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








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