;将内存ffff:0~ffff:b单元中的数据复制到0:200~0:20b单元中
;分析  0:200~0:20b可以转化为0200:0~0200:b

assume cs:code
code segment
	;让bx指向0
	mov bx,0h
	mov cx,0ch
	
	mov ax,0ffffh
	mov ds,ax
	
	mov ax,0020h
	mov es,ax
	
s:	mov al,ds:[bx]
	mov es:[bx],al
	inc bx
	loop s
	
	mov ax,4c00h
	int 21h
code ends
end