寄存器冲突的问题(1005)

本文为《汇编语言程序设计》1005小节例程。点击链接…进课程主页。

问题：编程将data段中的字符串转化为大写。

assume cs:code
data segment
             db 'conversation'
data ends
code segment
start:   mov ax,data
         mov ds,ax
         mov si,0
         mov cx,12
         call capital
         mov ax,4c00h
         int 21h
capital: and byte ptr [si],11011111b
         inc si
         loop capital
         ret
code ends
end start

用0作为字符串的结束

assume cs:code
data segment
         db 'conversation',0
         data ends
code segment
start:    mov ax,data
          mov ds,ax
          mov si,0
          call capital
          mov ax,4c00h
          int 21h
capital:  mov cl, [si]
          mov ch, 0
          jcxz ok
          and byte ptr [si], 11011111b
          inc si
      ok: ret
code ends
end start

将以下字符串转为大写

assume cs:code
data segment
         db 'word',0
         db 'unix',0
         db 'wind',0
         db 'good',0
data ends

code segment
start:  mov ax,data
          mov ds,ax
          mov bx,0
          mov cx,4
       s: mov si,bx
          call capital
          add bx,5
          loop s
          mov ax,4c00h
          int 21h
capital: mov cl,[si]
           mov ch,0
           jcxz ok
           and byte ptr [si],11011111b
           inc si
           jmp short capital
     ok: ret
code ends
end start
;程序中cx既用于循环，又用于读取数据——冲突！

解决了寄存器冲突的程序

;正确的程序
assume cs:code
data segment
         db 'word',0
         db 'unix',0
         db 'wind',0
         db 'good',0
data ends

code segment
   start: mov ax,data
          mov ds,ax
          mov bx,0
          mov cx,4
       s: mov si,bx
          call capital
          add bx,5
          loop s
          mov ax,4c00h
          int 21h
 capital: push cx
          push si
  change: mov cl,[si]
          mov ch,0
          jcxz ok
          and byte ptr [si],11011111b
          inc si
          jmp short change
      ok: pop si
          pop cx
          ret
code ends
end start