Converting to binary - getting IndexOutOfBoundsException

Question

Below is some code I'm working on, I thought I'd make myself a binary calculator to make my life slightly easier. However, when I run it, I get an error telling me that there is a Java.lang.StringIndexOutofBoundsException. I don't really know how to fix it, because as far as I can tell, I've done everything correctly:

private static void ten()
{
    Scanner scan = new Scanner(System.in);

    System.out.println("What number would you like to convert to binary?");
    System.out.print("Enter the integer here:  ");
    int x = scan.nextInt();

    String bon = Integer.toString(x , 2);

    int myArrays [ ] = new int [ 7 ];

    myArrays[0] = bon.charAt(0); 
    myArrays[1] = bon.charAt(1); 
    myArrays[2] = bon.charAt(2); 
    myArrays[3] = bon.charAt(3); 
    myArrays[4] = bon.charAt(4); 
    myArrays[5] = bon.charAt(5); 
    myArrays[6] = bon.charAt(6); 
    myArrays[7] = bon.charAt(7); 

    for (int i = 0; i < myArrays.length; i++)
    {
        System.out.print(myArrays [ i ] + " ");
        int count = 0;
        count++;
        if (count == 10) {
            System.out.println();
            count = 0;
        }
    }

}

You need to perform basic debugging: Read the stack trace of your exception, as it tells you exactly which line is causing the problem. Then, add some `System.out.println` statements *before* that line, so you can see both your String and the index you're trying to access. — VGR, Feb 12 '16 at 22:24
I'm slightly confused here. Some people are saying that I should increase my array by one, and others say I should decrease my array by one. Which one do I do? — arsb48, Feb 12 '16 at 22:27
and debug your own code, which should have been the first thing you should have done before coming here — user1231232141214124, Feb 12 '16 at 22:31
I've tried debugging my code, and I've come across a different problem entirely. My array, should all eight integers of the array be filled, prints out "49" eight times. I assumed that the array would be printed, not something else. But it makes sense now, I'll try to do it by myself. Thanks for all the help! — arsb48, Feb 12 '16 at 22:35

score 1 · Answer 1 · answered Feb 12 '16 at 22:26

Depending on the number you enter, the length of the binary string will vary. If you enter 0 or 1, you'll get "0" or "1". But your code assumes your number has 8 bits. This means that it will only work if the eighth bit is set.

It looks like you are trying to print the bits separated by spaces, followed by a new line. It would be better to use a more direct approach:

String b = Integer.toString(x, 2);
for (int idx = 0; idx < b.length(); ++idx) {
  System.out.print(b.charAt(idx));
  System.out.print(' ');
}
System.out.println();

score 0 · Answer 2 · answered Feb 12 '16 at 22:23

0

Ok so you have an array of size 7 which only gives you 0-6 but you make a call to array[7] so increase your array size by one

answered Feb 12 '16 at 22:23

Arachnid Hivemind

504
2
14

I thought arrays started at element "0"? – arsb48 Feb 12 '16 at 22:24
You are correct about the array size (should be array[8]). However, that is not where he is getting an IndexOutOfBoundsException - it is at the line of code: myArrays[3] = bon.charAt(3); – pczeus Feb 12 '16 at 22:29

score 0 · Answer 3 · answered Feb 12 '16 at 22:25

0

myArrays is a 7 elements len array, if you do operations from 0 to 7 you are going outside the limit..

try from 0 to 6!

answered Feb 12 '16 at 22:25

ΦXocę 웃 Пepeúpa ツ

47,427
17
69
97

score 0 · Answer 4 · answered Feb 12 '16 at 22:25

0

Your array size needs to be larger... int myArrays [ ] = new int [ 8 ];

answered Feb 12 '16 at 22:25

Lencalot

385
2
19

score 0 · Answer 5 · answered Feb 12 '16 at 22:26

0

There are two things to note in your code:

1.

int myArrays [] = new int [7];

Should be changed to:

int myArrays [] = new int [8];

2.

You call bon.charAt() without checking that your variable bon has enough characters.

answered Feb 12 '16 at 22:26

thyago stall

1,654
3
16
30

score 0 · Answer 6 · answered Feb 12 '16 at 22:31

0

Java.lang.StringIndexOutofBoundsException Pretty much indicates string's length is less than the index. You should consider the length. Also there are other Out of bounds in arrays that others mentioned.

answered Feb 12 '16 at 22:31

madz

1,803
18
45

score 0 · Answer 7 · answered Feb 12 '16 at 22:49

Try this:

Replace these 2 lines:

String bon = Integer.toString(x , 2);
int myArrays [ ] = new int [ 7 ];

With these lines (1 just to display the initial value):

String byteString = Integer.toBinaryString(x & 0xFF);
byteString = String.format("%8s", byteString).replace(' ', '0');
int[] myArrays = new int[8];
System.out.println("(" + byteString  + ")");

Converting to binary - getting IndexOutOfBoundsException

7 Answers7