System.out.println repeating itself

Question

Would anyone point me in the right direction, of why when i use a for loop the println function comes up two times in the output. Thanks

public static void main(String[] args) {

    Scanner scan = new Scanner(System.in);
    System.out.println("Enter the number of employees to calculate:");
    int numberEmployees = scan.nextInt();

    for(int i=0; i<numberEmployees; i++){               
            System.out.println("Enter First Name:");
            name = scan.nextLine();
            System.out.println("Enter Last Name:");
            last = scan.nextLine();
            System.out.println("Enter Document #:");
            document = scan.nextInt();
            System.out.println("Enter Basic Salary");
            basicSalary = scan.nextInt();
            System.out.println("Enter # of Hours");
            hours = scan.nextInt();
    }
}

OUTPUT

Enter the number of employees to calculate:
1
Enter First Name:
Enter Last Name:
daniel
Enter Document #:

This has been asked so many times. See: http://stackoverflow.com/questions/7056749/scanner-issue-when-using-nextline-after-nextint — Eng.Fouad, Jul 31 '12 at 22:46

Dennis Meng · Accepted Answer · 2013-09-20T22:52:44.913

3

The problem is that when you entered 1 with a new line, the nextInt() function doesn't remove the newline that you had from entering in the 1. Change your calls to scan.nextInt() to Integer.parseInt(scan.nextLine()) and it should behave the way you want.

To further explain; here's stuff from the Java API.

A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. The resulting tokens may then be converted into values of different types using the various next methods.

and

The next() and hasNext() methods and their primitive-type companion methods (such as nextInt() and hasNextInt()) first skip any input that matches the delimiter pattern, and then attempt to return the next token. Both hasNext and next methods may block waiting for further input.

So, what evidently happens (I didn't see anything on the page to confirm it) is that after next(), hasNext(), and their related methods read in a token, they immediately return it without gobbling up delimiters (in our case, whitespace) after it. Thus, after it read in your 1, the newline was still there, and so the following call to nextLine() had a newline to gobble and did so.

edited Sep 20 '13 at 22:52

answered Jul 31 '12 at 22:42

Dennis Meng

5,109
14
33
36

Alternatively, you can add extra calls to `scan.nextLine()` or find more elegant ways to ignore the newline character. – Code-Apprentice Jul 31 '12 at 22:45
Personally I try not to interleave `nextLine()` with stuff like `nextInt()` because then I have to worry about where the Scanner will read its next token. Since we're expecting only one thing per line anyway, it's much easier to just go ahead and read/parse one full line at a time. – Dennis Meng Jul 31 '12 at 22:48
Good point. I just wanted to give some alternatives. I haven't looked at the API, but you would think the developers/designers would have added a convenient way to flush out extraneous characters, especially whitespace, for a case like this. – Code-Apprentice Jul 31 '12 at 22:50
I have the API pulled up; I'll update the answer with stuff from it to further explain what's going on. – Dennis Meng Jul 31 '12 at 22:52

score 0 · Answer 2 · answered Jul 31 '12 at 22:42

It appears that the newline character remains in your input after the first entry. When the next input is requested, the Scanner sees a newline character and interprets it as the end of the input. This makes it appear to skip every other input request. I would suggest checking out the Java API docs as to the exact behavior of Scanner's methods.

System.out.println repeating itself

2 Answers2