System.IndexOutOfRangeException : Index was outside the bounds of the array

Question

I have the following array:

public string reArrange(string s)
{
    char[] array = s.ToCharArray();
    int length = array.Length;
    char[] arranged = new char[length];

    for (int i = 0; i < length; i++)
    {
        int newposition = length - i;
        arranged[newposition] = array[i];
    }

    return new string(arranged);
}

But the above method raises the following error:

System.IndexOutOfRangeException: Index was outside the bounds of the array.

So what might be wrong?

score 5 · Accepted Answer · answered Oct 19 '12 at 01:51

5

When i is zero, you access the array at index newposition which equals length; that's one beyond the last valid index for the array, which is 0 through length-1.

This will fix the problem:

int newposition = length - i - 1;

answered Oct 19 '12 at 01:51

Sergey Kalinichenko

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score 2 · Answer 2 · edited May 23 '17 at 12:13

2

Say the length is 10 characters. On the first iteration of your loop, newposition = 10 - 0 = 10. This index is out of the bounds of the arranged array.

Also, see this post about reversing a string. From that post...

public static string Reverse( string s )
{
    char[] charArray = s.ToCharArray();
    Array.Reverse( charArray );
    return new string( charArray );
}

edited May 23 '17 at 12:13

Community

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1

answered Oct 19 '12 at 01:53

TylerOhlsen

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+1 Beat me to the "better" way to reverse. =) – Jim D'Angelo Oct 19 '12 at 01:57

score 1 · Answer 3 · answered Oct 19 '12 at 01:52

1

You are going to far,

for (int i = 0; i < length-1; i++)

answered Oct 19 '12 at 01:52

Orn Kristjansson

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but in this way i can not reach the last element in the array ... – John John Oct 19 '12 at 01:53
It should be `i = 1` to `length` instead of `i = 0` to `length - 1` – mlorbetske Oct 19 '12 at 01:53
ah, missed the second array access – mlorbetske Oct 19 '12 at 01:56

mlorbetske · Answer 4 · 2012-10-19T02:04:41.213

1

public string reArrange(string s)
{
    char[] array = s.ToCharArray();
    int length = array.Length;
    char[] arranged = new char[length];

    for (int i = 0; i < length; i++)
    {
       int newposition = length - i - 1;
       arranged[newposition] = array[i];
    }
    return new string(arranged);
}

edited Oct 19 '12 at 02:04

answered Oct 19 '12 at 01:58

mlorbetske

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but in this case i will not be able to reach the last elemnet in the array[] – John John Oct 19 '12 at 01:59
The last element in the array is at index `length - 1`. Let's say the array has a length of 3, at `i = 0`, `newposition = 3 - 0 - 1 = 2` (length - 1), at `i = 2` (length - 1), `newposition = 3 - 2 - 1 = 0` at `i = 1`, `newposition = 3 - 1 - 1 = 1`, covering all the valid indexes – mlorbetske Oct 19 '12 at 02:01
if the array lenght is 5 then using (i < length -1) i will never reach the array[4]!!! – John John Oct 19 '12 at 02:04
oops, typo, editing now, thanks – mlorbetske Oct 19 '12 at 02:04

System.IndexOutOfRangeException : Index was outside the bounds of the array

4 Answers4