;-----------------------------------------------------;
; Title : Final Project ;
; COS 230 ;
; Programmer : Hani AL Banai ;
; ;
; This program is to simulate Traffic Light System ;
; It will display three bars of three different ;
; colors , a delay is going to control the time each ;
; bar will be visible ;
;-----------------------------------------------------;
.model small
.stack 100h
.code
main proc
mov ax,@data
mov ds,ax
mov si,3 ; counter
bars: dec si
mov cx,8555 ; delay
L1: push cx ;
mov cx,2 ;
L2:push cx ;
mov cx,8555 ;
L3:push cx ;
pop cx ;
loop L3 ;
pop cx ;
loop L2 ;
pop cx ;
loop L1 ;
mov ah,6 ; scroll window down
mov al,0 ; all lines
mov ch,2 ; upper left row for bar 1
mov cl,11 ; upper left column for bar 1
mov dh,22 ; lower right row for bar 1
mov dl,21 ; lower right column for bar 1
mov bh,01000000b ; red background color
int 10h ; do it now
mov cx,8555 ; delay
L4: push cx
mov cx,2
L5:push cx
mov cx,8555
L6:push cx
pop cx
loop L6
pop cx
loop L5
pop cx
loop L4
mov ah,6 ; scroll window down
mov al,0 ; all lines
mov ch,0 ; upper left row for whole window
mov cl,0 ; upper left column for whole window
mov dh,24 ; lower right row for whole window
mov dl,79 ; lower right column for whole window
mov bh,00000000b ; back background color
int 10h ; do it now
mov ah,6 ; scroll window down
mov al,0 ; all lines
mov ch,2 ; upper left row for bar 2
mov cl,31 ; upper left column for bar 2
mov dh,22 ; lower right row for bar 2
mov dl,41 ; lower right column for bar 2
mov bh,01111111b
int 10h ; do it now
mov cx,8555 ; delay
L10: push cx ;
mov cx,2 ;
L11:push cx ;
mov cx,8555 ;
L12:push cx ;
pop cx ;
loop L12 ;
pop cx ;
loop L11 ;
pop cx ;
loop L10 ;
mov ah,6 ; scroll window down
mov al,0 ; all lines
mov ch,0 ; upper left row for whole window
mov cl,0 ; upper left column for whole window
mov dh,24 ; lower right row for whole window
mov dl,79 ; lower right column for whole window
mov bh,00000000b ; back background color
int 10h ; do it now
mov ah,6 ; scroll window down
mov al,0 ; all lines
mov ch,2 ; upper left row for bar 3
mov cl,51 ; upper left column for bar 3
mov dh,22 ; lower right row for bar 3
mov dl,61 ; lower right column for bar 3
mov bh,00100100b ; green background color
int 10h ; do it now
mov cx,8555 ; delay
L7: push cx ;
mov cx,2 ;
L8:push cx ;
mov cx,8555 ;
L9:push cx ;
pop cx ;
loop L9 ;
pop cx ;
loop L8 ;
pop cx ;
loop L7 ;
mov ah,6 ; scroll window down
mov al,0 ; all lines
mov ch,0 ; upper left row for whole window
mov cl,0 ; upper left column for whole window
mov dh,24 ; lower right row for whole window
mov dl,79 ; lower right column for whole window
mov bh,00000000b ; back background color
int 10h ; do it now
cmp si,0 ; check if the counter reached zero
jne bars ; if not repeat
mov ax,4c00h ; else exit
int 21h
main endp
end main
;------------------------------------------------------------------------------;
; Programmer : Hani AL Banai ;
; COS 230 ;
; Final Exam program number 5 using 32-bit registers ;
; Write a program that accepts two integers from the keyboard and display their;
; greatest common divisor on standard output, using a recursive assembler ;
; function to calculate GCD.The program should continue accepting pairs of ;
; integers until a zero is entered for one of the integers ;
;------------------------------------------------------------------------------;
.model small
.386
.data
Prompt1 db 0dh,0ah,0dh,0ah
db " Enter two integers to find their greatest common divisor (0 to exit)"
db 0dh,0ah
db " after entering each integer hit the ENTER key "
db 0dh,0ah,0
Prompt2 db 0dh,0ah
db " The greatest common divisor is : ",0
db 0dh,0ah,0
IntOne dd ?
IntTwo dd ?
.code
extrn Crlf:proc,Readint:proc,Writeint:proc
main proc
mov ax,@data
mov ds,ax
; Display the first message to ask the user to enter 2 integers
L0:
mov dx,offset Prompt1
call Writestring
; Read what the user entered
mov edx,offset IntOne ; first integer
call Readint
cmp eax,0
je L3 ; if it is equal to zero exit
call Crlf
push eax
mov edx,offset IntTwo ; second integer
call Readint
cmp eax,0 ; if it is equal to zero exit
je L3
call Crlf
push eax
; display the answering message
mov dx,offset Prompt2
call Writestring
; do the calculations
pop eax
mov ebx,eax
pop eax
mov edx,0
idiv ebx ; for a signed division
call GCD ; call the gcd procedure
; display the gcd for the two integers processed
mov eax,ebx
mov bx,10
mov edx,eax
call Writeint
Loop L0 ; repeat until the user enters a zero
L3:
mov ax,4c00h
int 21h
main endp
; greates common divisor procedure
GCD proc
cmp edx,0 ; do not divide by zero !
je L8
mov eax,ebx
mov ebx,edx
mov edx,0
idiv ebx
jmp GCD ; call the function recursivily
L8: ; exit
ret
GCD endp
; write string procedure
Writestring proc
push ds ; set ES to DS
pop es
mov di,dx ; let ES:DI point to the string
call Str_length ; get length of string in AX
mov cx,ax ; CX = number of bytes to write
mov ah,40h ; write to file or device
mov bx,1 ; choose standard output
int 21h ; call DOS
ret
Writestring endp
; string lenght procedure
Str_length proc
push cx
push di ; save pointer to string
mov cx,0FFFFh ; set CX to maximum word value
mov al,0 ; scan for null byte
cld ; direction = up
repnz scasb ; compare AL to ES:[DI]
dec di ; back up one position
mov ax,di ; get ending pointer
pop di ; retrieve starting pointer
sub ax,di ; subtract start from end
pop cx
ret ; AX = string length
Str_length endp
end main
;------------------------------------------------------------------------------;
; Programmer : Hani AL Banai ;
; COS 230 ;
; Assignment 7 ;
; This is a user interactive program , the user should enter a variavle's name ;
; and the program will search for it in this computer's envornment . Messages ;
; will display as the program runs to feedback the user . ;
;------------------------------------------------------------------------------;
.model small
.stack 100h
.data
Prompt1 db 0dh,0ah,0dh,0ah
db "Enter the variabl's name you want to search for (blank to exit):-"
db 0dh,0ah,0dh,0ah,0
Prompt2 db 0dh,0ah
db "You entered :"
db 0dh,0ah,0
Prompt3 db 0dh,0ah,0dh,0ah
db "Sorry ! we couldnt find this variable. Please try again"
db 0dh,0ah,0dh,0ah,0
Prompt4 db 0dh,0ah,0dh,0ah
db "Yes ! we found this variable it is:"
db 0dh,0ah,0
VarName db 50 dup(0),0dh,0ah,0
.code
extrn Crlf:proc
main proc
mov ax,@data
mov ds,ax
; Display the first message to ask the user to enter a name
L0:
mov dx,offset Prompt1
call Writestring
; Read what the user entered
mov dx,offset VarName
call Readstring
call Crlf
mov al,0
mov di,offset VarName
cmp al,[di]
je L3
; Display the second message to let the user know that we read the input string
mov dx,offset Prompt2
call Writestring
mov dx,offset VarName
call Writestring
; Search for the variable in the envirnment
mov si,offset VarName
mov si,dx ; let ES:DI point to the string
call Str_length ; get length of string in AX
mov cx,ax ; CX = number of bytes
mov es,es:[2ch]
sub di,di
sub al,al
mov cx,7FFFh
L99:
repne scasb
cmp byte ptr [di],0
jne L99
mov cx,di
cmp cx,dx
call compare
jz L1
jmp L2
; If we find it , display the variable again
L1:
mov dx,offset Prompt4
call Writestring
mov dx,offset VarName
call Writestring
Loop L0
; Else , display a message to say the variable is not found
L2:
mov dx,offset Prompt3
call Writestring
Loop L0
; If the user did not enter a name exit
L3:
mov ax,4c00h
int 21h
main endp
; write string procedure
Writestring proc
push ds ; set ES to DS
pop es
mov di,dx ; let ES:DI point to the string
call Str_length ; get length of string in AX
mov cx,ax ; CX = number of bytes to write
mov ah,40h ; write to file or device
mov bx,1 ; choose standard output
int 21h ; call DOS
ret
Writestring endp
; string lenght procedure
Str_length proc
push cx
push di ; save pointer to string
mov cx,0FFFFh ; set CX to maximum word value
mov al,0 ; scan for null byte
cld ; direction = up
repnz scasb ; compare AL to ES:[DI]
dec di ; back up one position
mov ax,di ; get ending pointer
pop di ; retrieve starting pointer
sub ax,di ; subtract start from end
pop cx
ret ; AX = string length
Str_length endp
; string comparison procedure
compare proc
push si
push di
L10:
inc di
inc si
mov al,[si]
mov al,[di]
loopz L10
pop di
pop si
ret
compare endp
; Readstring procedure :
Readstring proc
push cx
push si
push cx
mov si,dx
A1:
mov ah,1
int 21h
cmp al,0dh
je A2
mov [si],al
inc si
Loop A1
A2:
mov byte ptr [si],0
pop ax
sub ax,cx
pop si
pop cx
ret
Readstring endp
end main
