Why is allocating a byte array throwing an Overflow exception?

Question

I am receiving an error on my C# Console Application. I am trying to do some code but its throwing an exception whenever I do so, not all the time just some of the time, by the way it is not just me using this program there is 30+ members on it using it also.

Code:

int MessageLength = Base64Encoding.DecodeInt32(new byte[] { data[pos++], data[pos++], data[pos++] });
int MessageId = Base64Encoding.DecodeInt32(new byte[] { data[pos++], data[pos++] });

byte[] Content = new byte[MessageLength - 2];

The Exception is:

Exception logged 1/23/2015 10:54:54 AM in packet handling: System.OverflowException: Arithmetic operation resulted in an overflow.

The exception is happening on line 37, the following line...

byte[] Content = new byte[MessageLength - 2];

Here is a image of the lengths changing of 'pos' and 'MessgaeLength'

I will run some code to check, ill reply with the length in 2 minutes. — LiamHardyLovesKody
– LiamHardyLovesKody, Commented Jan 23, 2015 at 11:06

Patrick Hofman · Accepted Answer · 2015-01-23 11:09:09Z

3

Make sure MessageLength > 2.

For example :

if (MessageLength < 3)
    throw new Exception("Incorrect MessageLength");

Otherwise, you are trying to instantiate an array with Length 0 / -X.

edited Jan 23, 2015 at 11:09

Patrick Hofman

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answered Jan 23, 2015 at 11:05

vincentp

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2 Comments

Lasse V. Karlsen Over a year ago

0 is an ok value, it might not be logical in this context, but you can certainly construct an array with a length of 0 in .NET.

LiamHardyLovesKody Over a year ago

I choose this method and it has stopped the exceptions so far, thank you.

Patrick Hofman · Accepted Answer · 2015-01-23 11:07:53Z

1

The problem is most probably in the length of MessageLength. If the length is less than 2, you will get an overflow.

Try this failing code:

int x = 0;
byte[] Content = new byte[x - 2];

You should check before allocating that x-2 is more than or equals to 0:

if (MessageLength >= 2)
{
    byte[] Content = new byte[MessageLength - 2];

    ...
}

answered Jan 23, 2015 at 11:07

Patrick Hofman

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