用汇编语言打印一个数字？

Question

mov al,10 
add al,15 

如何打印'al'的值？

Answer 1

你试过int 21h service 2？ DL是要打印的字符。

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

要打印的整数值，你必须写一个循环来分解整数单个字符。如果你可以用十六进制打印这个值，这很简单。

如果您不能依靠DOS服务，您可能也可以使用BIOS int 10h和AL设置为0Eh或0Ah。

Answer 2

调用WinAPI函数（如果你正在开发win应用程序）

Answer 3

汇编语言没有直接打印任何东西的方法。您的汇编程序可能会或可能不会提供一个提供这种设施的库，否则您必须自己编写它，这将是一个相当复杂的功能。您还必须决定在哪里打印东西 - 在窗口中，在打印机上？在汇编中，这些都不是为你完成的。

Answer 4

虽然Win16是否支持该特定方法是为了让别人回答，但您可能会有一些运气调用Win32 API的MessageBoxA。

Answer 5

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 6

mov al,3 ;print ♥ 


mov dl,al 

;call print service(2) to print from dl 


mov ah,2 
int 21h 

;return to DOS 


mov ah,76 ;76 = 4ch 

int 21h ;call interrupt 

Answer 7

mov al,4 ; input 

; print AL 
    or al,30h ; Important! =>Convert an integer from 0-9 to its ASCII encoding 
    mov i,al 

    MOV AH, 2 ; 
    MOV DL, i ; Print Character. 
    INT 21H ; 

Answer 8

;  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 

Answer 9

存储在EAX与十六进制输出（用于80386+）
（在64位操作系统使用DOSBOX）

.code 
    mov ax,@DATA  ; get the address of the data segment 
    mov ds,ax   ; store the address in the data segment register 
;----------------------- 
    mov eax,0FFFFFFFFh ; 32 bit value (0 - FFFFFFFF) for example 
;----------------------- 
; convert the value in EAX to hexadecimal ASCIIs 
;----------------------- 
    mov di,OFFSET ASCII ; get the offset address 
    mov cl,8   ; number of ASCII 
P1: rol eax,4   ; 1 Nibble (start with highest byte) 
    mov bl,al 
    and bl,0Fh   ; only low-Nibble 
    add bl,30h   ; convert to ASCII 
    cmp bl,39h   ; above 9? 
    jna short P2 
    add bl,7   ; "A" to "F" 
P2: mov [di],bl   ; store ASCII in buffer 
    inc di    ; increase target address 
    dec cl    ; decrease loop counter 
    jnz P1    ; jump if cl is not equal 0 (zeroflag is not set) 
;----------------------- 
; Print string 
;----------------------- 
    mov dx,OFFSET ASCII ; DOS 1+ WRITE STRING TO STANDARD OUTPUT 
    mov ah,9   ; DS:DX->'$'-terminated string 
    int 21h    ; maybe redirected under DOS 2+ for output to file 
         ; (using pipe character">") or output to printer 

    ; terminate program... 

.data 
ASCII DB "00000000",0Dh,0Ah,"$" ; buffer for ASCII string 

替代

DOS打印32位值字符串直接输出到videobuffer而不使用软件中断：

;----------------------- 
; Print string 
;----------------------- 
    mov ax,0B800h  ; segment address of textmode video buffer 
    mov es,ax   ; store address in extra segment register 

    mov si,OFFSET ASCII ; get the offset address of the string 

; using a fixed target address for example (screen page 0) 
; Position`on screen = (Line_number*80*2) + (Row_number*2) 

    mov di,(10*80*2)+(10*2) 
    mov cl,8   ; number of ASCII 
    cld     ; clear direction flag 

P3: lodsb ; get the ASCII from the address in DS:SI + increase si 
    stosb ; write ASCII directly to the screen using ES:DI + increase di 
    inc di ; step over attribut byte 
    dec cl ; decrease counter 
    jnz P3 ; repeat (print only 8 ASCII, not used bytes are: 0Dh,0Ah,"$") 

; Hint: this directly output to the screen do not touch or move the cursor 
; but feel free to modify.. 

Answer 10

AH = 09 DS：DX =指向字符串 “$”

returns nothing 


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

裁判结束：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