已知21000h处字单元的内容为 BE00H,对于如下程序执行后,内存中字单元2000:1005中的内容为0000
以下数据均为16进制,为了看得清楚点,不加H了
mov ax,2000h
mov ds,ax
mov bx,1000h ;ds:[bx]=2000:1000
mov ax,[bx] ;2000:1000 00
;2000:1001 be
;ax=ds:[bx]=be00,ah=be,al=00
inc bx ;bx=1001
inc bx ;bx=1002
mov [bx],ax ;2000:1000 00
;2000:1001 be
;2000:1002 00
;2000:1003 be
inc bx ;bx=1003
inc bx ;bx=1004
mov [bx],ax ;2000:1000 00
;2000:1001 be
;2000:1002 00
;2000:1003 be
;2000:1004 00
;2000:1005 be
inc bx ;bx=1005
mov [bx],al ;2000:1000 00
;2000:1001 be
;2000:1002 00
;2000:1003 be
;2000:1004 00
;2000:1005 00
inc bx ;bx=1006
mov [bx],al ;2000:1000 00
;2000:1001 be
;2000:1002 00
;2000:1003 be
;2000:1004 00
;2000:1005 00
;2000:1006 00
每一步操作后内存单元的内容的我都写了,LZ应该能看清楚了
你最初的ax值推算错误 |