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匯編語(yǔ)言第二版習(xí)題答案(全)王爽編寫-wenkub.com

2025-04-06 01:53 本頁(yè)面
   

【正文】 或mov sp, 36 。Microsoft(R) Windows DOS(C)Copyright Microsoft Corp 19902001.C:\DOCUME~1\ADMINI~1debug rAX=0000 BX=0000 CX=0016 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000DS=0C69 ES=0C69 SS=0C79 CS=0C79 IP=0000 NV UP EI PL NZ NA PO NC0C79:0000 B80020 MOV AX,2000d 0c69:00C69:0000 CD 20 FF 9F 00 9A F0 FE1D F0 4F 03 80 06 8A 03 . ........O.....0C69:0010 80 06 17 03 80 06 6F 0601 01 01 00 02 FF FF FF ......o.........0C69:0020 FF FF FF FF FF FF FF FFFF FF FF FF 2D 0C 4C 01 .............0C69:0030 40 0B 14 00 18 00 69 0CFF FF FF FF 00 00 00 00 .....i.........0C69:0040 05 00 00 00 00 00 00 0000 00 00 00 00 00 00 00 ................0C69:0050 CD 21 CB 00 00 00 00 0000 00 00 00 00 20 20 20 .!...........0C69:0060 20 20 20 20 20 20 20 2000 00 00 00 00 20 20 20 .....0C69:0070 20 20 20 20 20 20 20 2000 00 00 00 00 00 00 00 ........q (1)下面的程序?qū)崿F(xiàn)依次用內(nèi)存0:0~0:15單元中的內(nèi)容改寫程序中的數(shù)據(jù),完成程序:assume cs:codesg codesg segment dw 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h start: 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 21h codesg ends end startC:\DOCUME~1\ADMINI~1debug u0C79:0010 B80000 MOV AX,00000C79:0013 8ED8 MOV DS,AX0C79:0015 BB0000 MOV BX,00000C79:0018 B90800 MOV CX,00080C79:001B 8B07 MOV AX,[BX]0C79:001D 2E CS:0C79:001E 8907 MOV [BX],AX0C79:0020 83C302 ADD BX,+020C79:0023 E2F6 LOOP 001B0C79:0025 B8004C MOV AX,4C000C79:0028 CD21 INT 210C79:002A 7503 JNZ 002F0C79:002C E97BFF JMP FFAA0C79:002F 5E POP SIg 0025AX=0680 BX=0010 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000DS=0000 ES=0C69 SS=0C79 CS=0C79 IP=0025 NV UP EI PL NZ AC PO NC0C79:0025 B8004C MOV AX,4C00d 0:0 f0000:0000 68 10 A7 00 BB 13 80 0616 00 A5 03 B1 13 80 06 h...............d 0c79:0 f0C79:0000 68 10 A7 00 BB 13 80 0616 00 A5 03 B1 13 80 06 h...............tAX=4C00 BX=0010 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000DS=0000 ES=0C69 SS=0C79 CS=0C79 IP=0028 NV UP EI PL NZ AC PO NC0C79:0028 CD21 INT 21pProgram terminated normallyqC:\DOCUME~1\ADMINI~1 (2)下面的程序?qū)崿F(xiàn)依次用內(nèi)存0:0~0:15單元中的內(nèi)容改寫程序中的數(shù)據(jù),數(shù)據(jù)的傳送用棧來(lái)進(jìn)行。這個(gè)可能在你跟蹤一些程序的時(shí)候會(huì)遇到,到時(shí)候有個(gè)思考方向。sp= 00FEH ,修改的內(nèi)存單元地址是 2200:00FE 內(nèi)容為 31F0 push [6] 。sp= 00FEH ,修改的內(nèi)存單元地址是 2200:00FE 內(nèi)容為 5CCAH push bx 。ax= 5BEAH add ax,[2] 。③ 再次體會(huì):數(shù)據(jù)和程序有區(qū)別嗎?如何確定內(nèi)存中的信息哪些是數(shù)據(jù),哪些是程序?指令序列CSIPDSAXBX初始值2000h00001mov ax,6622h2000h3h06622h02jmp 0ff0:0100ff0h100h06622h03mov ax,2000hff0h103h02000h04mov ds,axff0h105h2000h2000h05mov ax,[8]ff0h108h2000hc389h06mov ax,[2]ff0h10bh2000hea66h0(1)補(bǔ)全下面的程序,使其可以將10000H1000FH中的8個(gè)字,逆序拷貝到20000H2000FH中。....。下面內(nèi)容摘自于網(wǎng)上還有另一種情況,如果你發(fā)現(xiàn)你能修改ROM中的生產(chǎn)日期,那么原因如下:每個(gè)計(jì)算機(jī)的結(jié)構(gòu)都不一樣,教材考慮的是普通8086PC機(jī)上的效果,個(gè)別計(jì)算機(jī)的效果可能不同。機(jī)器碼 匯編指令 寄存器b8 20 4e mov ax,4E20H ax=4E20H 05 16 14 add ax,1416H ax=6236H bb 00 20 mov BX,2000H bx=2000H 01 d8 add ax,bx ax=8236H 89 c3 mov bx,ax bx=8236H 01 d8 add ax,bx ax=046CH b8 1a 00 mov ax,001AH ax=001AH bb 26 00 mov bx,0026H bx=0026H 00 d8 add al,bl ax=0040H 00 dc add ah,bl ax=2640H 00 c7 add bh,al bx=4026H b4 00 mov ah,0 ax=0040H 00 d8 add al,bl ax=0066H 04 9c add al,9CH ax=0002H Microsoft(R) Windows DOS(C)Copyright Microsoft Corp 19902001.C:\DOCUME~1\ADMINI~1debuga0C1C:0100 mov ax,4e200C1C:0103 add ax,14160C1C:0106 mov bx,20000C1C:0109 add ax,bx0C1C:010B mov bx,ax0C1C:010D add ax,bx0C1C:010F mov ax,001a0C1C:0112 mov bx,00260C1C:0115 add al,bl0C1C:0117 add ah,bl0C1C:0119 add bh,al0C1C:011B mov ah,00C1C:011D add al,bl0C1C:011F add al,9c0C1C:0121rAX=0000 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=0100 NV UP EI PL NZ NA PO NC0C1C:0100 B8204E MOV AX,4E20tAX=4E20 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=0103 NV UP EI PL NZ NA PO NC0C1C:0103 051614 ADD AX,1416tAX=6236 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=0106 NV UP EI PL NZ NA PE NC0C1C:0106 BB0020 MOV BX,2000tAX=6236 BX=2000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=0109 NV UP EI PL NZ NA PE NC0C1C:0109 01D8 ADD AX,BXtAX=8236 BX=2000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=010B OV UP EI NG NZ NA PE NC0C1C:010B 89C3 MOV BX,AXtAX=8236 BX=8236 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=010D OV UP EI NG NZ NA PE NC0C1C:010D 01D8 ADD AX,BXtAX=046C BX=8236 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=010F OV UP EI PL NZ NA PE CY0C1C:010F B81A00 MOV AX,001AtAX=001A BX=8236 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=0112 OV UP EI PL NZ NA PE CY0C1C:0112 BB2600 MOV BX,0026tAX=001A BX=0026 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000DS=0C1C ES=0C1C SS=0C1C CS=0C1C IP=0115 OV UP EI PL NZ NA PE CY0C1C:0115 00D8 ADD AL,BLtAX=0040 BX=0026 CX=0000 DX=0000 SP
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