1.思路:入栈-出栈存放
assume cs:code
code segment
dw 0123H,0456H,0789H,0ABCH,0DEFH,0FEDH,0CBAH,0987H
dw 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
start: mov ax,cs
mov ss,ax
mov sp,30H
mov bx,0
mov cx,8
s: push cs:[bx]
add bx,2
loop s
mov bx,0
mov cx,8
s0: pop cs:[bx]
add bx,2
loop s0
mov ax,4c00H
int 21H
code ends
end start