Printing out a number in assembly language?

Question

mov al,10
add al,15

How do I print the value of \'al\'?

Answer 1

Call WinAPI function (if u are developing win-application)

Answer 2

Have you tried int 21h service 2? DL is the character to print.

mov dl,'A' ; print 'A'
mov ah,2
int 21h

To print the integer value, you'll have to write a loop to decompose the integer to individual characters. If you're okay with printing the value in hex, this is pretty trivial.

If you can't rely on DOS services, you might also be able to use the BIOS int 10h with AL set to 0Eh or 0Ah.

Answer 3

PRINT_SUM PROC NEAR
 CMP AL, 0
 JNE PRINT_AX
 PUSH AX
 MOV AL, '0'
 MOV AH, 0EH
 INT 10H
 POP AX
 RET 
    PRINT_AX:    
 PUSHA
 MOV AH, 0
 CMP AX, 0
 JE PN_DONE
 MOV DL, 10
 DIV DL    
 CALL PRINT_AX
 MOV AL, AH
 ADD AL, 30H
 MOV AH, 0EH
 INT 10H    
    PN_DONE:
 POPA  
 RET  
PRINT_SUM ENDP

Answer 4

AH = 09 DS:DX = pointer to string ending in "$"

returns nothing


- outputs character string to STDOUT up to "$"
- backspace is treated as non-destructive
- if Ctrl-Break is detected, INT 23 is executed

ref: http://stanislavs.org/helppc/int_21-9.html

---

.data  

string db 2 dup(' ')

.code  
mov ax,@data  
mov ds,ax

mov al,10  
add al,15  
mov si,offset string+1  
mov bl,10  
div bl  
add ah,48  
mov [si],ah  
dec si  
div bl  
add ah,48  
mov [si],ah  

mov ah,9  
mov dx,string  
int 21h

Answer 5

Assuming you are writing a bootloader or other application that has access to the BIOS, here is a rough sketch of what you can do:

• Isolate the first digit of the hex byte
• If it is greater than 9 (i.e. 0x0A to 0x0F), subtract 10 from it (scaling it down to 0 to 5), and add 'A' (0x41).
• If it is less than or equal to 9 (i.e. 0x00 to 0x09), add '0' to it.
• Repeat this with the next hex digit.

Here is my implementation of this:

; Prints AL in hex.
printhexb:
    push ax
    shr al, 0x04
    call print_nibble
    pop ax
    and al, 0x0F
    call print_nibble
    ret
print_nibble:
    cmp al, 0x09
    jg .letter
    add al, 0x30
    mov ah, 0x0E
    int 0x10
    ret
.letter:
    add al, 0x37
    mov ah, 0x0E
    int 0x10
    ret   

Answer 6

;        good example of      unlimited num print

.model small

.stack 100h

.data

number word 6432

string db 10 dup('$')

.code


main proc

mov ax,@data 

mov ds,ax



mov ax,number

mov bx ,10

mov cx,0

l1:

mov dx,0

div bx

add dx,48

push dx

inc cx

cmp ax,0

jne l1


mov bx ,offset string 

l2:

pop dx           

mov [bx],dx

inc bx



loop l2



mov ah,09

mov dx,offset string

int 21h
mov ax,4c00h

int 21h


main endp

end main