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