String index out of range error in java

Question

I am trying to make different strings print based off of the number of letters in the word, but my second "if" statement is breaking the code.

import java.util.Scanner;
public class practice {


public static void main(String[] args) {
     Scanner in = new Scanner(System.in);
     System.out.printf("Enter a string: ");       
     String aString = in.next();
     int length = aString.length();

   char first =   aString.charAt(0);
   char second =  aString.charAt(1);
   char third =   aString.charAt(2);
   char fourth =  aString.charAt(3);
   char fifth =   aString.charAt(4);
   char sixth =   aString.charAt(5);
   char seventh = aString.charAt(6);

  if (length == 7) {
     System.out.println(String.format("Line 1: %s %s %s %s ", first, third, fifth, sixth));
     System.out.println(String.format("Line 2: %s %s %s ", second, fourth, sixth));
     System.out.println(String.format("Line 3: %s %s %s %s ", seventh, fifth, third, first));
  }
  else {
   if (length < 7) {
     System.out.println("the length is less than 7");
   }
      }
}

}

If I enter a word that has less than 7 letters I just get an error that reads "String index out of range: 6". How can I fix this?

From a very cursory look, I believe it's the line char seventh = aString.charAt(6) that is breaking if your input does not have a 7th character. — Ambidextrous
– Ambidextrous, Commented Feb 15, 2016 at 3:57
put char first = aString.charAt(0); char second = aString.charAt(1); char third = aString.charAt(2); char fourth = aString.charAt(3); char fifth = aString.charAt(4); char sixth = aString.charAt(5); char seventh = aString.charAt(6); inside if — wijaya
– wijaya, Commented Feb 15, 2016 at 3:59

prasad · Accepted Answer · 2016-02-15 03:59:30Z

since i don't know the exact requirement i just resolved the error check below coede

import java.util.Scanner;
public class practice {


public static void main(String[] args) {
 Scanner in = new Scanner(System.in);
 System.out.printf("Enter a string: ");
 String aString = in.next();
 int length = aString.length();


  if (length == 7) {
   char first =   aString.charAt(0);
   char second =  aString.charAt(1);
   char third =   aString.charAt(2);
   char fourth =  aString.charAt(3);
   char fifth =   aString.charAt(4);
   char sixth =   aString.charAt(5);
   char seventh = aString.charAt(6);
 System.out.println(String.format("Line 1: %s %s %s %s ", first, third, fifth, sixth));
 System.out.println(String.format("Line 2: %s %s %s ", second, fourth, sixth));
 System.out.println(String.format("Line 3: %s %s %s %s ", seventh, fifth, third, first));
}
else {
if (length < 7) {
 System.out.println("the length is less than 7");
}
  }
}

}

Elliott Frisch · Accepted Answer · 2016-02-15 03:58:54Z

1

First check the length, then access the characters. Also, you can use printf (instead of String.format and println). Something like

if (length >= 7) {
     char first =   aString.charAt(0);
     char second =  aString.charAt(1);
     char third =   aString.charAt(2);
     char fourth =  aString.charAt(3);
     char fifth =   aString.charAt(4);
     char sixth =   aString.charAt(5);
     char seventh = aString.charAt(6);

     System.out.printf("Line 1: %s %s %s %s%n", first, third, fifth, sixth);
     System.out.printf("Line 2: %s %s %s%n", second, fourth, sixth);
     System.out.printf("Line 3: %s %s %s %s%n", seventh, fifth, third, first);
 } else {
     System.out.println("the length is less than 7");
 }

answered Feb 15, 2016 at 3:58

Elliott Frisch

202k20 gold badges166 silver badges265 bronze badges

Comments

unmultimedio · Accepted Answer · 2016-02-15 04:01:12Z

I bet you entered a 6 letter word.

This is because of your line:

char seventh = aString.charAt(6); // It tries to look for a 7th position in a String that has only six positions

I'd put all of the char assignments into the if:

public static void main(String[] args) {
 Scanner in = new Scanner(System.in);
 System.out.printf("Enter a string: ");       
 String aString = in.next();
 int length = aString.length();


  if (length == 7) {
       char first =   aString.charAt(0);
       char second =  aString.charAt(1);
       char third =   aString.charAt(2);
       char fourth =  aString.charAt(3);
       char fifth =   aString.charAt(4);
       char sixth =   aString.charAt(5);
       char seventh = aString.charAt(6);

     System.out.println(String.format("Line 1: %s %s %s %s ", first, third, fifth, sixth));
     System.out.println(String.format("Line 2: %s %s %s ", second, fourth, sixth));
     System.out.println(String.format("Line 3: %s %s %s %s ", seventh, fifth, third, first));
  }
  else {
    if (length < 7) {
        System.out.println("the length is less than 7");
    }
  }
}

Brijesh · Accepted Answer · 2016-02-15 04:15:04Z

Try below code, it is more generic.

            import java.util.Scanner;
            public class practice {


            public static void main(String[] args) {
                 Scanner in = new Scanner(System.in);
                 System.out.printf("Enter a string: ");       
                 String aString = in.next();
                 char[] charArr=new char[10];
                 int length = aString.length();
            for(int i=0;i<length;i++){
                charArr[i] =   aString.charAt(i);
            }

              if (length == 7) {
                 System.out.println(String.format("Line 1: %s %s %s %s ", charArr[0], charArr[2], charArr[4], charArr[5]));
                 System.out.println(String.format("Line 2: %s %s %s ", charArr[1], charArr[3], charArr[5]));
                 System.out.println(String.format("Line 3: %s %s %s %s ", charArr[6], charArr[4], charArr[2], charArr[0]));
              }
              else {
               if (length < 7) {
                 System.out.println("the length is less than 7");
               }
                  }
            }

            }

Collectives™ on Stack Overflow

String index out of range error in java

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related