-r
AX=0000 BX=0000 CX=0016 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=1431 ES=1431 SS=1441 CS=1441 IP=0000 NV UP EI PL NZ NA PO NC
1441:0000 B80020 MOV AX,2000
-t
AX=2000 BX=0000 CX=0016 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=1431 ES=1431 SS=1441 CS=1441...
- [tomato] MOV BX,0B800H MOV ES,BX MOV 04/17 15:26
- [游客] 程序有问题,你的返回指令处 MOV AH,4C 这条指令丢了H了,所以数据就不对了。 04/15 11:28
- [游客] 结论正确。第一题的原因是? 04/15 11:20
- [游客] 回答OK。 04/15 10:24
- [游客] 没问题。想显示完全年份的话,可以搜搜博客中的文章,有提到的。 04/15 10:10
- [游客] 理解OK。 04/15 09:00
- [bslg123] 支持一下! 04/13 10:01
- [caroline0712] 3ks 已经改正 04/08 15:18
- [crazyman] 监测点16.2第二个空不对。仔细看看assume关联。 04/07 21:25
- [caroline0712] ;数值显示 改进版()显示部分的改进) ;将DATA中的数据以十进制显示; ;将二进制信息存储 03/30 15:47
[2009-03-21 21:33] 第四章实验三
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[2009-03-21 16:08] 第三章实验
(1)ax=5BEA,AX=00E0;BX=30F0,BX=6028
SP=00FE,2200:00FE 00E0
SP=00FC,2200:00FC 6028
SP=00FE,6028
SP=0100,00E0
SP=00FE,2200:00FE 30F0
SP=00FC,2200:00FE 2F38
-a 0b39:0100
0B39:0100 mov ax,ffff
0B39:0103 mov ds,ax
0B39:0105 mov ax,2200
0B39:0108 mov ...
SP=00FE,2200:00FE 00E0
SP=00FC,2200:00FC 6028
SP=00FE,6028
SP=0100,00E0
SP=00FE,2200:00FE 30F0
SP=00FC,2200:00FE 2F38
-a 0b39:0100
0B39:0100 mov ax,ffff
0B39:0103 mov ds,ax
0B39:0105 mov ax,2200
0B39:0108 mov ...
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[2009-03-20 10:52] 第三章检测点
检测点3.1
(1)
[1]AX= 2662H,
[2]BX= E626H,
[3]AX= E626H,
[4]AX= 2662H,
[5]BX= D6E6H,
[6]AX= FD48H,
[7]AX= 2C14H,
[8]AX= 0H,
[9]AX=00E6H,
[10]BX=0,
[11]BX=0026H,
[12]AX=000C
(2)
[1]写出CPU执行的指令序列:
[2]写出CPU执行每条指令后,CS和IP和相关寄存器的值
解:
MOV AX,6622H;CS=2000H,IP=3
JMP OFFO:0100;CS=2000H,IP=8...
(1)
[1]AX= 2662H,
[2]BX= E626H,
[3]AX= E626H,
[4]AX= 2662H,
[5]BX= D6E6H,
[6]AX= FD48H,
[7]AX= 2C14H,
[8]AX= 0H,
[9]AX=00E6H,
[10]BX=0,
[11]BX=0026H,
[12]AX=000C
(2)
[1]写出CPU执行的指令序列:
[2]写出CPU执行每条指令后,CS和IP和相关寄存器的值
解:
MOV AX,6622H;CS=2000H,IP=3
JMP OFFO:0100;CS=2000H,IP=8...
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[2009-03-20 09:48] 第二章实验任务
(1)使用E命令和A命令均可将指令写入内存
-e 13c3:00d0 b8 20 4e 05 16 14 bb 00 20 01 d8 89 c3 01 d8 b8 1a 00 bb 26 00 00 d8 00 dc 00 c7 b4 00 00 d8 04 9c
-u 13c3:00d0 00f0
13C3:00D0 B8204E MOV AX,4E20
13C3:00D3 051614 ADD AX,1416
13C3:00D6 BB0020 MOV BX,2000
13C3:00D9 01D8 ...
-e 13c3:00d0 b8 20 4e 05 16 14 bb 00 20 01 d8 89 c3 01 d8 b8 1a 00 bb 26 00 00 d8 00 dc 00 c7 b4 00 00 d8 04 9c
-u 13c3:00d0 00f0
13C3:00D0 B8204E MOV AX,4E20
13C3:00D3 051614 ADD AX,1416
13C3:00D6 BB0020 MOV BX,2000
13C3:00D9 01D8 ...
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[2009-03-19 20:38] 第二章检测点
检测点2.1
1.F4A3H,31A3H,3123H,6246H,826CH
CX=6246H,AX=826CH,AX=04D8H,AX=0482H,AX=6C82H,AX=D882H,
AX=D888H,AX=D810H,AX=4246H
2.计算2的4次方
MOV AX,2
ADD AX,AX
ADD AX,AX
ADD AX,AX
检测点2.2
1.0001H:0---0001H:FFFFH(0~64K)
2.1001H,2000H
检测点2.3
1.CPU修改IP4次
MOV AX,BX;修改IP的值为0003
SUB...
1.F4A3H,31A3H,3123H,6246H,826CH
CX=6246H,AX=826CH,AX=04D8H,AX=0482H,AX=6C82H,AX=D882H,
AX=D888H,AX=D810H,AX=4246H
2.计算2的4次方
MOV AX,2
ADD AX,AX
ADD AX,AX
ADD AX,AX
检测点2.2
1.0001H:0---0001H:FFFFH(0~64K)
2.1001H,2000H
检测点2.3
1.CPU修改IP4次
MOV AX,BX;修改IP的值为0003
SUB...
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[2009-03-16 14:57] 检测点12
检测点12.1
(1)0070:018B
(2)4*N; 4*N+2...
(1)0070:018B
(2)4*N; 4*N+2...
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[2009-03-15 21:37] 检测点11
检测点11.1
有符号数/无符号数:
(1)ZF=1,PF=1,SF=0
(2)ZF=0;PF=0,SF=0
(3)ZF=0;PF=0,SF=0
(4)ZF=0,PF=0,SF=0
(5)ZF=0,PF=0,SF=0
(6)ZF=0,PF=1,SF=0
(7)ZF=0,PF=1,SF=1/0
检测点11.2
CF OF SF ZF PF
0 0 0 1 1
0 0 0 0 0
0 0 1/0 0 1
0 0 1/0 0 ...
有符号数/无符号数:
(1)ZF=1,PF=1,SF=0
(2)ZF=0;PF=0,SF=0
(3)ZF=0;PF=0,SF=0
(4)ZF=0,PF=0,SF=0
(5)ZF=0,PF=0,SF=0
(6)ZF=0,PF=1,SF=0
(7)ZF=0,PF=1,SF=1/0
检测点11.2
CF OF SF ZF PF
0 0 0 1 1
0 0 0 0 0
0 0 1/0 0 1
0 0 1/0 0 ...
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[2009-03-15 16:52] 第十章实验十程序一
DATAS SEGMENT
;此处输入数据段代码
db 'welcome to masm',0
DATAS ENDS
STACKS SEGMENT
;此处输入堆栈段代码
DW 40 DUP(0)
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV DH,8
MOV DL,3
MOV CL,2;颜色
MOV AX,DATAS
MOV DS,AX
;此处输入代码段代码
MOV S...
;此处输入数据段代码
db 'welcome to masm',0
DATAS ENDS
STACKS SEGMENT
;此处输入堆栈段代码
DW 40 DUP(0)
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV DH,8
MOV DL,3
MOV CL,2;颜色
MOV AX,DATAS
MOV DS,AX
;此处输入代码段代码
MOV S...
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[2009-03-13 14:06] 检测点10
检测点10.1
answer:(1)1000H;(2)0H
analysis:as in essence the RETF instruction is composited by two operations:first push the top element as the address of IP,then push the continue top element as the address of CS。
检测点10.2
answer:(1)ax=6
analysis: as the following operational sequence:
(1)mov ax...
answer:(1)1000H;(2)0H
analysis:as in essence the RETF instruction is composited by two operations:first push the top element as the address of IP,then push the continue top element as the address of CS。
检测点10.2
answer:(1)ax=6
analysis: as the following operational sequence:
(1)mov ax...
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