4

我刚开始学习汇编语言，我已经被困在“在屏幕上显示存储在寄存器中的十进制值”的部分。我正在使用emu8086，任何帮助将不胜感激！:)

.model small  ;Specifies the memory model used for program to identify the size of code and data segments

org 100h      ;allocate 100H memory locations for stack

.data         ;the segment of the memory to declare/initialze the variables

var1 db 0006
var2 db 0002
var3 db 0001


.code           ;start of the code segment

main proc       ;start of the first procedure

mov bl, var1
add bl, var2
add bl, var3


mov ah, 00h   ; display  function here?
mov dl, bl    ; output the bl register's value?  
int 21h

mov ah, 4ch  ;exit DOS function                                           
int 21h

endp         ;end of the first procedure

end main     ;end of the complete assembly program

ret
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2 回答 2

4 
mov ah, 00h   ; display  function here?

不，单字符显示功能在AH=2 /int 21h

由于您的 BL 寄存器仅包含一个很小的值 (9)，因此它所需要的只是：

mov  ah, 02h
mov  dl, bl
add  dl, "0"   ; Integer to single-digit ASCII character
int  21h

如果值变得更大但不超过 99，您可以通过：

mov  al, bl       ; [0,99]
aam               ; divide by 10: quotient in ah, remainder in al (opposite of DIV)
add  ax, "00"
xchg al, ah
mov  dx, ax
mov  ah, 02h
int  21h
mov  dl, dh
int  21h

不使用AAM指令的解决方案：

mov  al, bl       ; [0,99]
cbw               ; Same result as 'mov ah, 0' in this case
mov  dl, 10
div  dl           ; Divides AX by 10: quotient in al, remainder in ah
add  ax, "00"
mov  dx, ax
mov  ah, 02h
int  21h
mov  dl, dh
int  21h
于 2015-12-13T21:48:54.813 回答
2

在 emu8086 中，您可以为此目的使用现成的宏和过程。

例子：

include 'emu8086.inc'       ; Include useful macros and procedures

.model small

.stack

.data

var1 db 6
var2 db 2
var3 db 7

.code

DEFINE_PRINT_NUM         ; Create procedure PRINT_NUM          
DEFINE_PRINT_NUM_UNS     ; Create procedure PRINT_NUM_UNS

crlf proc
    mov ah, 2
    mov dl, 13
    int 21h
    mov dl, 10
    int 21h
    ret
crlf endp

main proc

    mov ax, @data
    mov ds, ax

    ; test output: 54321 & -11215 
    mov ax, 54321
    call PRINT_NUM_UNS   ; Print AX as unsigned number
    call crlf
    mov ax, 54321
    call PRINT_NUM       ; Print AX as signed number
    call crlf

    mov bl, var1
    add bl, var2
    add bl, var3

    mov ax, bx           ; AX contains the number for PRINT_NUM
    xor ah, ah           ; Could contain crap
    call PRINT_NUM
    call crlf

    mov ax, 4c00h
    int 21h

main endp

end main
于 2015-12-13T20:30:44.813 回答