|
主题 : : 实验五(5)(6)自己做的程序,请高手鉴定 [待解决] |
回复[ 11次 ]
点击[ 603次 ] | |
|
|
|
|
[帖 主] [ 发表时间:2011-02-12 17:23 ] | |
荣誉值:0
信誉值:2
注册日期:2011-02-09 21:07 |
(5)程序:
assume cs:code
a segment
db 1,2,3,4,5,6,7,8
a ends
b segment
db 1,2,3,4,5,6,7,8
b ends
c segment
db 0,0,0,0,0,0,0,0
c ends
code segment
start:
mov bx,0
mov cx,8
s:
;read a[bx]
mov ax,a
mov ds,ax
mov dl,ds:[bx]
;add b[bx]
mov ax,b
mov ds,ax
add dl,ds:[bx]
;write c[bx]
mov ax,c
mov ds,ax
mov ds:[bx],dl
inc bx
loop s
mov ax,4c00h
int 21h
code ends
end start
(6)程序
assume cs:code
a segment
dw 1,2,3,4,5,6,7,8,9,0ah,0bh,0ch,0dh,0eh,0fh,0ffh
a ends
b segment
dw 0,0,0,0,0,0,0,0
b ends
code segment
start:
mov ax,a
mov ds,ax
mov ax,b
mov ss,ax
mov sp,16
mov cx,8
s:
push [bx]
add bx,2
loop s
mov ax,4c00h
int 21h
code ends
end start | | |
|
|
|
|
[第1楼] [ 回复时间:2011-02-13 14:41 ] | |
荣誉值:268
信誉值:12
注册日期:2010-06-18 22:19 |
第一哦用计算好偏移地址就可以了。
mov ax,a
mov ds,ax
mov bx,0
mov cx,8
s:
mov al,[bx]
add al,[bx+8]
mov type ptr [bx+16],al
inc bx
loop s | | |
|