Homwork 1 here is an example of an input file and an output file. Note that the length of the payload is 3 bytes, the payload bytes are

AB CD 01 their exor is 67 frame is error-free

here are the test cases for hw1

Homework 2

Here is an example input file. It contains the code ABC, the name MOKHTAR and the integer 24.

Not that 24 (20 decimal) is not the digit 2 followed by the digit 4. It is the way the computer store integers in the memory.

Take a look at the file using od -c and you will see how every byte is displayed.

In order to read an integer, you have to read a number of bytes equal to the length on integer (4) and sote it in a location where the computer will interppret it as the integer number 24. the tared test casses and the datafile

 

here is another file that contains 2 records the two records are

ABC MOKHTAR 20 and

XYZ ABOELAZE 30

A file with records with KEY, name, mark1 and mark2 is here

its contents as follows:

ABC MOKHTAR 20 21

XYZ ABOELAZE 22 23

Not ethat both records ends with \0 \0 \0. That is because the integer is stored in 4 bytes, however the values that I have here is small and can fit in only one byte. The other three contains 0's which is the definition of \0 here is manother file which is similar to the previou sone but I changed the 23 to 10987.

Not that the last 4 bytes are 353 * \0 0\

10987 in binary is 00000000000000000010101011101011

Now when you do od on this the resul is

00000000 00000000 00101010 11101011

tha is 0(\0) another 0 (\0) then 00101010 whic is ASCII '*" finally 11101011 is octal 353. In the memory it is stored as 353 "*" \0 \0

 

 

homeowrk 3

here is a routing table example an input example, soon an output

test cases for hw3, and routing table