.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
替代字符串直接输出到视频缓冲器而不使用软件中断：
;-----------------------
; 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,"$")
; Hint: this directly output to the screen do not touch or move the cursor
; but feel free to modify..
returns nothing
- outputs character string to STDOUT up to "$"
- backspace is treated as non-destructive
- if Ctrl-Break is detected, INT 23 is executed
参考：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
; 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
; good example of unlimited num print
.model small
.stack 100h
.data
number word 6432
string db 10 dup('$')
.code
main proc
mov ax,@data
mov ds,ax
mov ax,number
mov bx ,10
mov cx,0
l1:
mov dx,0
div bx
add dx,48
push dx
inc cx
cmp ax,0
jne l1
mov bx ,offset string
l2:
pop dx
mov [bx],dx
inc bx
loop l2
mov ah,09
mov dx,offset string
int 21h
mov ax,4c00h
int 21h
main endp
end main
10 个回复
漂截嘘
服务2吗？是要打印的字符。要打印整数值，您必须编写一个循环来将整数分解为单个字符。如果你可以用十六进制打印值，这非常简单。 如果您不能依赖DOS服务，您也可以使用BIOS并将设置为或。揽芳僵迷仇
拟蓬
.code mov ax,@DATA ; get the address of the data segment mov ds,ax ; store the address in the data segment register ;----------------------- mov eax,0FFFFFFFFh ; 32 bit value (0 - FFFFFFFF) for example ;----------------------- ; convert the value in EAX to hexadecimal ASCIIs ;----------------------- mov di,OFFSET ASCII ; get the offset address mov cl,8 ; number of ASCII P1: rol eax,4 ; 1 Nibble (start with highest byte) mov bl,al and bl,0Fh ; only low-Nibble add bl,30h ; convert to ASCII cmp bl,39h ; above 9? jna short P2 add bl,7 ; "A" to "F" P2: mov [di],bl ; store ASCII in buffer inc di ; increase target address dec cl ; decrease loop counter jnz P1 ; jump if cl is not equal 0 (zeroflag is not set) ;----------------------- ; Print string ;----------------------- mov dx,OFFSET ASCII ; DOS 1+ WRITE STRING TO STANDARD OUTPUT mov ah,9 ; DS:DX->'$'-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,"$" ; buffer for ASCII string;----------------------- ; Print string ;----------------------- mov ax,0B800h ; segment address of textmode video buffer mov es,ax ; store address in extra segment register mov si,OFFSET ASCII ; get the offset address of the string ; using a fixed target address for example (screen page 0) ; Position`on screen = (Line_number*80*2) + (Row_number*2) mov di,(10*80*2)+(10*2) mov cl,8 ; number of ASCII cld ; clear direction flag P3: lodsb ; get the ASCII from the address in DS:SI + increase si stosb ; write ASCII directly to the screen using ES:DI + increase di inc di ; step over attribut byte dec cl ; decrease counter jnz P3 ; repeat (print only 8 ASCII, not used bytes are: 0Dh,0Ah,"$") ; Hint: this directly output to the screen do not touch or move the cursor ; but feel free to modify..疾很毋悲
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苏髓骗撩
撕吠
参考：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掀辟髓观粟
; 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峨躬坎抬焚
硕歌沙
; 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踩什不