assume cs:code
code segment
start:mov ax, 2000h
mov ds, ax
mov bx, 0
s:mov cl, [bx] ;如2000:0003内存单元中为0 ,cl=0
mov ch, 0
inc cx ;cx=0001
inc bx ;加1 2000:0004
loop s ;cx=cx-1
ok:dec bx ;上面加了1这里就减1 bx=0003
mov dx, bx ;DS:偏移地址 dx=0003
mov ax, 4c00h
int 21h
code ends
end start |