Một số Ví dụ Lập trình Hợp ngứ (Assembly)

; Ví dụ 1: In ra màn hình : 'CHAO CAC BAN!'

.MODEL TINY
.CODE

ORG 100H

START:
JMP LoiChao

CRLF DB 13,10,'$'
CHAO DB 'CHAO CAC BAN! $'

LoiChao:

MAIN PROC

;Xuong dong moi
MOV AH,9
LEA DX,CRLF
INT 21H

;Hien thi loi chao
MOV AH,9
LEA DX,CHAO
INT 21H

;Xuong dong moi
MOV AH,9
LEA DX,CRLF
INT 21H

;Tro ve DOS
INT 4cH

MAIN ENDP

END START

;------------------------------------------------------------------------------------------------------------

; Ví dụ 2: Tính tổng 2 số nhập vào từ bàn phím

.MODEL SMALL
.DATA
TB1 DB 'NHAP SO THU NHAT: $'
TB2 DB 'NHAP SO THU HAI: $'
TB3 DB 'KET QUA: $'
A DB ?
B DB ?
.CODE
MAIN PROC


;IN RA XAU1
MOV AX,@DATA
MOV DS,AX
LEA DX,TB1
MOV AH,9
INT 21H


;NHAP SO THU NHAT
MOV AH,1
INT 21H
MOV BL,AL
SUB BL,30H
MOV A,BL


;XUONG DONG VE DAU DONG
MOV AH,2
MOV DL,0DH
INT 21H
MOV DL,0AH
INT 21H

;IN RA XAU2
MOV AX,@DATA
MOV DS,AX
LEA DX,TB2
MOV AH,9
INT 21H


;NHAP SO THU HAI
MOV AH,1
INT 21H
MOV BL,AL
SUB BL,30H
MOV B,BL


;TINH TONG
MOV BL,B
ADD BL,A


;XUONG DONG VE DAU DONG
MOV AH,2
MOV DL,0DH
INT 21H
MOV DL,0AH
INT 21H

;IN XAU 3
MOV AX,@DATA
MOV DS,AX
LEA DX,TB3
MOV AH,9
INT 21H

MOV AH,2
;SUB BL,30H
MOV DL,BL
INT 21H

;KET THUC
MOV AH,4CH
INT 21H


END MAIN

;-------------------------------------------------------------------------------------------------

; Ví dụ 3: In ra màn hình Xâu nghịch đảo

org 100h

jmp start

; tao xau nghich dao

string1 db ' toi khong mo hoa $'

start: lea bx, string1

mov si, bx

next_byte: cmp [si], 'a'
je found_the_end
inc si
jmp next_byte

found_the_end: dec si

; now bx points to beginning,
; and si points to the end of string.

; do the swapping:

do_reverse: cmp bx, si
jae done

mov al, [bx]
mov ah, [si]

mov [si], al
mov [bx], ah

inc bx
dec si
jmp do_reverse

; reverse complete, print out:
done: lea dx, string1
mov ah, 09h
int 21h

; wait for any key press....
mov ah, 0
int 16h

ret

;-------------------------------------------------------------------------------------------------------------

; Ví dụ 4: Đổi số ra xâu

org 100h

jmp start

; text data:
msg1 db 0Dh,0Ah, " Nhap vao so nguyen bat ky tu -32768 -> 65535, nhan so O de dung: $"
msg2 db 0Dh,0Ah, " Dang so thap phan : $"

buffer db 7,?, 5 dup (0), 0, 0

;
binary dw ?

start:

; xuat dong thong bao 1:
mov dx, offset msg1
mov ah, 9
int 21h

; nhap chuoi tu ban phim:
mov dx, offset buffer
mov ah, 0ah
int 21h

; chan chan chuoi la so 0 thi dung chuong trinh:
mov bx, 0
mov bl, buffer[1]
mov buffer[bx+2], 0

lea si, buffer + 2 .

call tobin ; goi thu tuc doi chuoi thanh nhi phan.

mov binary, cx

jcxz stop ; dung lai neu cx = 0

; xuat thong bao thu 2:
mov dx, offset msg2
mov ah, 9
int 21h

mov ax,cx
call outdec

jmp start ; lap lai tu dau

stop:

ret ;

; Thu tuc doi string thanh so nhi phan co dau ('-').
; ket qua luu vao thanh ghi cx

tobin proc near
push dx
push ax
push si

jmp process

;==== local variables ====
make_minus db ? ; used as a flag.
ten dw 10 ; used as multiplier.
;=========================

process:

; reset the accumulator:
mov cx, 0

; reset flag:
mov cs:make_minus, 0

next_digit:

; read char to al and
; point to next byte:
mov al, [si]
inc si

; check for end of string:
cmp al, 0 ; end of string?
jne not_end
jmp stop_input
not_end:

; check for minus:
cmp al, '-'
jne ok_digit
mov cs:make_minus, 1 ; set flag!
jmp next_digit

ok_digit:

; multiply cx by 10 (first time the result is zero)
push ax
mov ax, cx
mul cs:ten ; dx:ax = ax*10
mov cx, ax
pop ax

; it is assumed that dx is zero - overflow not checked!

; convert from ascii code:
sub al, 30h

; add al to cx:
mov ah, 0
mov dx, cx ; backup, in case the result will be too big.
add cx, ax

; add - overflow not checked!

jmp next_digit

stop_input:

; check flag, if string number had '-'
; make sure the result is negative:
cmp cs:make_minus, 0
je not_minus
neg cx

not_minus:

pop si
pop ax
pop dx
ret
tobin endp

; Thu tuc nay xuat ra man hinh chuoi so .
; ke ca dau ('-') tu thanh ghi ax .

outdec proc near

mov cx, ax

push ax
push bx
push cx
push dx

or ax,ax ; kiem tra ax<0?
jge @end_if1

push ax
mov dl,'-'
mov ah,2
int 21h
pop ax
neg ax ; ax = -ax
@end_if1:
xor cx,cx
mov bx,10d ; bx chua so chia

@repeat1:
xor dx,dx
div bx
push dx
inc cx
or ax,ax
jne @repeat1
mov ah,2

@print_loop:
pop dx
or dl,30h
int 21h
loop @print_loop ;lap lai cho den khi xong

pop dx ; Phuc hoi lai cac thanh ghi
pop cx
pop bx
pop ax
ret
outdec endp