getting an input by scanner and creating an array only using for loop not

Question

private static int[] arr;
public static void inputarrays() {
    Scanner scan = new Scanner(System.in);

    System.out.println("length of an array");

    int x = scan.nextInt();
    arr = new int[x];

    System.out.println("values of an array");
    for(int n = 0; n < x; x++) {
        arr[n] = scan.nextInt();
    }
}

I looked up how to write code for getting an user input and creating an array like this one: Java - Creating an array from user input

It looks like my code is same as on the link's one which means it should work correctly. However, the scanner never closes when I enter the values of an array. I tried scan.close(), but it did not work. Also, I cannot use try&exception for this one.

Add this line: scan.nextLine(); after each scan.nextInt(); — forpas
– forpas, Commented Dec 16, 2018 at 18:30
and thank you for editing, I would follow the format next time — 판다하나만
– 판다하나만, Commented Dec 16, 2018 at 18:30
it sill does not work should I make a new array inside of the method instead of declaring it in the data field? — 판다하나만
– 판다하나만, Commented Dec 16, 2018 at 18:34

forpas · Accepted Answer · 2018-12-16 19:16:09Z

1

This method is working, try it:

private static int[] arr;
public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    System.out.println("length of an array");
    int x = scan.nextInt();
    scan.nextLine();
    arr = new int[x];
    System.out.println("values of an array");
    for(int n=0; n<x; n++){
        arr[n]=scan.nextInt();
        scan.nextLine();
    }
}

changed x++ to n++, because n is the index of the loop
added scan.nextLine(); after each call to scan.nextInt() so that each input line is fully consumed.

edited Dec 16, 2018 at 19:16

answered Dec 16, 2018 at 18:37

forpas

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6 Comments

판다하나만 Over a year ago

I had to declare arr in the data field so that is why I put it outside of the method. If I keep in that way, does it gonna keep that error?

Ivar Over a year ago

The nextLine()s are redundant here.

Ivar Over a year ago

The arr is declared in the question on the first line.

forpas Over a year ago

@Ivar I thought so, but 1 out of 2 times I run the code without the nextLine(); in InteliJ, the loop does not stop.

forpas Over a year ago

@Ivar there is a problem for sure when you mix inputs with nextLine() and nextInt(). In the case where only nextInt() is used I also never had no problems. But it could happen and just to be on the safe side...

|

PUNdorasbox · Accepted Answer · 2018-12-16 19:30:59Z

-2

You used n for your loop so you have to go for n++ and not x++.

private static int[] arr;
public static void inputarrays() {
    Scanner scan = new Scanner(System.in);

    System.out.println("length of an array");

    int x = scan.nextInt();
    arr = new int[x];
    System.out.println("values of an array");
    for(int n = 0; n < x; n++) {
        arr[n] = scan.nextInt();
    }
}

It should work this way.

edited Dec 16, 2018 at 19:30

answered Dec 16, 2018 at 19:09

PUNdorasbox

171 bronze badge

1 Comment

tung Over a year ago

this answer has nothing that hasn't already been said in other answers.

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