Question

我似乎遇到了一个问题。我自学了用这段代码打印ascii表的所有字符：

.MODEL SMALL
 .STACK 100H

 .DATA
    PROMPT  DB  'The 256 ASCII Characters are : $'

 .CODE
   MAIN PROC
     MOV AX, @DATA                ; initialize DS 
     MOV DS, AX

     LEA DX, PROMPT               ; load and print PROMPT 
     MOV AH, 9
     INT 21H

     MOV CX, 256                  ; initialize CX

     MOV AH, 2                    ; set output function  
     MOV DL, 0                    ; initialize DL with first ASCII character

     @LOOP:                       ; loop label
       INT 21H                    ; print ASCII character

       INC DL                     ; increment DL to next ASCII character
       DEC CX                     ; decrement CX
     JNZ @LOOP                    ; jump to label @LOOP if CX is 0

     MOV AH, 4CH                  ; return control to DOS
     INT 21H
   MAIN ENDP
 END MAIN

然而，我有点过头了，如何在前进后向后打印，然后打印每五个ASCII字符。打印0 ASCII字符，后跟逗号，然后打印255 ASCii字符输出继电器 0（ASCII），255（ASCII） 1（ASCII），254（ASCII）

如果有人愿意为我提供建议或添加我的代码，我将非常感激。

Answer 1

我改变了你的代码很多，我希望它可以帮助你学习，下一个小程序（用EMU8086制作）打印所有ASCII字符（你的原始算法），然后向后打印它们（你的原始算法有一些变化），然后打印每五个字符，最后打印替换它们（char 0，char 255，char 1，char 254等）：

.MODEL SMALL
.STACK 100H

.DATA   

PROMPT  DB  'The 256 ASCII Characters are : $'
PROMPT2 DB  13,10,13,10,'Now backwards : $'
PROMPT3 DB  13,10,13,10,'Now every fifth character : $'
PROMPT4 DB  13,10,13,10,'Now alternated : $'

.CODE
MAIN PROC

 MOV AX, @DATA                ; initialize DS 
 MOV DS, AX

;ONWARDS---------------------------------------------.

 LEA DX, PROMPT               ; load and print PROMPT 
 MOV AH, 9
 INT 21H

 MOV CX, 256                  ; initialize CX

 MOV AH, 2                    ; set output function  
 MOV DL, 0                    ; initialize DL with first ASCII character

 @LOOP:                       ; loop label
   INT 21H                    ; print ASCII character

   INC DL                     ; increment DL to next ASCII character
   DEC CX                     ; decrement CX
 JNZ @LOOP                    ; jump to label @LOOP if CX is 0

;BACKWARDS-----------------------------------------------.

 LEA DX, PROMPT2              ; load and print PROMPT2
 MOV AH, 9
 INT 21H

 MOV CX, 256                  ; initialize CX

 MOV AH, 2                    ; set output function  

 @LOOP2:                       ; loop label
   mov dl, cl ;USE CL AS CHARACTER BECAUSE IT'S GOING BACKWARDS.
   INT 21H                    ; print ASCII character
   DEC CX                     ; decrement CX
 JNZ @LOOP2                    ; jump to label @LOOP if CX is 0

;EVERY FIFTH-----------------------------------------------.

 LEA DX, PROMPT3              ; load and print PROMPT3
 MOV AH, 9
 INT 21H

 MOV CX, 256                  ; initialize CX

 MOV AH, 2                    ; set output function  
 MOV DL, 0                    ; initialize DL with first ASCII character

 @LOOP3:                       ; loop label
   INT 21H                    ; print ASCII character

   add dl, 5 ; increment DL BY 5 to next ASCII character
   sub CX, 5 ; decrement CX BY 5 TOO
   cmp cx, 0 ; IN CASE CX WON'T REACH EXACT ZERO.
   jge @LOOP3                   ; jump to label @LOOP if CX is 0

;ALTERNATED-------------------------------------------------------.

 LEA DX, PROMPT4              ; load and print PROMPT 
 MOV AH, 9
 INT 21H

 MOV CX, 256                  ; initialize CX

 MOV AH, 2                    ; set output function  
 MOV bl, 0                    ; initialize DL with first ASCII character
 MOV bh, 255                  ; initialize DL with last ASCII character

 @LOOP4:
   mov dl, bl  ;FIRST CHARS.
   INT 21H                    ; print ASCII character
   mov dl, bh ;LAST CHARS.
   INT 21H                    ; print ASCII character

   INC bl                     ; increment DL to next ASCII character
   dec bh
   sub CX, 2 ; decrement CX BY 5 TOO
   cmp cx, 0 ; IN CASE CX WON'T REACH EXACT ZERO.
   jge @LOOP4                   ; jump to label @LOOP if CX is 0



;WAIT FOR ANY KEY.    
  mov  ah, 7
  int  21h


 MOV AH, 4CH                  ; return control to DOS
 INT 21H

MAIN ENDP
END MAIN

Answer 2

您编写的代码不会显示所有256个ASCII字符。 DOS不会显示ASCII代码7,8,9,10和13的字符.DOS将执行Beep，Backspace，Tab，Linefeed和Carriage返回。您可以使用BIOS功能0Ah输出任何字符，但您必须自己前进光标我写了一个小程序改编你的程序来证明你没有必要在CX中使用explicite计数器，因为增加DL寄存器已经提供了足够的标志信息。请注意，DOS终止功能4Ch也使用AL作为参数。

.MODEL SMALL
.STACK 100H

.DATA
  PROMPT  DB  'The ASCII Characters from 32 to 255 are : $'

.CODE
  MAIN PROC
    MOV AX, @DATA              ; initialize DS 
    MOV DS, AX
    LEA DX, PROMPT             ; load and print PROMPT 
    MOV AH, 9
    INT 21H

    ; Here you can insert my second code snippet

    MOV DL, 32                 ; initialize DL with first ASCII character
    @LOOP:                     ; loop label
    MOV AH, 2                  ; set output function
    INT 21H                    ; print ASCII character
    INC DL                     ; increment DL to next ASCII character
    JNZ @LOOP                  ; jump to label @LOOP if DL is 0

    MOV AX, 4C00H              ; return control to DOS
    INT 21H
  MAIN ENDP
END MAIN

以下是显示0到31之间ASCII码的代码。

Again:
 mov cx, 1
 mov bx, 0007h
 mov al, dl
 mov ah, 0Ah
 int 10h      ;WriteCharacter
 push dx
 mov ah, 03h
 int 10h      ;GetCursorPostion
 inc dl       ;Advance column
 mov ah, 02h
 int 10h      ;SetCursorPosition
 pop dx
 inc dl       ;Next character
 cmp dl, 32
 jb Again

汇编语言：自学，遇到一点麻烦

2 个答案: