Question

A network protocol contains a four-byte integer, specifying the length of the payload in
the packet. The implementation of this protocol has a mistake in it. When a packet is
received, the protocol implementation needs to copy the payload to a buffer. It first copies
the length field from the packet header to a variable, but the program forgets to convert
the number into the host order. Assume the value of this variable is X. The program then
allocates X bytes of memory to hold a copy of the payload. On a Little-Endian machine,
if the payload of a received packet is 255 bytes, how much memory will be allocated?
What is a likely consequence of this mistake?

Answer 1

Since the host machine works on little-endian system, it will store the length in reverse order.

Thus 255 Bytes which could be written as 000000FF in hexadecimal form will be the stored in payload length field at sender side. But at receiver side this field gets stored in reverse order in variable x.

x = FF000000

= 2^32 - 2^24 B

= 4080 MB

= 4 GB(approax)

Hence the host machine will end up allocating approaximately 4 GB memory for a payload of size 255 B.

This will lead to wastage of memory as payload occupies only 255 B , remaining memory is not useful.