Java: console skipping input

Question

I'm trying to parse a char from console input using in.nextLine() and from there take charAt(0). My problem is after asking the user to enter a string to perform the in.nextLine() on, it skips the input and yields an error due to trying to get the first character of a null string.

System.out.print("Select an operator (+, -, *, /), 'c' or 'C' to clear, or 'q' to quit: ");
String temp = in.nextLine();
char tempOperator = temp.charAt(0);

the error is

java.lang.StringIndexOutOfBoundsException: String index out of range: 0
at java.lang.String.charAt(Unknown Source)

full program is available here

General comments and suggestions always welcome. Thanks in advance.

Yogendra Singh · Accepted Answer · 2012-10-31 16:25:34Z

4

When you do cValue = in.nextDouble();, it reads the next token(full value) and parses it to double. If the return key was pressed at this time, \n is the next token in the buffer to read.

When you do: String temp = in.nextLine();, it reads the \n from the buffer of the previous enter and charAt(0) fails as it has read only empty("") string.

To overcome the issue, either skip the previous buffer by adding an addition in.nextLine(), add \n \r as skip pattern as below (this is the pattern defined in Scanner.class as LINE_SEPARATOR_PATTERN):

 in.skip("\r\n|[\n\r\u2028\u2029\u0085]");

or put a small while loop as below:

   String temp = "";
   while((temp.length() < 0){
      temp = in.nextLine();
   }

edited Oct 31, 2012 at 16:25

answered Oct 31, 2012 at 16:10

Yogendra Singh

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3 Comments

zerodeficit Over a year ago

I wish I had 15 rep so I could up-vote you. Thanks for explaining rather than just providing the solution. Would you mind explaining why the while loop would be better than using a additional in.nextLine()?

Rohit Jain Over a year ago

Why would you need that while loop?

Yogendra Singh Over a year ago

@zerodeficit: I prefer while loop just for the reason that user may press return key more than once after entering the first input. This loop protects you from exceptions as before until the program hasn't received the genuine input.

Andrew Backes · Accepted Answer · 2012-10-31 16:10:06Z

1

One thing you might consider is instead of doing this:

String temp = in.nextLine();

Do this:

String temp = in.next();

That of course is assuming that you only want a single character from the user input, which it looks like that is all you are asking the user for. This will get rid of the exception you are receiving. However, you may want to read the whole line first to see what the length is, if you would like to throw an error when the user types more than one character.

answered Oct 31, 2012 at 16:10

Andrew Backes

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Comments

Rohit Jain · Accepted Answer · 2012-10-31 16:17:16Z

1

Well, the problem is in the line in.nextDouble().

Scanner#nextDouble reads the next token from the user as double. So, when you pass a value 4.5 as input, then the nextDouble just reads the token 4.5 and skips the linefeed after that input that user have entered. Now this linefeed(Since this is a new line), will be regarded as input for the next Scanner.nextLine. And hence your in.nextLine after in.nextDouble is reading the newline left over by the previous in.nextDouble.

So, the workaround is the same that @tjg184 has pointed out in his answer. Add an empty in.nextLine after in.nextDouble that will read that newline left over. So, that your coming in.nextLine reads the input actually passed.

cValue = in.nextDouble();
in.nextLine();  // Add this before your `while` which will read your `newline`

while (true) {
    //continue;
}

Whereas, Scanner.nextLine reads the complete input till the newline. So, it will not leave any newline to be read by the next read by user.

Although this answer was not needed, as the @tjg184's answer does the same. I just added it to give a better explanation of what exactly is happening.

edited Oct 31, 2012 at 16:17

answered Oct 31, 2012 at 16:10

Rohit Jain

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1 Comment

Rohit Jain Over a year ago

@tjg184. Thanks :). Your answer is also correct. But I added this answer just to clear it up little bit more.

tjg184 · Accepted Answer · 2012-10-31 15:58:33Z

0

Looking at your program, it's because of String temp = in.nextLine();. When the user hits the "Enter" key, it essentially is skipping this statement since the user typed hit "Enter". Try the following. Note, I only added the following line String enter = in.nextLine();

import java.util.Scanner;

public class Calculator
{    
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        char operator = 'z'; //initialize operator
        double cValue; //current value
        double rhValue; //right hand value
        boolean cont = true;

        System.out.print("Enter starting value: ");
        cValue = in.nextDouble();

        // disregard the "Enter" key
        String enter = in.nextLine();

        while(true)
        {
            System.out.print("Select an operator (+, -, *, /), 'c' or 'C' to clear, or 'q' to quit: ");
            String temp = in.nextLine();
            char tempOperator = temp.charAt(0);

            if (tempOperator == 'c' || tempOperator == 'C')
            {
                cValue = 0.0;
                System.out.println("Current value is: " + cValue);
                System.out.println();
            }
            else if(tempOperator == 'q')
            {
                System.out.println("Final result: " + cValue);
                System.exit(1);
            }
            else if(tempOperator == '+' || tempOperator == '-' || tempOperator == '*' || tempOperator == '/')
            {
                operator = tempOperator;
                break;
            }
            else
                throw new IllegalArgumentException("operator not valid");
        }


        System.out.println("Enter a right hand value (type double): ");
        rhValue = in.nextDouble();

        System.out.println("Math expression: answer " + operator + "= " + rhValue);
        switch(operator)
        {
            case '+': cValue =+ rhValue;
                      break;
            case '-': cValue =- rhValue;
                      break;
            case '*': cValue = cValue * rhValue;
                      break;
            case '/': cValue = cValue / rhValue;
                      break;
        }

        System.out.println("Current value is: " + cValue);
    }
}

answered Oct 31, 2012 at 15:58

tjg184

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4 Comments

Rohit Jain Over a year ago

But you didn't gave the reason, why the line is skipped when user hits Enter. That's the whole point. User will anyhow hit enter right?

tjg184 Over a year ago

@RohitJain if you look at the JavaDoc for Scanner, it states that nextLine() "Advances this scanner past the current line and returns the input that was skipped." The Enter key being pressed is forcing a new line, right after they just entered the first double value. Hope this helps.

tjg184 Over a year ago

@RohitJain Let me know what part is unclear. I updated the code to accommodate the user hitting the Enter key by disregarding the input.

zerodeficit Over a year ago

so in.nextDouble() only takes the double and the enter is then consumed by the next console.in BUT this doesn't apply to in.nextLine as the enter is the delimiter for the required input, right?

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