HW # 5: Theme: Data Definitions, Addressing Modes, Arrays

Description

Comp 3350: Computer Organization & Assembly Language
HW # 5:  Theme: Data Definitions, Addressing Modes, Arrays
All main questions carry equal weight.
(Credit awarded to only those answers for which work has been shown.)
1.    [Memory Map] Fill in the following memory diagram with the data provided below. Please assume that the data segment begins at 0x0038E000.
.data
Alpha        DWORD            1A2B3C4Dh, 12AB34CCh
Beta         BYTE            7Ch
Gamma     DWORD      9ABCDEF0h
Delta        BYTE            12h
Address    Variable    Data
0038E000    ALPHA    4Dh
0038E001    ALPHA    3Ch
0038E002    ALPHA    2Bh
0038E003    ALPHA    1Ah
0038E004    ALPHA    CCh
0038E005    ALPHA    34h
0038E006    ALPHA    ABh
0038E007    ALPHA    12h
0038E008    BETA    7Ch
0038E009    GAMMA    F0h
0038E00A    GAMMA    DEh
0038E00B    GAMMA    BCh
0038E00C    GAMMA    9Ah
0038E00D    Delta    12h
2.    [Addressing Modes] Copy the following code into your assembly development environment and single-step through it.  For each single step execution, submit the screenshot.  For those instructions referencing memory, do the linear address computation (typewritten).
TITLE Addressing Modes            (main.asm)
INCLUDE Irvine32.inc
.data
alpha     DWORD        65219751h, 24875139h
beta         DWORD        3B2C791Ah, 0A57716Dh
gamma      DWORD     0C58Bh
.code
main PROC
mov eax, 1111h;            Immediate
mov ecx, eax;            Register to Register
mov edi, OFFSET beta;        Immediate
mov [gamma], eax;         Direct
mov esi, gamma;            Direct
mov esi, 4;              Immediate
mov eax, beta[esi];        Indirect-offset
mov ebx, OFFSET alpha;        Immediate
mov eax, [ebx];              Indirect
mov eax, 4[ebx];             Indirect-displacement
mov eax, 4[ebx][esi];          Base-Indirect-displacement
exit
main ENDP
END main
STEP 1:
STEP 2:
STEP 3:
STEP 4:
STEP 5:
STEP 6:
STEP 7:
STEP 8:
STEP 9:
STEP 10:
STEP 11:
STEP 12:
Linear Address Calculations:
mov eax, 1111h;            Immediate
mov ecx, eax;            Register to Register
mov edi, OFFSET beta;        Immediate
mov [gamma], eax;         Direct; 0x010B4010 = 0000C58B
mov esi, gamma;            Direct; 0x010B4010 = 00001111
mov esi, 4;              Immediate
mov eax, beta[esi];    Indirect-offset; 0x010B400C = 0A57716D
mov ebx, OFFSET alpha;        Immediate
mov eax, [ebx];              Indirect; 0x010B4000 = 65219751
mov eax, 4[ebx];     Indirect-displacement; 0x010B4004 = 24875139
mov eax, 4[ebx][esi];      Base-Indirect-displacement; 0x010B4008 = 3B2C791A
invoke ExitProcess,0;
main ENDP
END main
3.    [Indirect addressing] Write a program that subtracts the corresponding even indexed elements of Array2 from Array1 and stores the results in Array3; e.g. for the 8th element, Array3 [7]  Array1 [7] – Array2 [7].  Note that Array3 will have about half the number of elements of the other two arrays.  Include commands to display the elements of all the arrays.  Submit screenshot of the displays of the elements of all the arrays.  You can use WriteInt or WriteHex to display the elements of the arrays.  Fill in Array1 and Array2 each by your own ten numbers.
.data
Array1        WORD 7h, …
Array2     WORD 4h, …
Array3     WORD 10 DUP (?)
For example, if we have:
Array1 WORD 9h, 8h, 7h, 6h
Array2 WORD 1h, 2h, 3h, 4h
Then the expected result stored in Array3 should be:
8h,4h
OUTPUT:
4.    [Loops] Write a program to compute the sum of first n integers of the series:  Sum = 1 + 2 – 3 + 4 -5 + 6 -7 +8 – 9… Your program must:
a.    Prompt user for integer n,
b.    Read the value of n from user input
c.    Calculate Sum, and;
d.    Print Sum on screen.
Please use the “WriteInt” procedure, not “DumpRegs”. Other relevant procedures: “ReadInt” and “WriteString.” The calculation can be done in many ways, and all submissions that evidence proper programming practice are acceptable. In your homework submission, please embed both the code and one screen shot for n = 10.
Line41: END sumInt