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主题 : : 检测点9.2这样做好吗?希望你们能告诉我! [待解决] |
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[帖 主]
[ 发表时间:2008-12-17 17:39 ]
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注册日期:2007-11-28 19:42 |
assume cs:code
code segment
start: mov ax,2000h
mov ds,ax
mov bx,0
s:
mov cl,[bx]
mov ch,0h
jcxz ok
inc bx
jmp short s
ok:mov dx,bx
mov ax,4c00h
int 21h
code ends
end start | | |
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[第1楼]
[ 回复时间:2008-12-18 15:18 ]
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注册日期:2008-01-01 17:48 |
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[第2楼]
[ 回复时间:2009-03-01 21:43 ]
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注册日期:2008-12-18 16:07 |
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[第3楼]
[ 回复时间:2009-03-19 21:39 ]
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注册日期:2008-09-10 11:50 |
assume cs:code
code segment
start:
mov ax,2000h
mov ds,ax
mov bx,0
s:
mov ch,0
mov cl,ds:[bx]
add bx,1
jcxz ok
jmp short s
ok:
mov dx,bx
mov ax,4c00h
int 21h
code ends
end start
我是这样做的。 | | |
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[第4楼]
[ 回复时间:2009-03-19 21:41 ]
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注册日期:2008-09-10 11:50 |
assume cs:code
code segment
start:
mov ax,2000h
mov ds,ax
mov bx,0
s:
mov ch,0
mov cl,[bx]
jcxz ok
add bx,1
jmp short s
ok:
mov dx,bx
mov ax,4c00h
int 21h
code ends
end start
不好意思,顺序写错了。 | | |
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[第5楼]
[ 回复时间:2009-05-15 15:02 ]
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注册日期:2009-04-22 12:29 |
mov cl,[bx]
mov ch,0
jcxz ok
inc bx | | |
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[第6楼]
[ 回复时间:2009-07-05 13:50 ]
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注册日期:2009-06-27 15:52 |
sub cx,cx
mov cl,[bx]
jcxz ok
inc bx | | |
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[第7楼]
[ 回复时间:2009-10-23 21:35 ]
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注册日期:2009-10-23 21:32 |
mov ah ,[bx]
mov cx,ax
jcxz ok
inc bx | | |
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[第8楼]
[ 回复时间:2009-11-14 10:57 ]
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注册日期:2009-11-12 11:56 |
mov cx,[bx]
sub cx,0
jcxz ok
inc bx | | |
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[第9楼]
[ 回复时间:2009-12-26 16:28 ]
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注册日期:2008-10-12 21:48 |
jcxz ok
inc bx
和
inc bx
jcxz ok
可谓一字师啊! | | |