Scanner class fails to take string input

Question

It is a simple java code.. but Scanner class isn't taking the string as input. why?

public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int x=sc.nextInt();
        double y=sc.nextDouble();
        String s =sc.nextLine();

        System.out.println("String: "+s);
        System.out.println("Double: "+y);
        System.out.println("Int: "+x);
}

Use `String s =sc.next();` or add `sc.nextLine();` after reading `double`. — SatyaTNV, Feb 07 '16 at 11:15
Possible duplicate of [Skipping nextLine() after using next(), nextInt() or other nextFoo() methods](http://stackoverflow.com/questions/13102045/skipping-nextline-after-using-next-nextint-or-other-nextfoo-methods) — SatyaTNV, Feb 07 '16 at 11:21

score 2 · Accepted Answer · answered Feb 07 '16 at 11:17

Because the sc.nextInt() and sc.nextDouble() method does not consume the newline character of your input, so that newline is consumed in the next call to sc.nextLine()

public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int x=sc.nextInt();
        sc.nextLine(); 
        double y=sc.nextDouble();
        sc.nextLine();
        String s =sc.nextLine();

        System.out.println("String: "+s);
        System.out.println("Double: "+y);
        System.out.println("Int: "+x);
}

score 1 · Answer 2 · edited May 23 '17 at 12:15

1

Use nextLine() method to read all values and then parse them into the corresponding type (Integer, Double, etc). See why here: Integer.parseInt(scanner.nextLine()) vs scanner.nextInt()

edited May 23 '17 at 12:15

Community

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answered Feb 07 '16 at 11:15

Baderous

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But why does it fail to take input as string? – Darpanjbora Feb 07 '16 at 11:17
Because of the newline that remains in the input buffer. – Baderous Feb 07 '16 at 11:18

Scanner class fails to take string input

2 Answers2