以8086汇编语言存储和检索内存中的数字

4 回答

• -1

  上面的代码有两个错误

  • $ 符号必须在引号中，否则可以生成错误 .

  • 在输入后，我们必须在 dl 中 al 的值 al ，以在屏幕上打印输入的值 .

  回复于 2024-04-28T18:03:51+08:00
• 0

  代码中有一些错误 . 要使用DOS函数0Ah（BUFFERED INPUT）输入字符串，我们必须使用输入缓冲区 . 此DOS输入缓冲区的格式如下所示（表01344） .

  Ralph Browns x86 / MSDOS中断列表：
  http://www.pobox.com/~ralf/files.html
  ftp://ftp.cs.cmu.edu/afs/cs.cmu.edu/user/ralf/pub/

  RBIL->inter61b.zip->INTERRUP.F
  --------D-2109-------------------------------
  INT 21 - DOS 1+ - WRITE STRING TO STANDARD OUTPUT
          AH = 09h
      DS:DX -> '$'-terminated string
  Return: AL = 24h (the '$' terminating the string, despite official docs which
          state that nothing is returned) (at least DOS 2.1-7.0 and NWDOS)
  Notes:  ^C/^Break are checked, and INT 23 is called if either pressed
      standard output is always the screen under DOS 1.x, but may be
        redirected under DOS 2+
      under the FlashTek X-32 DOS extender, the pointer is in DS:EDX
  SeeAlso: AH=02h,AH=06h"OUTPUT"
  --------D-210A-------------------------------
  INT 21 - DOS 1+ - BUFFERED INPUT
      AH = 0Ah
      DS:DX -> buffer (see #01344)
  Return: buffer filled with user input
  Notes:  ^C/^Break are checked, and INT 23 is called if either detected
      reads from standard input, which may be redirected under DOS 2+
      if the maximum buffer size (see #01344) is set to 00h, this call returns
        immediately without reading any input
  SeeAlso: AH=0Ch,INT 2F/AX=4810h
  
  Format of DOS input buffer:
  Offset  Size    Description (Table 01344)
   00h    BYTE    maximum characters buffer can hold
   01h    BYTE    (call) number of chars from last input which may be recalled
          (ret) number of characters actually read, excluding CR
   02h  N BYTEs   actual characters read, including the final carriage ret
  

  为了使用DOS函数09h显示ASCII字符串（来自用户输入），我们必须将缓冲区2的偏移地址输入DX，另外我们必须在调用函数之前在字符串后面放置一个“$” . 可以通过添加上次输入的字符数来计算在字符串后存储“$”的地址 .

  BUFFER DB 1, ?, " ", 0Dh, "$" ; Input buffer (only for one ASCII)
  
  mov bh, 0                     ; clear high byte of BX
  lea si, BUFFER+1              ; get the offset address+1 of the buffer
  mov bl, [si]                  ; get the number of byte from the last input
  mov BYTE PTR[si+bx+1], "$"    ; store "$" after the end of the string+CR
  

  如果输入的字节数大于1，则上面的示例更有意义，因此用户可能没有将最大数量的ASCII填充到缓冲区中，因此我们不知道字符串的结尾，所以我们必须从最后一个输入中获取ASCII数 . （但总而言之，我们必须确保缓冲区的尺寸足够大 . ）

  lea dx, BUFFER+2
  mov ah, 9
  int 21h
  

  短剑

  回复于 2024-04-28T18:03:51+08:00
• 0

  ; multi-segment executable file template.
  
  data segment
      ; add your data here!
      pkey db "press any key...$"
  ends
  
  stack segment
      dw   128  dup(0)
  ends
  
  code segment
  start:
  ; set segment registers:
      mov ax, data
      mov ds, ax
      mov es, ax
  
      ; add your code here
      mov cx,4
      input:
      mov ah,1
      int 21h
      push ax
      loop input 
  
      mov dx,13d
      mov ah,2
      int 21h
  
      mov dx,10d
      mov ah,2
      int 21h
  
      mov cx,4  
  
      output:
      pop bx
      mov dl,bl
      mov ah,2
      int 21h
      loop output
  
      exit:        
      lea dx, pkey
      mov ah, 9
      int 21h        ; output string at ds:dx
  
      ; wait for any key....    
      mov ah, 1
      int 21h
  
      mov ax, 4c00h ; exit to operating system.
      int 21h    
  ends
  
  end start ; set entry point and stop the assembler.
  

  这是8086编译器中堆栈的一个例子 . 谢谢

  回复于 2024-04-28T18:03:51+08:00
• 0

  .model small
  .stack 32h
  .data 
  Message db " enter a no:" ,'$'
  .code
  Mov ax,@data
  Mov ds,ax
  Mov dx,offset message; display message
  Mov ah,01h; enter no
  Int 21h
  Mov dl,al
  Mov ah,02h; print no on screen
  Int 21h
  Mov ah,4ch
  Int 21h
  End
  

  回复于 2024-04-28T18:03:51+08:00