why loop iteration range given by keyboard count itself as an iteration?

Question

public class P7 {
    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int range=scanner.nextInt();
        for(int i=1;i<=range;i++){
            String stringInput =scanner.nextLine();
            String[] total =stringInput.split("\\s");
            int length=total.length;
            System.out.println(length);
        } 
    }
}

If you input range as 3 it will print its length as 1. But first iteration should print the length after getting stringInput. Moreover when you input range as a String, you will get InputMismatchException. Which is correct. So where is the problem? Thanks.

Possible duplicate of [Scanner is skipping nextLine() after using next() or nextFoo()?](https://stackoverflow.com/questions/13102045/scanner-is-skipping-nextline-after-using-next-or-nextfoo) — Jacob G., Jul 19 '18 at 13:22

Shar1er80 · Accepted Answer · 2018-07-19T14:27:07.263

Scanner.nextInt() only consumes up to the number that you input and leaves the "\n" in the Scanner buffer. You can clear that buffer by using Scanner.next() right after you call Scanner.nextInt().

import java.util.Scanner;

public class StackOverflow {
    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int range=scanner.nextInt();
        scanner.next();

        for(int i=1;i<=range;i++){
            String stringInput =scanner.nextLine();
            String[] total =stringInput.split("\\s");
            int length=total.length;
            System.out.println(length);
        }

        scanner.close();
    }
}

Result:

3
The asdfj
2
asldkja sfaslkj asdfljk
3
asdfjk asdfjlkasdf alksjdf asdflkj
4

why loop iteration range given by keyboard count itself as an iteration?

1 Answers1