Java charAt() String index out of range: 5

Question

I am trying to figure out "what 5-digit number when multiplied by 4 gives you its reverse?" using this code but I get error: Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 5 at java.lang.String.charAt(String.java:658) at Digits.main(Digits.java:12)

 public class Digits{
  public static void main(String[] args) {
    int n = 0;
    int b = 0;
    String number = Integer.toString(n);
    String backwards = Integer.toString(b);

for (int x = 9999; x < 100000 ; x++ ) {
  n = x;
  b = x *4;

  if (number.charAt(0) == backwards.charAt(5 )&& number.charAt(1) == backwards.charAt(4)
  && number.charAt(2) == backwards.charAt(3) && number.charAt(3) == backwards.charAt(2)
  && number.charAt(4) == backwards.charAt(1) && number.charAt(5) == backwards.charAt(0)) {
    System.out.println(n);
    break;
  }
}

Any help would be grealy appreciated

Elliott Frisch · Accepted Answer · 2015-07-13 04:24:16Z

4

Correct. Because the first five characters are at indices 0, 1, 2, 3 and 4. I would use a StringBuilder (because of StringBuilder.reverse()). And, I would suggest you restrict variable visibility. Then remember to modify number and backwards when you change n and/or b. Something like

for (int x = 9999; x < 100000; x++) {
    int n = x;
    int b = x * 4;
    String number = Integer.toString(n);
    String backwards = Integer.toString(b);
    StringBuilder sb = new StringBuilder(number);
    sb.reverse();
    if (sb.toString().equals(backwards)) {
        System.out.printf("%s * 4 = %s", number, backwards);
    }
}

And I get

21978 * 4 = 87912

edited Jul 13, 2015 at 4:24

answered Jul 13, 2015 at 4:19

Elliott Frisch

202k20 gold badges166 silver badges265 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Tim Biegeleisen Over a year ago

You're too fast me for me, Frisch!

Juned Ahsan · Accepted Answer · 2015-07-13 04:17:23Z

0

backwards and number are String, which internally uses an array. And an array are indexed from 0 to size-1 . Hence such statements will throw ArrayIndexOutOfBoundsException:

backwards.charAt(5 )
number.charAt(5)

answered Jul 13, 2015 at 4:17

Juned Ahsan

68.9k12 gold badges101 silver badges138 bronze badges

1 Comment

Tim Biegeleisen Over a year ago

Your observation is correct, but this doesn't help with the problems in his code.

Maximillian Laumeister · Accepted Answer · 2015-07-13 04:18:19Z

At the time you create your strings, both of your ints are 0, so both of your strings are "0" for the duration of your program. What you really want is the strings to change every time your number changes. So your code should look more like this:

public class Digits{
  public static void main(String[] args) {
    int n = 0;
    int b = 0;
    String number;
    String backwards;

for (int x = 10000; x < 100000 ; x++ ) {
  n = x;
  b = x *4;

  number = Integer.toString(n);
  backwards = Integer.toString(b)

  . . .
}

In addition, arrays in Java are zero-indexed, so for instance for the string "10000", your program will throw the index out of bounds exception on backwards.charAt(5) because the string is indexed from character 0 to character 4.

Collectives™ on Stack Overflow

Java charAt() String index out of range: 5

3 Answers 3

1 Comment

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related