Suppose a packet contains 23 bytes of data, and the MTU of the network is 1024 bytes. We have to make sure the data is divisible by 8. But in this case, that is not possible since if the data is represented as 24 bytes,there will be an extra byte (where do we get it from?) and if the data is represented as 16 bytes, we will face a problem as 7 bytes are still remaining. What do we do.
May 18, 2009 5:33 AM
May 28, 2009 11:47 AM