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Thread: Exception in thread "main"?

  1. #1
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    Default Exception in thread "main"?

    Hello, I recently decided to use the skills I acquired so far to build a basic calculator (very basic). But for some reason I get this exception when I run the program.

    Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at java.util.Scanner.nextDouble(Unknown Source)
    at CalculatorBasic.main(CalculatorBasic.java:9)

    This is the code:

    import java.util.Scanner;
     
     
    class CalculatorBasic {
    	public static void main(String args[]) {
    		Scanner myscanner = new Scanner(System.in);
    		double fnum, second, answer;
    		System.out.println("Enter the first number: ");
    		fnum = myscanner.nextDouble();
    		System.out.println("Enter the second number: ");
    		second = myscanner.nextDouble();
    		answer = fnum + second;
    		System.out.println(answer);
     
    	}
    }

    I would like to know what's wrong with this code, thank you on advance.


  2. #2
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    Default Re: Exception in thread "main"?

    When the user enters a number, they press enter. So you have a double and the enter character "in" the Scanner.

    Then you call the nextDouble() function, which scans past the double. But you still have the enter character in the Scanner.

    You then call nextDouble() again, which is looking for another double, but what it sees is that enter character. The enter character is not a double, so it throws that exception.

    Use the nextLine() function to scan past the enter character.
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  3. #3
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    Default Re: Exception in thread "main"?

    Can you show me where should I place the .nextLine() please?

  4. #4
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    Default Re: Exception in thread "main"?

    Quote Originally Posted by Empower View Post
    Can you show me where should I place the .nextLine() please?
    Where have you tried placing it?
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    Default Re: Exception in thread "main"?

    I tried placing it under fnum = myscanner.nextDouble().

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    Default Re: Exception in thread "main"?

    Quote Originally Posted by Empower View Post
    I tried placing it under fnum = myscanner.nextDouble().
    And what happened when you did that? Can you post your updated code, along with the full text of your input and error?
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  7. #7
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    Default Re: Exception in thread "main"?

    First, I typed fnum = myscanner.nextLine() under the first one (fnum = myscanner.nextDouble()) but it really sounded stupid because I already defined fnum as a double. So I did something crazy and placed "myscanner.nextLine()" alone under the first one.

    import java.util.Scanner;
     
     
    class CalculatorBasic {
    	public static void main(String args[]) {
    		Scanner myscanner = new Scanner(System.in);
    		double fnum, second, answer;
    		System.out.println("Enter the first number: ");
    		fnum = myscanner.nextDouble();
                    myscanner.nextLine();
    		System.out.println("Enter the second number: ");
    		second = myscanner.nextDouble();
    		answer = fnum + second;
    		System.out.println(answer);
     
    	}
    }

    The same error:

    Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at java.util.Scanner.nextDouble(Unknown Source)
    at CalculatorBasic.main(CalculatorBasic.java:10)

  8. #8
    Crazy Cat Lady KevinWorkman's Avatar
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    Default Re: Exception in thread "main"?

    The code you just posted works fine for me. Can you post the exact input you're using?
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    Default Re: Exception in thread "main"?

    What do you mean by input? If you mean code, that's exactly the one I'm using. Sorry if I misunderstood.

    --- Update ---

    I apologize - it was a failure at my end. Thank you for the nextLine() suggestion!

  10. #10
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    Default Re: Exception in thread "main"?

    I mean, what are you typing into the command line to give input to the Scanner? When the Scanner asks for a number, what are you giving it?

    Are you sure this is the code you're running?

    Edit- out of curiosity, what was the failure on your end?
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  11. #11
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    Default Re: Exception in thread "main"?

    I realized that it works even without the nextLine(). So I believe it's all good now.

    My failure: Used a number that is not a double.

  12. #12
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    Default Re: Exception in thread "main"?

    Ah I see that now. I was thinking of the following situation:

    import java.util.Scanner;
     
    public class Main{
     
    	public static void main(String args[]) {
    		Scanner myscanner = new Scanner(System.in);
     
    		System.out.println("Enter a number and press enter:");
     
    		double one = myscanner.nextDouble();
     
    		System.out.println("Enter some text and press enter:");
     
    		String text = myscanner.nextLine();
     
    		System.out.println("The number you entered: " + one);
    		System.out.println("The text you entered: " + text);
     
    	}
    }

    The above program won't work correctly because the nextDouble() function doesn't scan past the newline character *after* the number.

    However, your program works because nextDouble() does actually scan past the newline character *before* the number.

    Anyway, glad you got it figured out.
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  13. The Following User Says Thank You to KevinWorkman For This Useful Post:

    Empower (January 8th, 2014)

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