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主题 : : 第六章实验(六)作业 请高手指教 [待解决] |
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[帖 主]
[ 发表时间:2011-09-18 09:35 ]
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注册日期:2011-08-18 06:18 |
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 ss,ax
mov sp,30h
mov bx,0
mov cx,8
s: push ds:[bx]
add bx,2
loop s
mov ax,4c00h
int 21h
code ends
end start
D:\>debug p5_6.exe
-r
AX=0000 BX=0000 CX=004C DX=0000 SP=0000 BP=0000 SI=0000 DI=00
DS=0B78 ES=0B78 SS=0B88 CS=0B8B IP=0000 NV UP EI PL NZ NA PO N
0B8B:0000 B8880B MOV AX,0B88
-d ds:100
0B78:0100 01 00 02 00 03 00 04 00-05 00 06 00 07 00 08 00 ................
0B78:0110 09 00 0A 00 0B 00 0C 00-0D 00 0E 00 0F 00 FF 00 ................
0B78:0120 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
0B78:0130 B8 88 0B 8E D8 BB 00 00-8E D0 BC 30 00 B9 08 00 ...........0....
0B78:0140 FF 37 83 C3 02 E2 F9 B8-00 4C CD 21 FF 36 24 21 .7.......L.!.6$!
0B78:0150 B8 E5 05 EB 12 90 FF 36-24 21 B8 ED 05 EB 08 90 .......6$!......
0B78:0160 FF 36 24 21 B8 F5 05 50-E8 09 5F 83 C4 04 FF 76 .6$!...P.._....v
0B78:0170 04 FF 36 24 21 E8 BC 44-83 C4 04 80 3E 60 08 00 ..6$!..D....>`..
-
-u
0B8B:0000 B8880B MOV AX,0B88
0B8B:0003 8ED8 MOV DS,AX
0B8B:0005 BB0000 MOV BX,0000
0B8B:0008 8ED0 MOV SS,AX
0B8B:000A BC3000 MOV SP,0030
0B8B:000D B90800 MOV CX,0008
0B8B:0010 FF37 PUSH [BX]
0B8B:0012 83C302 ADD BX,+02
0B8B:0015 E2F9 LOOP 0010
0B8B:0017 B8004C MOV AX,4C00
0B8B:001A CD21 INT 21
0B8B:001C FF362421 PUSH [2124]
-g 001a
AX=4C00 BX=0010 CX=0000 DX=0000 SP=0020 BP=0000 SI=0000 DI=0000
DS=0B88 ES=0B78 SS=0B88 CS=0B8B IP=001A NV UP EI PL NZ AC PO NC
0B8B:001A CD21 INT 21
-d ds:0
0B88:0000 01 00 02 00 03 00 04 00-05 00 06 00 07 00 08 00 ................
0B88:0010 09 00 0A 00 0B 00 0C 00-00 00 1A 00 8B 0B 7E 05 ..............~.
0B88:0020 08 00 07 00 06 00 05 00-04 00 03 00 02 00 01 00 ................
0B88:0030 B8 88 0B 8E D8 BB 00 00-8E D0 BC 30 00 B9 08 00 ...........0....
0B88:0040 FF 37 83 C3 02 E2 F9 B8-00 4C CD 21 FF 36 24 21 .7.......L.!.6$!
0B88:0050 B8 E5 05 EB 12 90 FF 36-24 21 B8 ED 05 EB 08 90 .......6$!......
0B88:0060 FF 36 24 21 B8 F5 05 50-E8 09 5F 83 C4 04 FF 76 .6$!...P.._....v
0B88:0070 04 FF 36 24 21 E8 BC 44-83 C4 04 80 3E 60 08 00 ..6$!..D....>`..
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[第1楼]
[ 回复时间:2011-10-05 16:56 ]
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注册日期:2011-10-03 23:41 |
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[第2楼]
[ 回复时间:2012-01-23 17:46 ]
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注册日期:2012-01-10 21:29 |
楼主的
mov sp,30h
为什么?不理解啊
为什么不能是16
我用16也行啊
我的代码
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 bx,0
mov cx,8
s: push [bx]
add bx,2
loop s
mov ax,4c00h
int 21h
code ends
end start | | |
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[第3楼]
[ 回复时间:2012-01-23 17:48 ]
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sorry没仔细看楼主的代码
楼主好像没有定义b段这样不行吧 | | |
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[第4楼]
[ 回复时间:2012-01-23 17:52 ]
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注册日期:2012-01-10 21:29 |
楼主把a,b合定义成一个栈段了啊。好像跟题目要求不符啊。如果a与b中间还有定义数据。那这样就不行了 | | |
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[第5楼]
[ 回复时间:2012-01-31 09:14 ]
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注册日期:2008-01-19 14:51 |
楼主的代码是不够灵活,但是,针对这道题目也不能算错。 | | |
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