Java-ArrayIndexOutOfBoundsException Error

Question

When i run the following program I get an error at line 20, and this is my code:

package J1;
import java.util.Scanner;

public class SpeedLimit {

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    int input = keyboard.nextInt();

    String[] tab = new String[2];
    String output="";
    int speed = 0;
    while(input!=-1){
        int last =0;
        for (int i=0; i<input ; i++){

            String pair = keyboard.next();

            tab = pair.split(" ");
            speed = speed + Integer.parseInt(tab[0])*(Integer.parseInt(tab[1])-last);
            last = Integer.parseInt(tab[1]);
        }
        output = output +speed + "miles" + "\n";
        speed =0;
     input = Integer.parseInt(keyboard.nextLine());
    }
   System.out.println(output);
}

}

when i run the code, I enter the following input from the keyboard:

to get this result as an output: 170 miles 180 miles 90 miles

but i get the following Error when i run the code

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
at J1.SpeedLimit.main(SpeedLimit.java:20)

In the line String pair = keyboard.next(); the .next() returns "the next complete token from this scanner." I think that @SanketMakani is suggesting that it might be returning only the first word instead of the whole line. If you replace the line with the one suggested, it will fetch the whole line. — dolan
– dolan, Commented May 5, 2017 at 17:20

Sanket Makani · Accepted Answer · 2017-05-05 17:45:41Z

2

String pair = keyboard.next(); This reads only one token which are separated by " " so when you split pair by " ". It will only have one element, The String itself. So you need to read the whole line and then split it by delimited " ".

Another mistake is that when you change that line with String pair = keyboard.nextLine(); , You will still get error because System considers Enter key as input of .nextLine() method. So you need to discard that extra unnecessary input.

while(input!=-1){
    int last =0;
    for (int i=0; i<input ; i++){

        int ip1=keyboard.nextInt();
        int ip2=keyboard.nextInt();

        speed = speed + ip1*(ip2-last);
        last = ip2;
    }
    output = output +speed + "miles" + "\n";
    speed =0;
 input = keyboard.nextInt();
}

edited May 5, 2017 at 17:45

answered May 5, 2017 at 17:24

Sanket Makani

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6 Comments

Med Ada Over a year ago

it get the same problem when i type 4 in the execution ` 4 ` and give me this Error Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1 at J1.SpeedLimit.main(SpeedLimit.java:21)

Sanket Makani Over a year ago

Works perfectly on my machine. I will update the code try that code!

Med Ada Over a year ago

with the use of token1 and token 2 i got another Error when i type 10 7 AND this is the Error :

Exception in thread "main" java.lang.NumberFormatException: For input string: "" 	at java.lang.NumberFormatException.forInputString(Unknown Source) 	at java.lang.Integer.parseInt(Unknown Source) 	at java.lang.Integer.parseInt(Unknown Source) 	at J1.SpeedLimit.main(SpeedLimit.java:27)

Sanket Makani Over a year ago

@MedAda, Its almost impossible to get this error while getting inputs.You have surely made other mistakes But now leave all this and take input directly in the integer. Check edited code.

Med Ada Over a year ago

can you explain to me how Scanner work, and the difference between .next() and .nextLine() ?

|

Oussama Ben Ghorbel · Accepted Answer · 2017-05-05 17:28:08Z

0

You are reading the variable pair the wrong way and then you split it and assign it to tab which fails to automatically to fetch index cause pair variable got a problem.

*nextLine(): reads the remainder of the current line even if it is empty.

        keyboard.nextLine(); //To avoid the exception you commented
        String pair = keyboard.nextLine(); //here is solved
        tab = pair.split(" ");

edited May 5, 2017 at 17:28

answered May 5, 2017 at 17:19

Oussama Ben Ghorbel

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2 Comments

Med Ada Over a year ago

But when I try nextLine() instead of next(), I got this Error:

Exception in thread "main" java.lang.NumberFormatException: For input string: ""at java.lang.NumberFormatException.forInputString(Unknown Source) 	at java.lang.Integer.parseInt(Unknown Source) 	at java.lang.Integer.parseInt(Unknown Source) 	at J1.SpeedLimit.main(SpeedLimit.java:20)

Oussama Ben Ghorbel Over a year ago

Ah you should do keyboard.nextLine(); when you start looping to eat the enter key.

SajeeshR · Accepted Answer · 2017-05-05 17:34:22Z

0

Keyboard.next() will only read the input till the space, so pair and the array will have only one number, so tab[1] results in arrayOutOfBound exception. Use the method nextLine() to read the inputs with space.

answered May 5, 2017 at 17:34

SajeeshR

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Comments

Nishesh Pratap Singh · Accepted Answer · 2017-05-05 17:36:51Z

You Can try below changes in your code :

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    int input = Integer.parseInt(keyboard.nextLine());

    String[] tab = new String[2];
    String output="";
    int speed = 0;
    while(input!=-1){
        int last =0;            
        for (int i=0; i<input ; i++){
            String pair = keyboard.nextLine();              
            tab = pair.split(" ");
            speed = speed + Integer.parseInt(tab[0].trim())*(Integer.parseInt(tab[1].trim())-last);
            last = Integer.parseInt(tab[1]);
        }
        output = output +speed + " miles " + "\n";
        speed =0;
     input = Integer.parseInt(keyboard.nextLine());
    }
   System.out.println(output);
}

AppleCiderGuy · Accepted Answer · 2017-05-05 17:37:26Z

0

i did'n really understand how you are providing the inputs. but, if "3" happens to be your first line then split(" ") would return an array of length 1. thus, tab[0] would return 3 and tab[1] will give you a nullPointerException.

try adding an check for the length of tab before executing your line 20.

this should do the trick:

if(tab.length() > 1){

  speed = speed + Integer.parseInt(tab[0])*(Integer.parseInt(tab[1])-last);

}

answered May 5, 2017 at 17:37

AppleCiderGuy

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1 Comment

Med Ada Over a year ago

i want to excute this line : speed = speed + Integer.parseInt(tab[0])*(Integer.parseInt(tab[1])-last); i dont want to skip it

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