CS401 Assignment#1 Solution Spring 2018

Assignment No. 1 Graded

Semester Spring 2018

Computer Architecture and Assembly Language Programming– CS401

Total Marks: 15 Due Date: 10/05/2018

Question: 1                                                                                                                                       [15 Marks]

You will write a program to find the sum of even and odd digits of your VU id, You will store the sum of even digits in AX and sum of odd digits in BX register.

For example, If your VU id is bc123456789 you will use two arrays as 2,4,6,8

1,3,5,7,9

Then save the sum of first array in AX and second array in BX in hexadecimal format.

Explain each instruction of program in comments and also provide/paste snapshot of your assembled and debugged program result which will be run in AFD (A Full Screen Debugger) window showing the executed code and final result i.e. value in AX.

You will submit code and two screen shots in word file.

First screenshot will contain the command to compile the code Second screenshot will contain the final values in AX and BX

Note 1: You will use conditional jumps to solve the above problem. Marks will not be awarded for using any other approach.

Note 2: keep name of assembly file according to your VU id , i.e. if your VU id is BC123456789 , you will keep name of the assembly code file as 23456789.asm i.e. you will use last 8 digits of your id for the code file name.

Note 3: If the arrays, files names in the screen shots will not be according to your VU id you will get zero marks.

Note 4: No need to include alphabet part of your id in array. i.e. if your id is BC123456789 you will use 123456789 to make the arrays , you will ignore the BC part.

Example Run

Let say your id is BC123456789 , You will use two array as shown below, Array 1 : 2,4,6,8 ,

Array 2 : 1,3,5,7,9

Your first screen shot will look like this

Your final screen shot will be like shown below AX and BX conataing the sum of even and odd digits in your VU id in HEX format.

If you have 64bit version OS and you are using DOXBOX then your screen shots should look like this

Your 2^nd screen shot will be like shown below AX and BX conataing the sum of even and odd digits in your VU id in HEX format

CS401 Assignment Solution

jmp start ; set control on start label.

Arr1: dw 7H,3H,5H ; Decleration of odd array.

Arr2: dw 3h,2h,0h ; Decleration of even array including 0. start: ; Start label

mov bx, 0 ; Set bx value 0 mov ax, 0 ; set ax value 0

mov cx,3 ; Set counter to 3 because our odd array length is 3 top1: ; A label for loop iterations.

cmp cx, bx ; compaire if counter is 0

jne even ; set control to even label if counter is not 0 even:   ; Even label

add ax,[Arr1+bx] ; Adding odd array elements one by one add bx,2 ; increasing array index number

loop top1 ; set control back to loop1 label.

mov dx,ax ; store ax value which is the sum of odd array elements. mov bx, 0 ; set back bx to 0

mov  ax, 0 ; set back ax to 0

mov cx,3 ; set counter again to 3 top2: ; a lebel for loop

cmp cx,bx ; compairing counter value if it is 0 jne odd ; jum on add label if not equal to 0 odd: ; odd label

add ax,[Arr2+bx] ; adding even array elements one by one add bx,2 ; increasing Array index number

loop top2 ; set control back to loop2 mov bx,ax ; store ax value in bx

mov ax,dx ; getting back value from dx mov dx,0 ; set dx to 0

INT 21h ; Return control to OS

OR

[org 0x0100]

jmp start

message: db 'your IDs'            ; string to be printed

length: dw 20                        ; length of string

clrscr: push es

push ax

push cx

push di

mov ax, 0xb800

mov es, ax                          ; point es to video base

xor di, di                          ; point di to top left column

mov ax, 0x0720                      ; space char in normal attribute

mov cx, 2000                        ; number of screen locations

cld                                 ; auto increment mode

repstosw                            ; clear the whole screen

pop di

pop cx

pop ax

pop es

ret

; subroutine to print a string

; takes the x position, y position, attribute, address of string and

; its length as parameters

printstr: push bp

mov bp, sp

push es

push ax

push cx

push si

push di

mov ax, 0xb800

mov es, ax                        ; point es to video base

mov al, 80                        ; load al with columns per row

mul byte [bp+10]                  ; multiply with y position

add ax, [bp+12]                   ; add x position

shl ax, 1                         ; turn into byte offset

mov di,ax                         ; point di to required location

mov si, [bp+6]                    ; point si to string

mov cx, [bp+4]                    ; load length of string in cx

mov ah, [bp+8]                    ; load attribute in ah

cld                               ; auto increment mode

nextchar: lodsb                   ; load next char in al

stosw                             ; print char/attribute pair

loopnextchar                      ; repeat for the whole string

pop di

pop si

pop cx

pop ax

pop es

pop bp

ret 10

start: call clrscr                ; call the clrscr subroutine

mov ax, 10

push ax                           ; push x position

mov ax, 15

push ax                           ; push y position

mov ax, 0x12                      ; blue on black attribute

push ax                           ; push attribute

mov ax, message

push ax                           ; push address of message

push word [length]                ; push message length

callprintstr                      ; call the printstr subroutine

mov ax, 0x4c00                    ; terminate program

int 0x21