The INT 21H instruction in the 8086 is a software interrupt to vector 21H. In order for it to be used for input/output, the programming that responds to INT 21H must be present. This is part of the Operating System.
Read the manual, please. If you don't know where the manual is, Ralph Brown's famous interrupt-list might help.
1. syntax error2. int 21h -- assembly statement in x86 to access a MS-DOS service
// declare a function int* function(int, int); or int* (function)(int, int); // declare a pointer to a function int* (*pointer_to_function)(int, int);
code segment assume cs:code,ds:code mov bx,1200h mov cx,[bx] mov ax,01h l1:mul cx dec cl jnz l1 mov[bx+2],ax mov ah,4ch int 21h code ends end
I will explain in the easiest way the difference between the function and recursive function in C language. Simple Answer is argument of the function is differ but in the recursive function it is same:) Explanation: Function int function(int,int)// function declaration main() { int n; ...... ...... n=function(a,b); } int function(int c,int d) { ...... ...... ...... } recursive Function: int recursive(int,int)// recursive Function declaration main() { int n; ..... ..... ..... ..... n=recursive(a,b); } int recursive(int a,int b) { ..... .... .... .... } Carefully see, In the recursive Function the function arguments are same.
start: jmp main option1 db 0ah, 0dh, "1. OPTION 1 $" option2 db 0ah, 0dh, "2. OPTION 2 $" exit db 0ah, 0dh, "3. EXIT $" str1 db 0ah, 0dh, "Press Key: $" x db 0ah, 0dh, "You Pressed option 1 $" y db 0ah, 0dh, "You Pressed option 2 $" z db 0ah, 0dh, "End $" nvl db 0ah, 0dh, "Invalid Option $" one db "1" two db "2" tre db "3" main proc mov ah,09h lea dx,option1 int 21h lea dx,option2 int 21h lea dx,exit int 21h again: mov ah,09h lea dx,str1 int 21h mov ah,01 int 21h mov bl,al cmp bl,"1" je disp1 cmp bl,"2" je disp2 cmp bl,"3" je dispexit cmp al,one jne n cmp al,two jne n cmp al,tre jne n n: mov ah,09h lea dx,nvl int 21h jmp again disp1: mov ah,09h lea dx,x int 21h jmp again disp2: mov ah,09h lea dx,y int 21h jmp again dispexit: mov ah,09h lea dx,z int 21h int 20h main endp end start
Code segment assume cs:code mov si,1500h mov di,1600h mov ax,[si] mov bx,[si+2] add ax,bx mov [di],ax mov [di+2],dx mov ah,4ch int 21h code ends end
.model small .stack .data m db 'the no is odd $' m1 db 'the no is even $' a db 04h b db 02h .code mov ax,@data mov ds,ax mov ah,0 mov al,a mov bl,b div bl cmp ah,00h je l1 mov dx,offset m jmp l l1: mov dx,offset m1 jmp l l: mov ah,09 int 21h mov ah,4ch int 21h end
What is the 4ch lick
.model small .stack .data .code main proc mov al,10 mov bl,15 add al, bl start: mov cx,8 mov bl,'1' test bl,10000000b mov ah,09h int 21h next: shl bl,1 mov dl,1 mov ah,09h int 21h mov ah,4ch int 21h main endp end start end
mov ax, @data ; Initialize data section mov ds, ax mov bx, offset lookup ; Load offset of lookup in bx mov al, key ; key no. in al xlat ; translate byte in al mov bh, al ; al = lookup(key) mov ch, 02h ; Count of digits to be displayed mov cl, 04h ; Count to roll by 4 bits l2: rol bh, cl ; roll bl so that msb comes to lsb mov dl, bh ; load dl with data to be displayed and dl, 0fH ; get only lsb cmp dl, 09 ; check if digit is 0-9 or letter A-F jbe l4 add dl, 07 ; if letter add 37H else only add 30H l4: add dl, 30H mov ah, 02 ; Function 2 under INT 21H (Display character) int 21H dec ch ; Decrement Count jnz l2 mov ah, 4cH ; Terminate Program int 21H end