Java - Scanner causes loop to run twice

Question

I'm fairly new to Java and programming in general. I'm trying to ask for names of celebrities and put them in an array. When I try to use scanner to read input from the user, it manages to store the name, but runs the line System.out.println("Name is too short, try again"); before the condition of the loop.

Here are the relevant parts from my code:

public static final Scanner scan = new Scanner(System.in);

public static void main(String[] args) {

    String celeb[];
    int number;
    System.out.println("Welcome to Celebrity Guesser!");
    System.out.println("How many celebrities do you want to name?");
    number = scan.nextInt();
    celeb = new String[number];

    for(int i = 0; i < number; i++) {   //asks for user input
        celeb[i] = celebInput(i + 1);
    }
    for(int i = 0; i < number; i++) {   //asks questions to user
        String tempCeleb = celeb[i];
        celebOutput(tempCeleb, i + 1);
    }
}

//Reads names of celebrities from the user
static String celebInput(int n) {
    System.out.println("\nType in celebrity #" + n);
    String c = scan.nextLine();
    while(c.length() < 6) {
        System.out.println("Name is too short, try again");
        c = scan.nextLine();
    }
    return c;
}

Here is the output:

Welcome to Celebrity Guesser!
How many celebrities do you want to name?
2

Type in celebrity #1
Name is too short, try again
Miley Cyrus

Type in celebrity #2
Katy Perry

Why does the issue occur only for "celebrity #1" and not "celebrity #2"? Thank you.

The code is incomplete - we are mising the `celebOutput` method — Andersnk, Feb 26 '15 at 00:22
the output works fine, all the information is saved to the array. The issue is that it says the name is too short before the user has a chance to enter a name. Thanks! — ipwnmice, Feb 26 '15 at 00:24
@AndersNK The `celebOutput()` method probably isn't relevant to the code's operation. — APerson, Feb 26 '15 at 00:24
@ipwnmice It looks like there's a newline hanging around in the buffer, and the Scanner is simply picking that up. Silly question: are you sure you're running the code without hitting enter too many times? — APerson, Feb 26 '15 at 00:25

Adrian Leonhard · Accepted Answer · 2015-02-26T00:30:35.950

0

scan.nextInt will remove only the "2" from the input stream. The "\n" (new line) you enter when you press enter is still in the input stream; the first nextLine call will therefore return an empty line.

To clarify:

    // assuming we enter "2\ntest\n" at the command line
    Scanner s = new Scanner(System.in); // s now contains "2\ntest\n"
    System.out.println(s.nextInt()); // s now contains "\ntest"
    System.out.println(s.nextLine()); // s now contains "test"
    System.out.println(s.nextLine()); // s is empty

Output:

"2"
""
"test"

The solution would be to call nextLine after nextInt to clear the empty line.

edited Feb 26 '15 at 00:30

answered Feb 26 '15 at 00:23

Adrian Leonhard

6,220
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i tried the code without \n and it still has the same problem. Thanks though. – ipwnmice Feb 26 '15 at 00:26
@ipwnmice I think you misunderstand. You have to enter a \n otherwise the scanner won't read. – Adrian Leonhard Feb 26 '15 at 00:27
@ipwnmice Try using number = Integer.parseInt(scan.nextLine()); – Prudhvi Feb 26 '15 at 00:28
I see what you're saying now. I added `scan.nextLine();` after `number = scan.nextInt();` and it works fine now. Thanks. – ipwnmice Feb 26 '15 at 00:33

score 0 · Answer 2 · answered Feb 26 '15 at 00:30

0

Try number = Integer.parseInt(scan.nextLine()); instead of number = scan.nextInt();

I hope it works. Let me know.

answered Feb 26 '15 at 00:30

Prudhvi

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Java - Scanner causes loop to run twice

2 Answers2