这一题逻辑很简单,请看代码:
assume cs:code
code segment
mov ax,20h
mov ds,ax
mov bx,0
mov cx,64
s:mov [bx],bl
inc bx
loop s
mov ax,4c00h
int 21h
code ends
end
[2018-05-07 14:09] 实验4(2)