計算機組成原理課后部分答案

第2章習(xí)題二(缺2-16)2-5.將二進(jìn)制數(shù)(101010.01)2轉(zhuǎn)換為十進(jìn)制數(shù)及BCD碼。解：(101010.01)2=(42.25)10=(01000010.00100101)BCD2-6.將八進(jìn)制數(shù)(37.2)8轉(zhuǎn)換為十進(jìn)制數(shù)及BCD碼.解:(37.2)8=(31.25)10=(00110001.00100101)BCD2-7.將十六進(jìn)制數(shù)(AC.E)轉(zhuǎn)換為十進(jìn)制數(shù)及BCD碼.解:(AC.E)16=(172.875)10=(0000.0101)BCD2-8.將十進(jìn)制數(shù)(75.34)10轉(zhuǎn)換為8位二進(jìn)制數(shù)及八進(jìn)制數(shù)、十六進(jìn)制數(shù)。解:(75.34)10=(01001011.01010111)2=(213.256)8=(4B.57)162-9.將十進(jìn)制數(shù)13/128轉(zhuǎn)換為二進(jìn)制數(shù).解:(13/128)10=(1101/)2=(0.0001101)2非0最小正數(shù)最大正數(shù)絕對值最小負(fù)數(shù)非0最小正數(shù)最大正數(shù)絕對值最小負(fù)數(shù)絕對值最大負(fù)數(shù)(1)0(2)-0(3)0.1010(4)-0.1010 (5)1010 (6)-1010解：原碼補碼00000000000000000-0000000000.10100.10100000.1010000-0.10101.10100001.011000010100000101000001010-10102-13.某定點小數(shù)字長(1)非0最小正數(shù)(3)絕對值最小負(fù)數(shù)解：16位，含1位符號，原碼表示，分別寫出下列典型值的二進(jìn)制代碼與十進(jìn)制真值。(2)最大正數(shù)(4)絕對值最大負(fù)數(shù)二進(jìn)制代碼 十進(jìn)制真值0.000000000000001 2-150. 1-2-15-2-151.000000000000001-2-151. -(1-2-15)16位，含1位符號，補碼表示，分別寫出下列典型值的二進(jìn)制代碼與十進(jìn)制真值。(1)非0最小正數(shù)(3)絕對值最小負(fù)數(shù)解：(2)最大正數(shù)(4)絕對值最大負(fù)數(shù)二進(jìn)制代碼 十進(jìn)制真值2-14.某定點小數(shù)字長非0非0最小正數(shù)最大正數(shù)絕對值最小負(fù)數(shù)絕對值最大負(fù)數(shù)0. 1-2-151. -2-151.000000000000000-1第三章(缺3-2)3-3.用變形補碼計算［X］補+［丫］補，并指出是否溢出，說明是正溢還是負(fù)溢。［X］補=00110011 ［Y］補=00101101+)0010110101100000[X]補+[Y]補=01100000,符號位為01,為正溢。⑵［X］補=00010110 ［Y］補=0010010100110110+)0010010101011011［X］補+［Y］補=01011011,符號位為01,為正溢。⑶［X］補=11110011 ［Y］補=1110110111110011+)1110110111100000［X］補+［Y］補=11100000,符號位為11,結(jié)果正確。⑷［X］補=11001101 ［Y］補=1101001111001101+)1101001110100000［X］補+［Y］補=10100000,符號位為10,為負(fù)溢。3-4.用變形補碼計算［X］補-［Y］補，并指出是否溢出，說明是正溢還是負(fù)溢。(1)［X］補=00110011 ［Y］補=00101101解：［-Y］補=1101001100110011+)1101001100000110兇補-［Y］補=00000110,符號位為00,結(jié)果正確。⑵［X］補=00110011 ［Y］補=11010011解：［-Y］補=0010110100110011+)0010110101000000兇補-［Y］補=01000000,符號位為01,為正溢。⑶［X］補=00100011 ［Y］補=00110100解：［-Y］補=1100110000110011+)1100110011111111兇補-［Y］補=11111111,符號位為11,結(jié)果正確。⑷［X］補=00101010 ［Y］補=11110111解：［-Y］補=0000100100101010+)00001001

兇補-[Y]補=00110011,符號位為00,結(jié)果正確。3-12.擬出下述指令的讀取與執(zhí)行流程：(1)MOVR0，R2解：FT0PSMARFT1MRMDR?IR,PC+1~PCST0R2rCET0 CRR0(2)MOVR1,(PC)+解：FT0PCRMARFT1MRMDRRIR，PC+1RPCST0PCRMARST1MRMDRRCST2PC+1RZST3ZRPCET0CRR0(3)MOV-(SP),-(R1)解:FT0PCRMARFT1MRMDRRIR，PC+1RPCST0R1-1RZST1ZRMAR,R1ST2MRMDRRCDT0SP-1RZDT2ZRMAR,SPET0CRMDRET1MDRRM(4)MOV(R0)+,X(R3)解:FT0PCRMARFT1MRMDRRIR，PC+1RPCST0PCRMARST1MRMDRRD,PC+1RPCST2D+R3RZST3ZRMARST4MRMDRRCDT0R0RMARDT1R0+1RZDT2ZRR0ET0CRMDRET1MDRRM(5)MOV(R0),(PC)+解:FT0PCRMARFT1MRMDRRIR，PC+1RPCST0PCRMARST1MRMDRRCST2PC+1RZST3ZRPCDTORORMARETOCRMDRET1MDRRM(6)MOVDI,(SP)+解：FTOPC^MARFT1MRMDRRIR，PC+1RPCSTOSPRMARST1MRMDRRCST2SP+1RZST3ZRSPDTOPCRMARDT1MRMDRRMAR ，PC+1RPCETOCRMDRET1MDRRM3-13.擬出下述程序的讀取與執(zhí)行過程：(1)ADDRO，X（R1）解：FTOPSMARFT1MRMDPIR,PC+㈠PCSTOPCrMARST1MRMDRRD,PC+1RPCST2D+R1RZST3ZRMARST4MRMDRRCDTORORDETOCADDDRZET1ZRROSUB(R1)+,(PC)+解：FTOPCR解：FT1MRMDRRIR ，PC+1RPCSTOPCRMARST1MRMDRRCST2PC+1RZST3ZRPCDTORORMARDT1MRMDRRDDT2RO+1RZDT3ZRROETOCSUBDRZET1ZRMDRET2MDRRMAND(R3)+,RO解：FTOPCRMAR

FT1MRMDPIR ,PC+㈠PCSTOR0rCDT0R3RMARDT1MRMDRRDDT2R3+1RZDT3ZRR3ETOCANDDRZET1ZRMDRET2MDRRM(4)ORRO,DI解:FTOPCRMARFT1MRMDRRIR，PC+1RPCSTOPCRMARST1MRMDRRMAR，PC+1RPCST2MRMDRRCDT1RORDETOCORDRZET1ZRRO(5)EOR-(R2),R1解:FTOPCRMARFT1MRMDRRIR，PC+1RPCSTOR1RCDTOR2-1RZDT1ZRMAR,R2DT2MRMDRRDETOCEORDRZET1ZRMDRET2MDRRM(6)INC-(R2)解:FTOPCRMARFT1MRMDRRIR，PC+1RPCDTOR2-1RZDT1ZRMAR,R2DT2MRMDRRDETOINCDRZET1ZRMDRET2MDRRM(7)DEC(R1)解:FTOPCRMARFT1MRMDRRIR，PC+1RPCDTOR1RMARDT1MRMDRRDETODECDRZET1ZRMDR

ET2 MDR?M(8)COM(R0)+解:FT0PCRMARFT1MRMDRRIR，PC+1RPCDT0R0RMARDT1MRMDRRDDT2R0+1RZDT3ZRR0ET0COMDRZET1ZRMDRET2MDRRM(9)NEGDI解:FT0PCRMARFT1MRMDRRIR，PC+1RPCDT0PSMARDT1MRMDR?MAR,PC+1fPCDT2MRMDRRDET0NEGDRZET1ZRMDRET2MDRRM(10)SALR1解:FT0PCRMARFT1MRMDRRIR，PC+1RPCDT1R1RDET0SALDRZET1ZRR1(11)SARR2解:FT0PCRMARFT1MRMDRRIR，PC+1RPCDT1R2RDET0SARDRZET1ZRR214.擬出下述程序的讀取與執(zhí)行過程：JMPR1解:FT0PCRMARFT1ET0MRMDRRIR ，R1RPCPC+1RPC(2)JMP(R0)解:FT0PCRMARFT1MRMDRRIR，PC+1RPCET0R0RMARET1MRMDRRPC(3)JMPX(PC)解:FT0PCRMARFT1MRMDPIR ,PC+㈠PCSTOPCrMARST1MRMDRRDST2D+PCRZST3ZRPCRST(SP)+解:FTOPCRMARFT1MRMDRRIR ，PC+1RPCETOSPRPCET1MRMDRRPCET2SP+1RZET3ZRPCJSRRO解:FTOPCRMARFT1MRMDRRIR ，PC+1RPCSTORORCETOSP-1RZET1ZRMAR,SPET2PCRMDRET3MDRRMET4CRPCJSR(R3)解:FTOPCRMARFT1MRMDRRIR ，PC+1RPCSTOR3RMARST1MRMDRRCETOSP-1RZET1ZRMAR,SPET2PCRMDRET3MDRRMET4CRPCJSR(R2)+解:FTOPCRMARFT1MRMDRRIR ，PC+1RPCSTOR2RMARST1MRMDRRCST2R2+1RZST3ZRR2ETOSP-1RZET1ZRMAR,SPET2PCRMDRET3MDRRMET4CRPC第四章習(xí)題4（缺4-174-19）

15.假設(shè)(DS)=091DH,(SS)=1E4AH,(AX)=1234H,(BX)=0024H,(CX)=5678H,(BP)=0024H,(SI)=0012H,(DI)=0032H,(09226H)=00F6H,(09228H)=1E40H,(1E4F6H)=091DH.試給出下列各指令或程序段執(zhí)行的分別執(zhí)行的結(jié)果?(1) (1)MOVCL,20H[BX][SI](DS)X10H+20H+(BX)+(SI)+)oo42(DS)X10H+20H+(BX)+(SI)+)oo42D221Ioo(09226H)=00F6H將09226H字節(jié)單元的內(nèi)容送CL,結(jié)果(CL)=00F6HMOV[BP][DI],CX解：目的操作數(shù)的物理地址：(SS)X0H+(BP)+(DI)1E4A002+) 003將CX寄存器的內(nèi)容傳送到 1E4F6H字單元，結(jié)果(1E4F6H)=(CX)=5678H(3)LEABX,20H[BX][SI]MOVAX,2[BX]解：執(zhí)行完第一句指令后,BX的內(nèi)容為：200024+)001200056第二句指令中，源操作數(shù)的物理地址為 EA=(DS)X10H+2+(BX)091D00056十) 209228將09228H字單元的內(nèi)容送AX寄存器，結(jié)果(AX)=1E40HLDSSI,[BX][DI]MOV[SI],BX解：第一句指令中源操作數(shù)的邏輯地址為 DS:(BX)+(DI)其偏移量為(BX)+(DI)=0024H+0032H=0056H,執(zhí)行完第一句指令后，(SI)=0056H第二句指令中，源操作數(shù)的物理地址為(DS)X10H+(SI)=091D0H+0056H=09226H將BX寄存器的內(nèi)容送09226單元，結(jié)果(09226H)=(BX)=0024H(5)XCHGCX,32H[BX]XCHG20H[BX][SI],AX解：解：第一句指令中源操作數(shù)的物理地址為(DS)X10H+32H+(BX)=09226H+) 002409226H字單元和CX寄存器的內(nèi)容交換，結(jié)果 (CX)=00F6H,(09226H)=5678H第二句指令中目的操作數(shù)的物理地址為 (DS)X10H++20H+(BX)+(SI)=09226H091D0200024+)00120922609226H字單元和寄存器AX的內(nèi)容交換,結(jié)果(09226H)=1234H,(AX)=5678H4-20.假設(shè)(AX)=0A5C6H,(CX)=0F03H,則下列4條指令執(zhí)行后，(AX)= 81C6H ,CF=1STC;CF=1RCLAX,CL;AX=0010111000110110 ,CF=1ANDAH,CH;AH=00001110,CF=0RCRAX,CL;AX=1000000111000110,CF=1CF4-21■假設(shè)(AX)=0FC77H,(CX)=504H，則下列4條指令執(zhí)行后，(AX)=CFCLC ;CF=0SAR AX,CL;(AX)=1111111111000111,CF=0XCHGCH,CL;(CX)=405HSHLAX,CL;(AX)=1111100011100000,CF=14-22.假設(shè)(AX):=0FFFFH,則下述程序段執(zhí)行后 ，(AX)=0001HINCAX;(AX)=0000HNEGAX;(AX)=0000000000000001B=0000HDECAX;(AX)=0FFFFHNEGAX;(AX)=0001H4-23.假設(shè)(BX):=12FFH,則下述程序段執(zhí)行后 ，(BX)=0012H,ZF= 0,CF=0MOVCL,8;(CL)=8ROLBX,CL;(BX)=1111111100010010B=:0FF12HANDBX,0FFH;(BX)=0012HCMPBX,0FFH;(BX)=0012H CF=0,ZF=0解： ①第一句指令執(zhí)行后，(CL)=8②第二句指令執(zhí)行后 ，(BX)=1111111100010010B=0FF12H

BX~~001emo11111111③第三句指令執(zhí)行后，（BX）=0012HmiliiioooiooioA)oooooooomimiA)oooooooomimioooooooo00010010④第四句指令執(zhí)行后,(BX)=0012H0000000000010010+)1111111100000001mimioooioonoooooooo00010010④第四句指令執(zhí)行后,(BX)=0012H0000000000010010+)1111111100000001mimioooioonCF=O,ZF=OSTC;CF=1MOVDX,96;(DX)=0060HXORDH,0FFH;(DX)=1111111101100000B=0FF60HSBBAX,DX;(AX)=04-25.假設(shè)(AL)=08H,(BL)=07H,則下列程序段執(zhí)行后 ，4-24.假設(shè)(AX)=0FF60H, 則下述程序段執(zhí)行后 ,(AX)=0 ,CF=0(AH)= 01H,(AL)= 05H,CF= 1ADDAL,BLAAA(AL)=00001000+)(BL)=00000111(AL)=00001111(AL)46 00000110(AL)=00010101AL高四位清0(AL)=00000101AH中Im(AH)=00000001?CF114-26.假設(shè)DF=0,(DS:0100H)=01A5H,則下述程序段執(zhí)行后，(AL)=0A5H,SI=MOVSI,0100HLODSW解:LODSW指令的功能：AX—(DS:(SI))SI—(SI)±1或2對標(biāo)志位無影響0102H.執(zhí)行后,(AX)=01A5H,即(AL)=0A5H,(SI)=0102H第五章習(xí)題五(只有5-1 5-14其余的缺)1、DATASEGMENTVAR1DB?,?VAR2DB?,?ADRDWVAR1,VAR2DATAENDS若要使ADR+2字單元的內(nèi)容為0022H,上述空白處應(yīng)填寫什么語句解：上述空白處應(yīng)填寫ORG0020H5-14、DB1DB4 DUP(2,4,6,8)LEABX,DB1MOVCX,10HMOVAX,0LOP:ADDAL,[BX]ANDAL,0FHCMPAL,8JBENEXTINCAHSUBAL,08HNEXT:LOOPLOP,(AX)=?(CX)=?上述程序段執(zhí)行后，(AX)=?如果LOOPNE指令替代LOOP指令，那么上述程序段執(zhí)行后,(AX)=?(CX)=?解:

循壞(ALF(AL),r(AL)(AH)(CX)102H02H02H015206H06H06H01430CH0CH04H11340CH0CH04H212506H06H06H21160AH0AH02H310708H08HOSH398L0HOOHOOH38902H02H02H371006H06H06H36110CHOCH04H45120CHOCH04H54506H06H06H2360AHOAH02H62708H08HOSH61810HOOHOOH60由上表可知道，上述程序段執(zhí)行后， (AX)=0600H,(CX)=0如果LOOPNE指令替代LOOP指令，那么上述程序段執(zhí)行后，(AX)=0300H(CX)=8第六章(缺6-36-7，其中6-2題目有變)6-2某半導(dǎo)體存儲器容量為 16KX8位，可選用RAM芯片