. The Following User Says Thank You to dlorde For This Useful Post:

    JavaPF (April 28th, 2011)

  6. #5
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    Default Re: Very basic code problem

    I see, I can understand it now. Thanks a lot dlorde!

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