检测点6.1第一题

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题目：实现依次用内存0:00:15单元中的内容改写程序中的数据

解答：

源代码；

assume cs:codesgcodesg segment dw 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987hstart: mov ax,0 mov ds,ax mov bx,0 mov cx,8 s: mov ax,[bx] mov cs:[bx],ax add bx,2 loop s mov ax,4c00h int 21hcodesg endsend start

调试结果：

-rAX=0000 BX=0000 CX=002A DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=143C ES=143C SS=144C CS=144C IP=0010 NV UP EI PL NZ NA PO NC 144C:0010 B80000 MOVAX,0000 -d 0:0 f0000:0000 68 10 A7 00 8B 01 70 00-16 00 AD 03 8B 01 70 00 h.....p.......p.-d cs:0144C:0000 23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 #.V.............144C:0010 B8 00 00 8E D8 BB 00 00-B9 08 00 8B 07 2E 89 07 ................144C:0020 83 C3 02 E2 F6 B8 00 4C-CD 21 00 00 00 00 00 00 .......L.!......144C:0030 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0040 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0050 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0060 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0070 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................-tAX=0000 BX=0000 CX=002A DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=143C ES=143C SS=144C CS=144C IP=0013 NV UP EI PL NZ NA PO NC 144C:0013 8ED8 MOVDS,AX -tAX=0000 BX=0000 CX=002A DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=0015 NV UP EI PL NZ NA PO NC 144C:0015 BB0000 MOVBX,0000 -tAX=0000 BX=0000 CX=002A DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=0018 NV UP EI PL NZ NA PO NC 144C:0018 B90800 MOVCX,0008 -tAX=0000 BX=0000 CX=0008 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=001B NV UP EI PL NZ NA PO NC 144C:001B 8B07 MOVAX,[BX] DS:0000=1068-tAX=1068 BX=0000 CX=0008 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=001D NV UP EI PL NZ NA PO NC 144C:001D 2E CS: 144C:001E 8907 MOV[BX],AX CS:0000=0123-tAX=1068 BX=0000 CX=0008 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=0020 NV UP EI PL NZ NA PO NC 144C:0020 83C302 ADDBX,+02 -tAX=1068 BX=0002 CX=0008 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=0023 NV UP EI PL NZ NA PO NC 144C:0023 E2F6 LOOP001B -d cs:0144C:0000 68 10 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 h.V.............144C:0010 B8 00 00 8E D8 BB 00 00-B9 08 00 8B 07 2E 89 07 ................144C:0020 83 C3 02 E2 F6 B8 00 4C-CD 21 00 00 00 00 00 00 .......L.!......144C:0030 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0040 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0050 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0060 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0070 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................-pAX=0E30 BX=0010 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=0025 NV UP EI PL NZ AC PO NC 144C:0025 B8004C MOVAX,4C00 -d cs:00144C:0000 68 10 A7 00 BB 13 30 0E-16 00 AD 03 B1 13 30 0E h.....0.......0.144C:0010 B8 00 00 8E D8 BB 00 00-B9 08 00 8B 07 2E 89 07 ................144C:0020 83 C3 02 E2 F6 B8 00 4C-CD 21 00 00 00 00 00 00 .......L.!......144C:0030 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0040 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0050 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0060 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................144C:0070 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................-pAX=4C00 BX=0010 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=0000 ES=143C SS=144C CS=144C IP=0028 NV UP EI PL NZ AC PO NC 144C:0028 CD21 INT21 -pProgram terminated normally-q