Java Array index out of bounds Exception when getting string input in java

Question

I got an Java ArrayIndexOutOfBoundsException when getting String input in Java. Please help me. This is my code: I edited my code to split using : it says

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1 at solution2.Solution.main(Solution.java:27)

public class Solution {

 static String search;

 public static void main(String[] args){

   String[] fn,ln,tp;
  boolean[] isSet;
 Scanner sc=new Scanner(System.in);      
 int no=sc.nextInt();

 String[][] temp=new String[no][3];
 fn=new String[no];
  ln=new String[no];
   tp=new String[no];
   isSet=new boolean[no];
   boolean flag=false;

     for(int i=0;i<no;i++){
   temp[i]=sc.nextLine().split(":");
   fn[i]=temp[i][0];
   ln[i]=temp[i][1];
   tp[i]=temp[i][2];
   isSet[i]=true;

     }

       System.out.println(temp.length);

      search=sc.nextLine();

Which one is line 27? What is the input for which you're getting the error? — Tudor, Oct 20 '12 at 15:26
temp[i]=sc.nextLine().split(":"); fn[i]=temp[i][0]; ln[i]=temp[i][1]; tp[i]=temp[i][2]; line we get error — chamara, Oct 20 '12 at 15:48

Reimeus · Answer 1 · 2012-10-20T16:00:55.593

1

The exception is occurring on this line:

ln[i] = temp[i][1];

so it appears

temp[i] = sc.nextLine().split(":");

is not receiving enough tokens in the :-delimited string to have create String array of size 3.

You will need to ensure temp[i].length == 3 to ensure that you can assign these tokens.

An example of valid input (note: no newline) is:

1 test:foo:bar

edited Oct 20 '12 at 16:00

answered Oct 20 '12 at 15:31

Reimeus

158,255
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score 0 · Answer 2 · answered Oct 20 '12 at 15:31

The ArrayIndexOutOfBoundsException occurs when you are accessing non-existent indexes on the array created from split(":").

In this piece of code, temp[i] is not ensured to have values at index 0, 1, or 2, because if the nextLine() is something such as "dog", there are no colon characters to split around.

temp[i]=sc.nextLine().split(":");
fn[i]=temp[i][0];
ln[i]=temp[i][1];
tp[i]=temp[i][2];

To fix the issue, you should verify that the array indeed has the index before trying to access it.

temp[i]=sc.nextLine().split(":");
if (temp[i].length >= 3) {
    fn[i]=temp[i][0];
    ln[i]=temp[i][1];
    tp[i]=temp[i][2];
}

score 0 · Answer 3 · edited May 23 '17 at 11:48

You problem is, that sc.nextLine() return new line String (e.g. "\n") after you press [Enter]. Second time it's wait for you input. SeeJava String Scanner input does not wait for info, moves directly to next statement. How to wait for info?

In your case try to invoke sc.nextLine() before you processing:

sc.nextLine()
temp[i]=sc.nextLine().split(":");

EDIT: you are right, you must insert it after nextInt(), because nextLine() consume complete line.

score 0 · Answer 4 · answered Oct 20 '12 at 15:47

I insert sc.nextLine(). below int no = sc.nextInt(); line.

import java.util.Scanner;

public class Solution {

static String search;

public static void main(String[] args) {

    String[] fn, ln, tp;
    boolean[] isSet;
    Scanner sc = new Scanner(System.in);
    int no = sc.nextInt();
    sc.nextLine(); // **********

    String[][] temp = new String[no][3];
    fn = new String[no];
    ln = new String[no];
    tp = new String[no];
    isSet = new boolean[no];
    boolean flag = false;

    for (int i = 0; i < no; i++) {
        temp[i] = sc.nextLine().split(":");
        fn[i] = temp[i][0];
        ln[i] = temp[i][1];
        tp[i] = temp[i][2];
        isSet[i] = true;

    }

    System.out.println(temp.length);

    search = sc.nextLine();
}

}

Java Array index out of bounds Exception when getting string input in java

4 Answers4