IBMP匯編語言課后復(fù)習(xí)題答案(第二版)

第一章.習(xí)題1.1用降冪法和除法將下列十進(jìn)制數(shù)轉(zhuǎn)換為二進(jìn)制數(shù)和十六進(jìn)制數(shù)：(1)369(2)10000(3)4095(4)32767(2)10000=10011100010000B=2710H(3)4095=111111111111B=FFFH(4)32767=111111111111111B=7FFFH1.2將下列二進(jìn)制數(shù)轉(zhuǎn)換為十六進(jìn)制數(shù)和十進(jìn)制數(shù)：(1)101101(2)10000000(3)1111111111111111(4)11111111(2)10000000B=80H=128(3)1111111111111111B=FFFFH=65535(4)11111111B=FFH=2551.3將下列十六進(jìn)制數(shù)轉(zhuǎn)換為二進(jìn)制數(shù)和十進(jìn)制數(shù)：(1)FA(2)5B(3)FFFE(4)1234(2)5BH=1011011B=91(3)FFFEH=1111111111111110B=65534(4)1234H=1001000110100B=46601.4完成下列十六進(jìn)制數(shù)的運(yùn)算，并轉(zhuǎn)換為十進(jìn)制數(shù)進(jìn)行校核：(1)3A+B7(2)1234+AF(3)ABCD-FE(4)7AB×6F(2)1234+AFH=12E3H=4835(3)ABCD-FEH=AACFH=43727(4)7AB×6FH=35325H=2178931.5下列各數(shù)均為十進(jìn)制數(shù)，請(qǐng)用8位二進(jìn)制補(bǔ)碼計(jì)算下列各題，并用十六進(jìn)制數(shù)表示其運(yùn)算結(jié)果。(1)(-85)+76(2)85+(-76)(3)85-76(4)85-(-76)(5)(-85)-76(6))(-85)+76=10101011B+01001100B=11110111B=0F7H；CF=0；OF=0(2)85+(-76)=01010101B+10110100B=00001001B=09H；CF=1；OF=0(3)85-76=01010101B-01001100B=01010101B+10110100B=00001001B=09H；CF=0；OF=0OF=01.6下列各數(shù)為十六進(jìn)制表示的8位二進(jìn)制數(shù)，請(qǐng)說明當(dāng)它們分別被看作是用補(bǔ)碼表示的帶符號(hào)數(shù)或D8(2)FF1.7下列各數(shù)均為用十六進(jìn)制表示的8位二進(jìn)制數(shù)，請(qǐng)說明當(dāng)它們分別被看作是用補(bǔ)碼表示的數(shù)或字(1)4F(2)2B(3)73(4)59Forexample,Thisisanumber3692.答：46H6FH72H20H65H78H61H6DH70H6CH65H2CH0AH0DH54H68H69H73H20H69H73H20H61H20H6EH75H6DH62H65H72H20H33H36H39H32H2EH0AH0DH第二章.習(xí)題存儲(chǔ)器存儲(chǔ)器2AH2AH答：存儲(chǔ)器里的存放情況如右上圖所示。2.2題的信息存放情況存儲(chǔ)器字單元的容為2.2題的信息存放情況存儲(chǔ)器ABH答：3017:000A、3015:002A和3010:007A的存儲(chǔ)單元的物理地址都是3017AH。母，則應(yīng)在其前加一個(gè)0)(IP)=2B40H，試問該程序的第一個(gè)字的物理2.3題的信息存放情況答：該程序的第一個(gè)字的物理地址是2.3題的信息存放情況C>debugAX=0000BX=0000CX=0079DX=0000SP=FFEEBP=0000SI=0000DI=0000DS=10E4ES=10F4SS=21F0CS=31FFIP=0100NVUPDIPLNZNAPONC(1)加法和減法數(shù)據(jù)寄存器等(2)循環(huán)計(jì)數(shù)CX(3)乘法和除法AX、DX，乘數(shù)和除數(shù)用其他寄存器或存儲(chǔ)器(4)保存段地址段寄存器(5)表示運(yùn)算結(jié)果為0ZF=1(6)將要執(zhí)行的指令地址CS:IP(7)將要從堆棧取出數(shù)據(jù)的地址SS:SP答：答案見題目的右邊。堆棧段代碼段2.6題的存儲(chǔ)器分段示意圖請(qǐng)將下列左邊的項(xiàng)和右邊的解釋聯(lián)系起來(把所選字母放在括號(hào)中)：(1)CPU(M)A.保存當(dāng)前棧頂?shù)刂返募拇嫫鳌?2)存儲(chǔ)器(C)B.指示下一條要執(zhí)行的指令的地址。(3)堆棧(D)C.存儲(chǔ)程序、數(shù)據(jù)等信息的記憶裝置，微機(jī)有RAM和ROM兩種。(4)IP(B)D.以后進(jìn)先出方式工作的存儲(chǔ)空間。(5)SP(A)E.把匯編語言程序翻譯成機(jī)器語言程序的系統(tǒng)程序。(6)狀態(tài)標(biāo)志(L)F.唯一代表存儲(chǔ)空間中每個(gè)字節(jié)單元的地址。(7)控制標(biāo)志(K)G.能被計(jì)算機(jī)直接識(shí)別的語言。(8)段寄存器(J)H.用指令的助記符、符號(hào)地址、標(biāo)號(hào)等符號(hào)書寫程序的語言。(9)物理地址(F)I.把若干個(gè)模塊連接起來成為可執(zhí)行文件的系統(tǒng)程序。SS、ES。(13)連接程序(I)M.分析、控制并執(zhí)行指令的部件，由算術(shù)邏輯部件ALU和寄存器等(14)指令(O)N.由匯編程序在匯編過程中執(zhí)行的指令。NOCPU執(zhí)行的操作(一般還要指出操作數(shù)地址)，在程序運(yùn)行時(shí)答：答案見題目的括號(hào)中。第三章.習(xí)題(7)相對(duì)基址變址尋址EABX7DH；(5)EA=(BX)+D=0D5B4H；(6)EA=(BX)+(SI)=8E18H；(7)EA=(BX)+(SI)+D=1004FH；超過了段的邊界，最高進(jìn)位位丟失，因此EA=004FH。3.2試根據(jù)以下要求寫出相應(yīng)的匯編語言指令(2)ADDAL,[BX][SI](3)ADD[BX+0B2H],CX(4)ADDWORDPTR[0524H],2A59HADDALBH器間接尋址器相對(duì)尋址MOVDX,[BX]MOVBXOFFSETBLOCKMOVBX1)*2MOVDXBLOCKBX(3)MOVBX,OFFSETBLOCKMOVDX,[BX][SI]MOVAX,1200HVAXBXBHH(4)MOVAX,[BX]1B00:2002H00H(5)MOVAX,1100[BX]1B00:2003H80H(7)MOVAX,1100[BX][SI]8000:FF11H？→(AL)AXH？→(AH)(2)(AX)=0100H(3)(AX)=4C2AH3.6題的作圖表示(4)(AX)=3412HHH(275B9H)=098AH，試為以下的轉(zhuǎn)移指令找出轉(zhuǎn)移的偏移地址。JMPWORDPTRBXIPDS10H+(BX))=0600H，PA=02600HJMPDBX；(IP)=((DS)*10H+(BX)+D)=098AH，PA=0298AH3.6設(shè)當(dāng)前數(shù)據(jù)段寄存器的容為1B00H，在數(shù)據(jù)段的偏移地址2000H單元，含有一個(gè)容為0FF10H和8000HAX的指令序列，并畫MOVAX,[2000H+2]MOVBX,2000HMOVES,AXLESBX,[BX]MOVAX,ES:[BXMOVAX,ES:[BX](2)OBJ=0624H+02H+6BH=0691H(3)OBJ=0624H+02H+0C6H=05ECH；C6H對(duì)應(yīng)的負(fù)數(shù)為-3AH(向上轉(zhuǎn)移，負(fù)位移量)H(1)MOVAX,0ABH(2)MOVAX,BX(3)MOVAX,[100H](4)MOVAX,VAL(5)MOVAX,[BX](6)MOVAX,ES:[BX](7)MOVAX,[BP](8)MOVAX,[SI](9)MOVAX,[BX+10](10)MOVAX,VAL[BX](11)MOVAX,[BX][SI](12)MOVAX,VAL[BX][SI]BX)=0100HARRAY著是名為ZERO的字單元，表示如下：ARRAYDW23,36,2,100,32000,54,0ZERODW?MOV[BX+(7)*2],AX(2)MOVAX,ARRAY[BX]MOVARRAY[BX+2],AXMOVAX,TABLE00HLEAAX,TABLE14HAX00HENTRYDW300HMOVBX,OFFSETTABLE00HADDBXENTRY3.11題的TABLEMOVAX,[BX]存儲(chǔ)方式CSTRINGDB‘BASEDADDRESSING’MOVDL,CSTRING+7-1出存儲(chǔ)單元的物理地址)。DSBH(ES)=2B00H，有關(guān)存儲(chǔ)單元的容如上右圖所示。請(qǐng)寫出兩條指令把字變量X裝入MOVAX,ES:[BX]OF(1)1234H(2)4321H(3)CFA0H(4)9D60H制數(shù)與十六進(jìn)制數(shù)4AE0H的差值，并根據(jù)結(jié)果設(shè)置標(biāo)志位SF、ZF、CF和OF的(1)1234H(2)5D90H(3)9090H(4)EA04H3.17寫出執(zhí)行以下計(jì)算的指令序列，其中X、Y、Z、R、W均為存放16位帶符號(hào)數(shù)單元的地址。SUBAX,XADDAX,WMOVZ,AXADDBX,6MOVCX,RADDCR,9MOVAX,WSUBAX,BXSUBAX,CXMOVZ,AXYMOVAX,WIMULXMOVMOVBCWDMOVYZ,AXR,DXAX,WAX,XBX,5BXYSHLAX((DX),(AX))*2RCLDX,1HCFOFINT20H？OFAXADDAXFFFH(AX)=7FFEH,CF=1，SF=0，ZF=0，OF=0ADDAX；(AX)=8000H,CF=0，SF=1，ZF=0，OF=1FHSUBAXFFFFHAXHCF=1，SF=1，ZF=0，OF=1SUBAXAXFFFFHCF=1，SF=1，ZF=0，OF=0ANDAXDH；(AX)=58D1H,CF=0，SF=0，ZF=0，OF=0SALAXAXBAH,CF=0，SF=1，ZF=0，OF=1SARAXAXDDH,CF=0，SF=1，ZF=0，OF=0NEGAX；(AX)=272FH,CF=1，SF=0，ZF=0，OF=0DATAXDW0148HDW2316HDATAYDW0237HDW4052H出指令序列：CMOVBX,DATAXADCBX,DATAYADDDATAY,AXMOVAX,DATAX+2ADDDATAY+2,AXMOVAX,DATAXADDDATAY,AXMOVAX,DATAX+2ADCDATAY+2,AXMOVDATAY+4,0；用于存放進(jìn)位位ADCDATAY+4,0(4)RESULT1DW0DW0RESULT2DW0DW0┇MOVAX,DATAXMULDATAYMOVRESULT1,AXMOVRESULT1+2,DXMOVAX,DATAX+2MULDATAY+2MOVRESULT2,AXMOVRESULT2+2,DXBBDW0CCDW0DDDW0┇MOVAX,DATAXMULDATAYMOVAA,AXMOVBB,DXMOVAX,DATAXMULDATAY+2ADDBB,AXADCCC,DXMOVAX,DATAX+2MULDATAYADDBB,AXADCCC,DXADCDD,0MOVAX,DATAX+2MULDATAY+2ADDCC,AXADCDD,DXMOVAX,DATAXMOVBL,23DIVBL(7)MOVDX,DATAX+2MOVAX,DATAXDIVDATAYNEGAXNOTAXSBBDX,0ADDAX,1ADCDX,0MOVAX,AMOVDX,A+2CMPDX,0NEGDXNEGAXSBBZHENSHU；不是負(fù)數(shù)則轉(zhuǎn)走DX,0ZHENSHU:MOVB,AXMOVB+2,DXINT20H執(zhí)行下列指令序列后BX寄存器的容。執(zhí)行前(BX)=6D16H。SHRBX,CL.25試用移位指令把十進(jìn)制數(shù)+53和-49分別乘以2。它們應(yīng)該用什么指令？得到的結(jié)果是什么？如果SALALAL3*2)=6AHMOVAL,-49SALALAL*2)=9EHMOVAL,53SARAL(AL)=(53/2)=1AHMOVAL,-49SARALAL(-49/2)=0E7HSHLDX,CLSHLAX,CLSHRBL,CLORDL,BLXROLBX,1設(shè)數(shù)據(jù)段定義如下：CONAMEDB‘SPACEEXPLORERSINC.’用串指令編寫程序段分別完成以下功能：XCLDMOVSI,SEGCONAMEMOVDS,SIMOVES,SILEASI,CONAMELEADI,PRLINEREPMOVSB(2)MOVCX,20DMOVSI,SEGCONAMEMOVDS,SIMOVES,SILEASI,CONAMEADDSI,20-1LEADI,PRLINEADDDI,20-1REPMOVSB(3)MOVAX,WORDPTRCONAME+3-1(4)MOVWORDPTRPRLINE+5,AXCLDMOVDI,SEGCONAMEMOVES,DILEADI,CONAMEREPNESCASBJNENEXTDECDIMOVBX,DINEXT:┇STRINGDB‘ThedateisFEB&03’MOVCX,18MOVAL,‘&’CLDMOVDI,SEGSTRINGMOVES,DILEAREPNEDI,STRINGSCASBNEXTDECDINEXT:┇3.31假設(shè)數(shù)據(jù)段中數(shù)據(jù)定義如下：STUDENT_NAMEDB30DUP(?)STUDENT_ADDRDB9DUP(?)PRINT_LINEDB132DUP(?)MOVDI,DSMOVES,DICX,132MOVAL.,‘’CLDLEADI,PRINT_LINEREPSTOSBCX,9MOVCLDLEADI,STUDENT_ADDRREPNESCASBJNENO_DASHDECDINO_DASH:┇(3)MOVCX,9DLEADI,STUDENT_ADDRADDDI,9-1REPNESCASBJNENO_DASHINCDINO_DASH:┇(4)MOVCX,30CLDLEAREPEDI,STUDENT_NAMESCASBNEXTMOVCX,30LEADI,STUDENT_NAMEREPSTOSBNEXT:┇CLDCX,30LEASI,STUDENT_NAMELEADI,PRINT_LINEREPMOVSBMOVCX,9DLEASI,STUDENT_ADDR+9-1LEADI,PRINT_LINE+132-1REPMOVSBMOVCX,5CLDMOVDI,SEGOLDSMOVDS,DIMOVES,DILEALEAREPESI,OLDSDI,NEWSCMPSBJNENEW_LESS┇NEW_LESS:┇OWJAEXCEEDJGEXCEEDJOOVERFLOWJLEEQ_SMAJBEEQ_SMA段：ADDAX,BXSUBAX,BXJNCL3JMPSHORTL5AXBXDCHHCHHDH(4)D023H9FD0H(5)94B7HB568H3.35指令CMPAX,BX后面跟著一條格式為J…L1的條件轉(zhuǎn)移指令，其中…可以是B、NB、BE、NBE、AXBX(1)1F52H1F52H(2)88C9H88C9H(3)FF82H007EH(4)58BAH020EH(5)FFC5HFF8BH(6)09A0H1E97H(7)8AEAHFC29H(8)D367H32A6H引起轉(zhuǎn)移到L1？2)JNB、JBE、JNL、JLE3)JNB、JNBE、JL、JLE(4)JNB、JNBE、JNL、JNLE5)JNB、JNBE、JL、JLEJBJBEJLJLE)JB、JBE、JNL、JNLE8)JNB、JNBE、JL、JLEMOVDX,X+2MOVAX,XADDAX,XADCDX,X+2CMPDX,Y+2JLL2JGL1CMPAX,YJBEL2L1：MOVAX,1JMPSHORTEXITL2：MOVAX,2EXIT：INT20HYEY程序開始程序開始MOVAL,STATUSBEN轉(zhuǎn)去執(zhí)行轉(zhuǎn)去執(zhí)行ROUTINE_1:┇YNYYROUTINE_2:┇JMPEXITROUTINE_4:┇轉(zhuǎn)去執(zhí)行N轉(zhuǎn)去執(zhí)行EXIT:INT20HROUTINE_2僅一位為1執(zhí)行ROUTINE_13.38在下列程序的括號(hào)中分別填入如下指令：(3)LOOPNEL20執(zhí)行EXIT程序結(jié)束試說明在三種情況下，當(dāng)程序執(zhí)行完后，AX、BX、CX、DX四個(gè)寄存器的容分別是什么？3.44題的程序流程圖TITLECODESGEXLOOP.COMSEGMENTASSUMECS:CODESG,DS:CODSEG,SS:CODSEGORG100HMOVBX,02MOVDX,03MOVCX,04L20:INCAXADDBX,AXSHRDX,1RETCODESGENDSENDBEGIN慮以下的調(diào)用序列：答：每次調(diào)用及返回時(shí)的堆棧狀態(tài)圖如下所示：(1)MAIN調(diào)0A0A0A0A0A0A0A)MOV[EAX+2*EBX]，CL(3)MOVDH，[EBX+4*EAX+1000H]答：(1)PA=(DS)*10H+EA=00100H+00001000H+00002000H=00003100H(2)PA=(DS)*10H+EA=00100H+00001000H+2*00002000H=00005100H(3)PA=(DS)*10H+EA=00100H+00002000H+4*00001000H+1000H=00007100HDSIEAXXECX3.42說明下列指令的操作(1)PUSHAX；將(AX)壓入堆棧POPESI；將堆棧中的雙字彈出到ESI寄存器中(3)PUSH[BX]；將((BX))對(duì)應(yīng)存儲(chǔ)單元中的字壓入堆棧)PUSHAD；32位通用寄存器依次進(jìn)棧3.43請(qǐng)給出下列各指令序列執(zhí)行完后目的寄存器的容。1)MOVEAX，299FF94H2)MOVEBX，40000000(3)MOVEAX，39393834HMOVEDX，9FE35DHH3.44請(qǐng)給出下列各指令序列執(zhí)行完后目的寄存器的容。MOVBX，-12MOVSXEBX，BX；(EBX)=0FFFFFFF4HMOVCL8MOVSXEDX，CL；(EDX)=0FFFFFFF8HMOVAH，7MOVZXECX，AH；(ECX)=00000007HXHMOVZXEBX，AX；(EBX)=00000099HMOVECX，307F455HMOVBX，98H3.47請(qǐng)編寫一程序段，要求把ECX、EDX和ESI的容相加，其和存入EDI寄存器中(不考慮溢出)。ADDEDI，ECXADDEDI，EDXADDEDI，EDXADDEDI，ESIADDEDI，ESIMOVAH，0；假定為無符號(hào)數(shù)，否則用CBW指令即可DIVCLMOVAH，0SHLAX，1MOVDX，AX址的存單元容為偏移地址的單元去執(zhí)行指令。UPDB0DOWNDB0存儲(chǔ)器TABLEDB100HDUP(?)；數(shù)組BEGIN：MOVCX，100HMOVBX，-1MOVSI，0MOVDI，0L1：INCBXCMPTABLE[BX]，42HL2：INCL3：INCL4：LOOPLLLLDILMOVUP，SIMOVDOWN，DI第四章.習(xí)題4.1指出下列指令的錯(cuò)誤：(1)MOVAH,BX；寄存器類型不匹配(2)MOV[BX],[SI]；不能都是存儲(chǔ)器操作數(shù)IDI(4)MOVMYDAT[BX][SI],ES:AX；AX寄存器不能使用段超越(5)MOVBYTEPTR[BX],1000；1000超過了一個(gè)字節(jié)的圍(6)MOVBX,OFFSETMYDAT[SI]；MYDAT[SI]已經(jīng)是偏移地址,不能再使用OFFSET(7)MOVCS,AX；CS不能用作目的寄存器(8)MOVECX,AX；兩個(gè)操作數(shù)的數(shù)據(jù)類型不同(1)CMP15,BX；錯(cuò)，立即數(shù)不能作為目的操作數(shù)CMPOP1,25(3)CMPOP1,OP2；錯(cuò)，不能都是存儲(chǔ)器操作數(shù)(4)CMPAX,OP1；錯(cuò)，類型不匹配，應(yīng)為CMPax,wordptrop14.3假設(shè)下列指令中的所有標(biāo)識(shí)符均為類型屬性為字的變量，請(qǐng)指出下列哪些指令是非法的？它們的(1)MOVBP,AL；錯(cuò)，寄存器類型不匹配(2)MOVWORD_OP[BX+4*3][DI],SP(3)MOVWORD_OP1,WORD_OP2；錯(cuò)，不能都是存儲(chǔ)器操作數(shù)(5)MOVSAVE_WORD,DS(6)MOVSP,SS:DATA_WORD[BX][SI]MOVBXSI；錯(cuò)，[BX][SI]未指出數(shù)據(jù)類型(8)MOVAX,WORD_OP1+WORD_OP2(9)MOVAX,WORD_OP1-WORD_OP2+100(10)MOVWORD_OP1,WORD_OP1-WORD_OP2(1)ADDVAR1,VAR2；不能都是存儲(chǔ)器操作數(shù)(2)SUBAL,VAR1；數(shù)據(jù)類型不匹配(3)JMPLAB[SI]；LAB是標(biāo)號(hào)而不是變量名，后面不能加(4)JNZVAR1；VAR1是變量而不是標(biāo)號(hào)(5)JMPNEARLAB；應(yīng)使用NEARPTR4.5畫圖說明下列語句所分配的存儲(chǔ)空間及初始化的數(shù)據(jù)值。(1)BYTE_VARDB‘BYTE’,12,-12H,3DUP(0,?,2DUP(1,2),?)(2)WORD_VARDW5DUP(0,1,2),?,-5,‘BY’,‘TE’,256H答：答案如下圖所示。BYTEVARHWORD_VAR00HDB50H,51H59H00H54H01H45H00HDB‘PQ’0DH02HEEH00HDW‘QP’00H將上面┇-內(nèi)容再┇ORG5150H01H重復(fù)4次┇DW$02H-4.7請(qǐng)?jiān)O(shè)置一個(gè)數(shù)據(jù)段DATASG，其中定義以下字符變量或數(shù)據(jù)變量。02H(6)(6)FLD6B為10個(gè)零的字節(jié)變量；02H54H(7)FLD7B為零件名(ASCII碼)及其數(shù)量(十進(jìn)制數(shù))的表格：-56HPART12PART2504.5題答案PART314FLDW為本段中字?jǐn)?shù)據(jù)變量和字節(jié)數(shù)據(jù)變量之間的地址差。FLD1BFLD2BFLD3BFLD4BFLD5BFLD6BFLD7BDB‘personalcomputer’DB32DB20HDB01011001BDB‘32654’DB10DUP(0)FLD1WFLD2WFLD3WFLD4WFLD5WFLD6WDATASGENDSDW0FFF0HDW01011001BDWFLD7BDW5,6,7,8,9DW5DUP(0)DWFLD1W-FLD1B4.8假設(shè)程序中的數(shù)據(jù)定義如下：PARTNODW?PNAMEDB16DUP(?)COUNTDD?PLENTHEQU$-PARTNO4.9有符號(hào)定義語句如下：BUFFDB1,2,3,‘123’EBUFFDB0LEQUEBUFF-BUFF4.10假設(shè)程序中的數(shù)據(jù)定義如下：LNAMEDB30DUP(?)ADDRESSDB30DUP(?)CITYDB15DUP(?)CODE_LISTDB1,7,8,3,2(2)MOVSI,WORDPTRCODE_LISTODELENGTHEQUCODELISTCODELISTNUMDATA_LISTMAXMINSEGMENTDB5DW-1,0,2,5,4,5DUP(?)DW?DW?DATA_SEGENDS；CODE_SEGSEGMENTMAINSTART:；PROCASSUMEPUSHSUBPUSHMOVAX,MOVDS,MOVCX,FARCS:CODE_SEG,DS:DATA_SEGDS；設(shè)置返回DOSAX,AXAXDATA_SEGAX4LEABX,DATA_LISTMOVAX,[BX]MOVMAX,AXMOVMIN,AXROUT1:ROUT2:ROUT3:ADDBX,2MOVAX,[BX]CMPAX,MAXJNGEROUT2MOVMAX,AXCMPAX,MINJNLEROUT3MOVMIN,AXLOOPROUT1RETMAINENDPCODE_SEGENDS；ENDSTART給出等值語句如下：ALPHAEQU100GAMMAEQU2(1)ALPHA*100+BETA；=2729H(2)ALPHAMODGAMMA+BETA；=19H(3)(ALPHA+2)*BETA–2；=9F4H(4)(BETA/3)MOD5；=3H(5)(ALPHA+3)*(BETAMODGAMMA)；=67H(6)ALPHAGEGAMMA；=0FFFFH(7)BETAAND7；=01H(8)GAMMAOR3；=03HTABLEADW10DUP(?)TABLEBDB10DUP(?)TABLECDB‘1234’┇MOVAX,LENGTHTABLEA；匯編成MOVAX,000AHMOVBL,LENGTHTABLEB；匯編成MOVBL,000AHMOVCL,LENGTHTABLEC；匯編成MOVCL,0001HTABLEADW20DUP(?)TABLEBDB‘ABCD’4.15指出下列偽操作表達(dá)方式的錯(cuò)誤，并改正之。(1)DATA_SEGSEG；DATA_SEGSEGMENT(偽操作錯(cuò))(2)SEGMENT‘CODE’；SEGNAMESEGMENT‘CODE’(缺少段名(3)MYDATASEGMENT/DATA；MYDATASEGMENT┇ENDS；MYDATAENDS(缺少段名字)(4)MAIN_PROCPROCFAR；刪除ENDMAIN_PROC也可以┇ENDMAIN_PROC；MAIN_PROCENDP；上下兩句交換位置MAIN_PROCENDP；ENDMAIN_PROC4.16按下面的要求寫出程序的框架HKDATA_SEGSEGMENTAT0E000HARRAY_BLABELBYTEARRAY_WDW50DUP(?)DATA_SEGENDS；以上定義數(shù)據(jù)段；STACK_SEGSEGMENTPARASTACK‘STACK’DW100HDUP(?)TOSLABELWORDSTACK_SEGENDS；以上定義堆棧段；CODE_SEGSEGMENTMAINPROCFARASSUMECS:CODE_SEG,DS:DATA_SEG,SS:STACK_SEGORG1000HSTART:MOVAX,STACK_SEGMOVSS,START:MOVSP,OFFSETTOSPUSHDSSUBAX,AXPUSHAXMOVAX,DATA_SEGMOVDS,AXMAIN┇MAINRETENDPCODE_SEGENDS；ENDSTARTD_SEGSEGMENTAUGWLABELWORDAUGENDDD99251SUMDD?D_SEGENDS；以上定義數(shù)據(jù)段；E_SEGSEGMENTADDWLABELWORDADDENDDD-15962E_SEGENDS；以上定義附加段；C_SEGSEGMENTMAINPROCASSUMEFARCS:C_SEG,DS:D_SEG,ES:E_SEGSTART:PUSHBPUSHDS；設(shè)置返回DOSAX,AXAXMOVAX,D_SEGMOVDS,AXMOVAX,E_SEGMOVES,AX；MOVAX,AUGWMOVBX,AUGW+2ADDAX,ES:ADDWADCBX,ES:ADDW+2MAINC_SEGMOVWORDMOVWORDRETENDPENDSPTRSUM,AXPTR[SUM+2],BX；ENDSTART4.18請(qǐng)說明表示程序結(jié)束的微操作和結(jié)束程序執(zhí)行的語句之間的差別。它們?cè)谠闯绦蛑袘?yīng)如何表示？答：表示程序結(jié)束的微操作是指示匯編程序MASM結(jié)束匯編的標(biāo)志，在源程序中用END表示；結(jié)束程序執(zhí)行的語句是結(jié)束程序運(yùn)行而返回操作系統(tǒng)的指令，在源程序中有多種表示方法，比如TDBDDBDWWVALAL，BVALCL，WVALAL，BVAL+1BYTEPTR[BX]，2CL，BYTEPTRWVAL第五章.習(xí)題5.1試編寫一個(gè)匯編語言程序，要求對(duì)鍵盤輸入的小寫字母用大寫字母顯示出來。BEGIN:MOVAH,1；從鍵盤輸入一個(gè)字符的DOS調(diào)用SEGMENTSEGMENTDBENDSMOVMOVLEAMOVANDMOVINCSHRDECJNZMOVMOVMOVMOVRETSTOP:CMPCMPBMOVMOVRETSTOPSTOPAL,20HDL,ALAH,2BEGIN5.2編寫程序，從鍵盤接收一個(gè)小寫字母，然后找出它的前導(dǎo)字符和后續(xù)字符，再按順序顯示這三個(gè)BEGIN:DISPLAY:STOP:MOVAH,CMPAL,BCMPAL,DECMOVDL,MOVCX,MOVAH,LOOPRET1‘a(chǎn)’STOP‘z’STOPALAL32DLDISPLAYDSEGSTOREDSEGBEGIN:A10:B10:STOP:4DUP(?)CL,4CH,4BX,STOREDX,AXDX,0FHBXAX,CLHA10DL,STORECL,STORE+1BL,STORE+2AL,STORE+35.4試編寫一程序，要求比較兩個(gè)字符串STRING1和STRING2所含字符是否完全相同，若相同則顯示DSEGSEGMENTSTRING1DBSTRING2DBYESDBNODBDSEGENDS‘Iamastudent.’‘Iamastudent!’；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEG,ES:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEG；BEGIN:LEASI,STRING1；設(shè)置串比較指令的初值LEADI,STRING2CLDMOVCX,STRING2-STRING1REPECMPSB；串比較JNEDISPNOLEADX,YES；顯示MATCHJMPDISPLAYDISPNO:LEADX,NO；顯示NOMATCHDISPLAY:MOVAH,9；顯示一個(gè)字符串的DOS調(diào)用INT21HRETMAINENDPCSEGENDS；ENDSTART5.5試編寫一程序，要求能從鍵盤接收一個(gè)個(gè)位數(shù)N，然后響鈴N次(響鈴的ASCII碼為07)。BEGIN:BELL:STOP:MOVAH,1INT21HSUBAL,‘0’JBSTOPCMPAL,9JASTOPCBWMOVCX,AXJCXZSTOPMOVDL,07HMOVAH,2CALLLOOPRETDELAY100msBELL這兩個(gè)數(shù)組中數(shù)據(jù)的個(gè)數(shù)顯示出來。DSEGSEGMENTCOUNTARRAYCOUNT1ARRAY1COUNT2ARRAY2ZHENFUCRLFDSEGEQU20DW20DUP(?)；存放數(shù)組DB0；存放正數(shù)的個(gè)數(shù)DW20DUP(?)；存放正數(shù)DB0；存放負(fù)數(shù)的個(gè)數(shù)DW20DUP(?)；存放負(fù)數(shù)DB0DH,0AH,‘$’ENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:MOVCX,COUNTLEABX,ARRAYLEASI,ARRAY1LEADI,ARRAY2BEGIN1:MOVAX,[BX]JSFUSHUMOV[SI],AX；是正數(shù)，存入正數(shù)數(shù)組INCCOUNT1；正數(shù)個(gè)數(shù)+1ADDSI,2JMPSHORTNEXTFUSHU:MOV[DI],AX；是負(fù)數(shù)，存入負(fù)數(shù)數(shù)組NEXT:MAININCCOUNT2ADDDI,2ADDBX,2LOOPBEGIN1LEADX,ZHENMOVAL,COUNT1CALLDISPLAYLEADX,FUMOVAL,COUNT2CALLDISPLAYRETENDP；DISPLAYPROCNEAR；顯示子程序OSINT21HAAM；將(AL)中的二進(jìn)制數(shù)轉(zhuǎn)換為二個(gè)非壓縮BCD碼MOVDL,AHMOVAH,2INT21HADDAL,‘0’MOVDL,ALMOVAH,2INT21HLEADX,CRLFMOVAH,9INT21HRETDISPLAYENDPCSEGENDS；ENDSTARTBEGIN:MOVBX,MOVCX,0COMPARE:MOVAX,ADDBX,DATA[BX]2；取數(shù)組的第一個(gè)偶數(shù)TESTLOOPNZJNZJCXZAX,01HCOMPARESTOPSTOP；不是，比較下一個(gè)數(shù)；最后一個(gè)數(shù)是偶數(shù)，即為最小偶數(shù)，退出COMPARE1:MOVDX,ADDBX,DATA[BX]2；取數(shù)組的下一個(gè)偶數(shù)TESTJNZCMPAX,JLEDX,01HNEXTDXNEXT；不是，比較下一個(gè)數(shù)MOVAX,DXNEXT:LOOPCOMPARE1STOP:RETBEGIN:MOVDL,MOVCX,08COMPARE:TESTAX,03HNOEQUALDLNOEQUAL:RORAX,RORAX,LOOP11COMPARE備判斷下一個(gè)數(shù)ADDDL,MOVAH,INT‘0’2STOP:RETBEGIN:AF:BINARY:DISPN:DISP:STOP:MOVBX,0MOVCH,4MOVCL,4SHLBX,CLMOVAH,1INT21HCMPAL,30HCMPAL,39HJAAFANDAL,0FHJMPBINARYANDAL,11011111BCMPAL,41HCMPAL,46HANDAL,0FHADDAL,9ORBL,ALDELCHMOVCX,16MOVDL,0ROLBX,1RCLDL,1ORDL,30HMOVAH,2INT21HLOOPDISPRET；不是‘0~F’的數(shù)重新輸入；是‘0~9’嗎？；不是，轉(zhuǎn)‘A~F’的處理；轉(zhuǎn)換為：0000B~1001B；不是‘A~F’的數(shù)重新輸入；不是‘A~F’的數(shù)重新輸入；轉(zhuǎn)換為：1010B~1111B；將鍵盤輸入的數(shù)進(jìn)行組合.10設(shè)有一段英文，其字符變量名為ENG，并以$字符結(jié)束。試編寫一程序，查對(duì)單詞SUN在該文中的出現(xiàn)次數(shù)，并以格式“SUN：xxxx”顯示出次數(shù)。DSEGSEGMENTENGDBDISPDBDATDBKEYWORDDBDSEGENDS‘Hereissun,sun,…,$’DH0AH,‘$’‘sun’；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEG,ES:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:MOVAX,0MOVDX,DISP-ENG-2；計(jì)算ENG的長(zhǎng)度(每次比較sun,因此比較次數(shù)-2)LEABX,ENGCOMP:MOVDI,BXLEASI,KEYWORDMOVCX,3REPECMPSBJNZNOMATCHINCAXADDBX,2NOMATCH:INCBXDECDXJNZCOMPDONE:MOVCH,4MOVCL,4LEABX,DATDONE1:ROLAX,CLMOVDX,AXANDDL,0FHADDDL,30HCMPDL,39HJLESTOREADDDL,07HSTORE:MOV[BX],DLINCBXDECCHJNZDONE1DISPLAY:LEADX,DISPMOVAH,09HINT21HRETMAINENDPCSEGENDS；是“A~F”所以要加7；ENDSTART5.11從鍵盤輸入一系列以$為結(jié)束符的字符串，然后對(duì)其中的非數(shù)字字符計(jì)數(shù)，并顯示出計(jì)數(shù)結(jié)果。DSEGBUFFCOUNTDSEGBEGIN:NEXT:SEGMENTDBDW50SEGMENTDBDW0ENDS┇LEAMOVLEAMOVMOVINTMOVINCCMPJNZLEAMOVINCCMPCOUNT,0AH,01BXBX,BUFFCL,[BX]BXDISP；從鍵盤輸入一個(gè)字符的功能調(diào)用；對(duì)非數(shù)字字符進(jìn)行計(jì)數(shù)；是$結(jié)束符，則轉(zhuǎn)去顯示CMPCL,30HJBNEXTCMPCL,39HNEXTCOUNEXTCOUNTNEXT┇DISP:DSEGSEGMENTMEMDW100DUP(?)DSEGENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:MOVSI,(100-1)*2MOVBX,-BEGIN:MOVCX,100COMP:ADDBX,COMP:CMPMEM[BX],0JZCONSLOOPCOMPCONS:CONS1:JMPFINISHMOVCONS:CONS1:CMPDI,SIJAENOMOVMOVAX,MEM[DI+2]MOVMEM[DI],AXADDDI,2JMPCONS1NOMOV:MOVWORDPTR[SI],0；后面的元素向前移位FINISH:MAINLOOPCOMPRETENDPCSEGENDS；ENDSTARTDSEGSEGMENTSTRINGDB100DUP(?)DSEGENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:MOVSI,0；(SI)作為地址指針的變化值MOVCX,100REPEAT:MOVAL,STRING[SI]CMPAL,30HJBGO_ONCMPAL,39HJAGO_ONJMPEXITGO_ON:INCSILOOPREPEATEXIT:RETMAINENDPCSEGENDS；以上定義代碼段；ENDSTART最多的數(shù)及其出現(xiàn)次數(shù)分別存放于AX和CX中。DSEGSEGMENTTABLEDW100HDUP(?)；數(shù)組中的數(shù)據(jù)是按增序排列的DATADW?COUNTDW0DSEGENDS；CSEGSEGMENTMAINSTART:BEGIN:NEXT:COMP:PROCFARASSUMECS:CSEG,DS:DSEGPUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGMOVCX,100H；循環(huán)計(jì)數(shù)器MOVSI,0MOVDX,0MOVAX,TABLE[SI]CMPTABLE[SI],AX；計(jì)算一個(gè)數(shù)的出現(xiàn)次數(shù)JNEADDRINCDXADDSI,2LOOPCOMPADDR:DONE:MAINCSEGCMPDX,COUNTJLEDONEMOVCOUNT,DXMOVDATA,AXLOOPNEXTMOVCX,COUNTMOVAX,DATARETENDPENDS；目前此數(shù)出現(xiàn)的次數(shù)最多，記下次數(shù)；出現(xiàn)最多的次數(shù)存入(CX)；出現(xiàn)最多的數(shù)存入(AX)；ENDSTART數(shù)據(jù)段的M+2n單元中，并將該數(shù)的偏移地址存放在M+2(n+1)單元中。DSEGSEGMENTnMDATAADDRDSEGEQU100HDWnDUP(?)DW?DW?ENDSM(n+1)單元；CSEGSEGMENTMAINPROCASSUMEFARCS:CSEG,DS:DSEGSTART:PUSHSUBPUSHMOVAX,MOVDS,DS；設(shè)置返回DOSAX,AXAXDSEGBEGIN:ZHEN:COMP:MOVCX,nLEADI,MMOVAX,[DI]MOVADDR,DICMPAX,0JNSZHENNEGAXMOVBX,[DI]CMPBX,0JNSCOMPNEGBXCMPAX,BXJAEADDRESSMOVAX,BXMOVADDR,DI；記下絕對(duì)值最大的數(shù)的地址；是正數(shù)，即為絕對(duì)值，轉(zhuǎn)去判斷下一個(gè)數(shù)；不是正數(shù)，變?yōu)槠浣^對(duì)值；是正數(shù)，即為絕對(duì)值，轉(zhuǎn)去比較絕對(duì)值大??；不是正數(shù)，變?yōu)槠浣^對(duì)值；記下絕對(duì)值最大的數(shù)的地址ADDRESS:ADDDI,2LOOPZHENMAINCSEGMOVDATA,AXRETENDPENDS；ENDSTARTAX寄存器中；并求出數(shù)組中有多少個(gè)數(shù)小于此平均值，將結(jié)果放在BX寄存器中。DSEGSEGMENTDATADW100HDUP(?)DSEGENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:MOVCX,100H；循環(huán)計(jì)數(shù)器MOVSI,0MOVBX,0；和((DI),(BX))的初始值MOVDI,0NEXT:MOVAX,DATA[SI]CWDADDBX,AX；求和ADCDI,DX；加上進(jìn)位位ADDSI,2LOOPNEXTMOVDX,DI；將((DI),(BX))中的累加和放入((DX),(AX))中MOVAX,BXMOVCX,100HIDIVCX；帶符號(hào)數(shù)求平均值，放入(AX)中MOVBX,0MOVSI,0COMP:CMPAX,DATA[SI]；尋找小于平均值的數(shù)JLENOINCBX；小于平均值數(shù)的個(gè)數(shù)+1NO:ADDSI,2LOOPCOMPRETMAINENDPCSEGENDS；以上定義代碼段；ENDSTARTDSEGSEGMENTMEMDB4DUP(?)NDW2A49HDSEGENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:ROTATE:NEXT:MOVCH,4MOVCL,4MOVAX,NLEABX,MEMMOVDL,ALANDDL,0FHADDDL,30HCMPDL,3AHJLNEXTADDDL,07HMOV[BX],DLINCBXRORAX,CLDECCHJNZROTATERETMAINENDPCSEGENDS；以上定義代碼段；ENDSTART1。)DSEGSEGMENTGRADEDW30DUP(?)；假設(shè)已預(yù)先存好30名學(xué)生的成績(jī)RANKDW30DUP(?)DSEGENDS；CSEGSEGMENTMAINSTART:BEGIN:LOOP1:PROCASSUMEPUSHSUBPUSHMOVAX,MOVDS,MOVDI,MOVCX,PUSHMOVCX,MOVSI,FARCS:CSEG,DS:DSEGDS；設(shè)置返回DOSAX,AXAXDSEG00；外循環(huán)計(jì)數(shù)器X；循環(huán)計(jì)數(shù)器0MOVAX,GRADE[DI]MOVDX,1LOOP2:CMPGRADE[SI],AXJBEGO_ONINCDXGO_ON:ADDSI,2LOOP2LOOP2XPOPMOVRNAK[DI],DXADDDI,2LOOPLOOP1RETMAINENDPCSEGENDS；以上定義代碼段；ENDSTARTDSEGSEGMENTADW15DUP(?)BDW20DUP(?)CDW15DUP(‘’)DSEGENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHDS；設(shè)置返回DOSSUBAX,AXPUSHAXMOVAX,DSEGBEGIN:LOOP1:LOOP2:NO:MAINCSEGMOVSI,0MOVBX,0MOVCX,15PUSHCXMOVCX,20MOVDI,0MOVAX,A[SI]CMPB[DI],AXJNENOMOVC[BX],AXADDBX,2ADDDI,2LOOPLOOP2ADDSI,2POPLPOPLOOPRETENDPLOOP1ENDS；ENDSTARTDSEGABCDDSEGSEGMENTDWDWDWDWENDS???0；CSEGSEGMENTMAINPROCASSUMEFARCS:CSEG,DS:DSEGSTART:PUSHSUBPUSHMOVAX,MOVDS,DS；設(shè)置返回DOSAX,AXAXDSEGBEGIN:CMPA,0JENEXTCMPB,0JENEXTCMPC,0JENEXTMOVAX,AADDAX,BADDAX,CMOVD,AXJMPSHORTEXITNEXT:MOVA,0MOVB,0MOVC,0EXIT:RETMAINENDPCSEGENDS；ENDSTARTDSEGSEGMENTARRAYDW3DUP(?)DSEGENDS；CSEGSEGMENTMAINPROCFARASSUMECS:CSEG,DS:DSEGSTART:PUSHSUBPUSHMOVAX,MOVDS,DS；設(shè)置返回DOSAX,AXAXDSEGBEGIN:LEASI,ARRAYMOVDX,0MOVAX,[SI]MOVBX,[SI+2]CMPAX,BXJNENEXT1INCDXNEXT1:CMP[SI+4],AXJNENEXT2INCDXNEXT2:CMP[SI+4],BXJNENUMINCDXNUM:CMP