Homework Assignment #2 [50 pts]

You are expected to do your own work on all homework assignments. You may (and are encouraged to) engage in general discussions with your classmates regarding the assignments, but specific details of a solution, including the solution itself, must always be your own work. (See the statement of Academic Dishonesty on the Syllabus.

How to turn in?

Assignments need to be turned in via Laulima. Check the Syllabus for the late assignment policy for the course.

What to turn in?

You should turn in a single plain text or PDF file named README.txt or README.pdf with your answers to the assignment’s questions.

A 25% grade penalty will be applied if the file format or file name is wrong.

Exercise #1 [24pts]: Memory Layout

Consider the following .data segment:

A	dd	1101b
B	db	043h, 01Ah, 0CEh
C	dw	-18
D	times   2 dw	012FAh
E	db	"c", 13, "aa",  0
F	dw      009h, 23o

Question #1 [18pts]

Show the contents of the memory bytes starting at address A, in hex, on a machine that uses Little Endian. Indicate labels as well. For instance, a (wrong) answer could look like:

A        B     C  ...
FF FF FA 02 03 06 ...

Question #2 [6pts]:

For each of the 6 labels, indicate whether it would lead to a different in-memory byte order on a Big Endian machine.

Exercise #2 [26pts]: Memory and Registers

Consider the following 13 bytes (in hex) declared by some .data segment on a Little Endian machine:

L1          L2    L3                L4
03 00 00 00 6C 6C 6F 00 A1 B2 C3 13 00

Consider now the following program fragment:

mov     eax, [L2]
inc     eax
mov     [L3], eax
mov     ebx, [L1]
mov     eax, L4
sub     eax, ebx
mov     word [eax], 01970h

After the code finishes executing, what are the contents, in hex, of the 13 memory bytes starting at address L1, on a machine using Little Endian?

Show your work for each instruction, as done for the examples in the lecture notes. Each instruction is worth 3 points. An instruction with no work shown will receive zero points.

Here is the way to show your work for each instruction:

• If an instruction updates the value of a register, show the value of that register (e.g., “ecx = 00AABBCC, bx = FF FF, eax = ‘the address of the 8th byte’, edx = ‘the address of the byte with value 42’);
• If an instruction updates memory, then show the full memory state after that instruction.

For instance, a partial answer (for a different program) could look like:

L1       L2       L3       L4       L5
3A FF 01 02 FF 6B B2 AA 42 41 61 62 D3 D5 D7

mov ebx, L1     ; ebx = address of the 1st byte

mov ecx, 0      ; ecx = 00 00 00 00

mov cx, [L3]    ; ecx = 00 00 AA B2

mov [L2], ecx   ; updates memory state:

L1       L2       L3       L4       L5
3A FF 01 B2 AA 00 00 AA 42 41 61 62 D3 D5 D7

mov ebx, L4     ; ebx = address of the byte with value 41

mov     bx, 12      ; ebx = ?? ?? 00 0C

...