st.asm

用汇编语言做的一些小程序例题

stack segment para stack 'stack'
    db 80 dup (?)
stack ends
;-----------------------------------------
dseg segment para 'data'

crlf     db 13,10,'$'

prompt     db   ' , ','$'


highmsg     db  ' The highest score is: ','$'
lowmsg      db  ' The lowest score is: ','$'
startmsg    db  ' Pleae input score: ','$'
dismsg      db  ' The Scores is : ','$'


dsc    db   0,0,0,0,0,0,0

outputmsg   db  ' The last score is: ','$'

score  db   0,0,0,0,0,0,0,0,0,0,0,0,0,0

dseg ends
;------------------------------------------

cseg segment para 'code'
begin proc  far
      assume cs:cseg,ds:dseg,ss:stack
      push ds
      sub ax,ax
      push ax
      mov ax,dseg
      mov ds,ax
      
;========start
      lea  dx,startmsg
      mov  ah,09h
      int  21h
      lea  dx,crlf
      int  21h

      call  input
      lea  dx,crlf
      mov  ah,09h
      int  21h

      call  changsc
      call  index

;========display high & low
;     high
      mov  al,dsc[6]     
      call  btobcd
      mov  bl,ah
      lea  dx,highmsg
      mov  ah,09h
      int  21h
       
      mov  dl,al
      mov  ah,02
      int  21h
      mov  dl,bl
      mov  ah,02
      int  21h
      lea  dx,crlf
      mov  ah,09h
      int  21h
;====low
      mov  al,dsc     
      call  btobcd
      mov  bl,ah
      lea  dx,lowmsg
      mov  ah,09h
      int  21h
       
      mov  dl,al
      mov  ah,02
      int  21h
      mov  dl,bl
      mov  ah,02
      int  21h
      lea  dx,crlf
      mov  ah,09h
      int  21h 
;=======average
      mov  cx,5
      lea  bx,dsc
      inc  bx
      mov  al,0
ave:
      add  al,[bx]
      inc  bx
loop  ave
;======display the last score
      sub  ah,ah
      mov  bl,5
      div  bl
      call  btobcd
      mov  bl,ah
      lea  dx,outputmsg
      mov  ah,09h
      int  21h
       
      mov  dl,al
      mov  ah,02
      int  21h
      mov  dl,bl
      mov  ah,02
      int  21h
      lea  dx,crlf
      int  21h

ret
begin endp
;=====================================================
input proc near 
  lea  dx,prompt
  lea  bx,score
  mov  cx,07
rev:
  mov  ah,1
  int  21h
  mov  [bx+1],al    ; 读入一个字节（十位数）
  int  21h
  mov  [bx],al     ;读入一个字节（个位数）
  add  bx,2
     
  mov  ah,09     ;输出一个分隔符逗号
  int  21h
loop  rev
  lea  dx,crlf
  mov  ah,09
  int  21h

ret
input  endp

tran proc near    ;将BCD码转换为二进制
  and  ax,0f0fh
  mov  bl,ah
  mov  cl,3
  shl  ah,cl
  mov  cl,1
  shl  bl,cl
  add  ah,bl
  add  al,ah
ret
tran endp

changsc proc near  ;将七个数转换为二进制
  lea  di,dsc
  lea  bx,score
  mov  cx,07
loop1:
  mov  al,[bx]
  inc  bx
  mov  ah,[bx]
  inc  bx
  call tran
  mov  [di],al
  inc  di
loop  loop1
ret
changsc  endp

index proc near  ;对DSC由小到大排序，用选择法
  mov cx,06
  lea bx,dsc
loop2:
  mov  al,[bx]
  mov  di,bx
  push  cx
  push  bx
loop3:
  inc  bx
  cmp  al,[bx]
  jle  nochg
  mov  al,[bx]
  mov  di,bx
nochg:
loop loop3
  pop  bx
  pop  cx
  cmp  di,bx
  je   noexch
  mov  ah,[bx]
  mov  [di],ah
  mov  [bx],al
noexch:
  inc  bx
loop  loop2
ret
index endp

btobcd proc near  ;将二进制转换为非压缩BCD码
   sub  ah,ah
   mov  dl,10
   div  dl
   add  al,30h     ;为十位数
   add  ah,30h     ;为个位数
ret
btobcd  endp

cseg  ends

     end  begin