How does the method Scanner.nextInt() works compare to Scanner.nextLine()

Question

I've just started Java recently. There's an exercise that ask me to split and display the number separated by spacebar. The standard input is basically like this:

2
2 2

The first line is the number of integers in the array. The second line is the array This is the first block of code that I use

   import java.util.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
    public static void main (String[] args) 
    {
        Scanner reader= new Scanner(System.in);
        int t=reader.nextInt();
        String []s=reader.nextLine().split(" ");
        for(int i=0;i<=t-1;i++)
        {
            System.out.println("The "+(i+1) +" number is "+s[i]);
        }
    }
}

When compile and run it gives me this error

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
    at Main.main(Main.java:15)

However when I change the the reader.nextInt() to reader.nextLine() and parse it into integer it works perfectly fine

import java.util.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
    public static void main (String[] args) 
    {
        Scanner reader= new Scanner(System.in);
        int t=Integer.parseInt(reader.nextLine());
        String []s=reader.nextLine().split(" ");
        for(int i=0;i<=t-1;i++)
        {
            System.out.println("The "+(i+1) +" number is "+s[i]);
        }
    }
}

This is what the output looks like

The 1 number is 2
The 2 number is 2

So why doesn't it work with reader.nextInt() ?

Edit about the reading the Line character, I still don't get it. It reads the string normally

score 0 · Answer 1 · edited May 23 '17 at 12:19

0

int t=reader.nextInt();

Reads a 2.

reader.nextLine()

Reads to the next newline character.

Calling it again will read 2 2, as expected.

Check the content of s. It's likely empty, or doesn't have 2 elements, thus the error

When you read the line twice, you consume the newline

For more info, see Scanner is skipping nextLine() after using next(), nextInt() or other nextFoo() methods

edited May 23 '17 at 12:19

Community

1
1

answered Oct 08 '16 at 22:56

OneCricketeer

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In the first scenario, If I simply change the s into String s=reader.nextLine(), it still shows 2 2 however when split, it is unable to split properly – kjfunh Oct 10 '16 at 14:36
You are welcome to post another question with your problem. `"2 2".split(" ")` works for me, though – OneCricketeer Oct 10 '16 at 14:38

How does the method Scanner.nextInt() works compare to Scanner.nextLine()

1 Answers1