用 SHR 指令除以 8

Question

Write a complete assembly program to read 8 digits, each digit separated by a single space from the keyboard (use single key input function). Convert them from character to numbers and calculate the average, lowest and highest score and display them on the screen.

Hint: Subtract 30h from each character to get actual digit. Before display, add 30h to each digit. For division operation, use appropriate shift instruction.

平均显示错误答案！我哪里做错了？请帮助我理解 SRH 指令的除法。

Include emu8086.inc
.model small
.stack 100h  
.data

 
.code
 mov bh,0
 mov bl,9  
 mov ah,1
 mov dh,0
 mov cx,0    
 
 input:    
 int 21H   
 putc 20h
 sub al,30h
 inc cx
 add dh,al
 cmp al,bh
 JA _max
 cmp al,bl
 JB _min  
 cmp cx,8
 JB input
 
 
 print:
 mov ah,2 
 printn
 mov dl,bh 
 add dl,30h
 printn "Max: "
 int 21h
   
 printn
 mov dl,bl
 add dl,30h
 printn "Min: "
 int 21h 
 
 AND dh, 0FH
 
 mov Cl,3
 shr dh,cl
 
 or dh,30H
 
 printn
 mov dl,dh
 printn "Avg: "
 int 21h 
 
 exit:
 mov ah,4ch
 int 21h
 
 _max:
 mov bh,al 
 cmp al,bl
 JB _min  
 cmp cx,8
 jae print
 jb input
 
 _min:
 mov bl,al
 cmp cx,10
 jae print
 jb input

Answer 1

无符号数除以 8 等同于将数字右移 3 次。在8086编程中我们需要在执行shr指令之前将移位计数放入CL寄存器中。因为您已将总和（您想要除法）放在 DH 寄存器中，并且您想要显示 DL 寄存器中的平均值（除法结果）（由 DOS.PrintChar 函数），只需 2 条指令即可完成此操作：

mov cl, 11   ; 11 = 3 (is divide DH by 8) + 8 (is move DH to DL)
shr dx, cl

_min:
mov bl,al
cmp cx,10

就像 _max 一样，这个 _min 应该使用 cmp cx, 8.

更好的是，您可能不应该重复那么多代码！下面的代码使用相反的条件并将所有内容很好地放在一起：

.code
  mov bx, 9    ; BH=0 is Max, BL=9 is Min
  xor dx, dx   ; DH=0 is Sum, DL=0 is Count
input:
  mov ah, 01h  ; DOS.GetCharacter
  int 21h  
  putc 20h
  sub al, 30h
  add dh, al   ; Add to Sum DH
  cmp al, bh
  jbe NotMax
  mov bh, al   ; Set new max BH
NotMax:
  cmp al, bl
  jae NotMin
  mov bl, al   ; Set new Min BL
NotMin:
  inc dx       ; Increment Count DL
  cmp dl, 8
  jb input

不要忘记在您的程序中添加一些有意义的注释，以便人们可以快速理解您所写的内容。

用 SHR 指令除以 8

Division by 8 with SHR instruction

assembly

dos

x86-16

emu8086