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Week1Supposeadatais376a8d43hinHexadecimal,howtoplaceitfromtheaddressthat20000hbylittleendianandbigendian?答：小端存储（littleendian）20000H到20003H地址分别存放的是43H,8DH,6AH和37H大端存储（bigendian）20000H到20003H地址分别存放的是37H,6AH,8DH和43HConvertthefollowdecimalintegerintoBCDandpackedBCDnumber.a)9b)76c)251答：BCD码：a)09Hb)0706Hc)020501H或a)00001001b)0000011100000110c)000000100000010100000001packedBCD:a)09Hb)76Hc)0251H或a)00001001b)01110110c)0000001001010001ConvertthefollowpackedBCDnumberintodecimalnumber:a)10001001b)00001001c)00110010d)00000001答：a)89b)9c)32d)1Convertthefollowbinarynumberintosigneddecimalnumber:a)10000001b)00110011c)10010010d)10001001答：a)-127b)51c)-110d)-119Week2Howmany8-bit,16-bitand32bitgeneral-purposeregistersarethereintheIA-32CPU?Whatarethem?RightAnsweris:8位通用寄存器有AL,AH,BL,BH,CL,CH,DL,DH共8个16位通用寄存器有AX,BX,CX,DX,SI,DI,BP,SP共8个32位通用寄存器有EAX,EBX,ECX,EDX,ESI,EDI,EBP,ESP共8个Intherealmode,findthememoryaddressofthenextinstructionexecutedbythemicroprocessorforthefollowingCS:IPcombinations:a)CS=1000HandIP=2000Hb)CS=1900HandIP=0200Hc)CS=1900HandIP=1000HD)CS=5679HandIP=CDEFHPleasegivethephysicalmemoryaddress.RightAnsweris:Physicalmemoryaddressis：a)12000Hb)19200Hc)1A000Hd)6357FHWillanoverflowoccurwhensignedFFHisaddedtoasigned01H?Howaboutsigned70Haddasinged40H?RightAnsweris:FFH+1，OF=0,不产生溢出70H+40H会产生溢出，OF=1Week3Supposethat(DS)=2000H,(ES)=2100H,(SS)=1500H,(SI)=00A0H,(BX)=0100H,(BP)=0010H,theoffsetaddressofVALis0050H。Determinethephysicaladdressaccessedbyeachofthefollowinginstructions,assumingrealmodeoperation.（1）MOVAX，[100H]（2）MOVVAL，BH（3）MOVAX，[BX]（4）MOVAX，ES：[BX]（5）MOVAX，[BP]（6）MOVAX，[SI]（7）MOVAX，[BX+50H]（8）MOVVAL[BX]，BP（9）MOVAH，[BX+SI]（10）MOVAL，VAL[BX][SI]RightAnsweris:（1）20100H（2）20050H（3）20100H（4）21100H（5）15010H（6）200A0H（7）20150H（8）20150H（9）201A0H（10）201F0HWritethedirectivetodefinethefollowingvariable:1)Declareanwordarraynamedbuffwhichhas800elementswithoutinitial.2)DeclareastringnamedSTRINGwithinitialvalue“MASM”,”1024”RightAnsweris:buffword800dup(?)或buffdw800dup(?)STRINGbyte“MASM”,“1024”或STRINGdb“MASM”,“1024”What’sdifferentbetweentheEQUoperatorand=operator?RightAnsweris:EQU定义的常量不能再赋值，=定义的可以再次定义Week4Whatisthepurposeofthe.386,.486p,.modeldirective?RightAnsweris:.386表示该程序使用386指令.486p表示该程序使用486保护模式指令What’swrongwiththefollowinginstructions:（l）MOVAH，BX（2）MOV[BX]，[SI]（3）MOVAX，[SI][DI]（4）MOVMYDAT[BX][SI]，ES:AX（5）MOVBL，1000（6）MOV2000，BX（7）MOVCS，AXRightAnsweris:1）目的操作数与源操作数不等宽2）两个操作数不能都是存储器操作数3）不能两个都是变址寄存器4）寄存器不能用段超越5）1000超出8位数范围6）立即数不能做目的操作数7）CS不能作为MOV的目的操作数Writeashortsequenceofinstructionsthatloadthedatasegmentregisterwith1000HRightAnsweris:MOVAX,1000HMOBDS,AXWhatisthedifferencebetweentheLEASI,NUMBinstructionandtheMOVSI,OFFSETNUMBinstruction?RightAnsweris:LEA指令的源操作数可以是任意寻址方式的存储器操作数，OFFSET形式的语句只能是变量名或标号名。OFFSET形式是立即数寻址，速度快。Supposethedatadeclaredasfollow:ORG26HNUM1BYTE12H,13HNUM2WORD$+3ORG$+4NUM3WORD325AH,5A6BHQuestion:Howdothedatastoreinmemory?Pleasedrawthememorymap.RightAnsweris:按存储地址顺序从低到高，各字节分别是:12H,13H,2BH,00,-,-,-,-,5AH,32H,6BH,5AH,Supposethedatadeclaredasfollow:ORG100HNUM1DB7,43H,‘AB’NUM2DW1,?NUM3DB2DUP(1,2DUP(5))Question:Howdothedatastoreinmemory?Pleasedrawthememorymap.RightAnsweris:按存储地址顺序从低到高，各字节分别是:7,43H,’A’,’B’,1,0,-,-,1,5,5,1,5,5Week5IfAX=1001HandDX=20FFH,listthesumandthecontentsofeachflagregisterbit(CF,AF,SF,ZFandOF).RightAnsweris:SUM=3100H,CF=0，AF=1，OF=0，SF=0，ZF=0ExplainthedifferencebetweentheSUBandCMP,ANDandTEST.RightAnsweris:CMP和SUB做同样操作，但不保存结果；TEST和AND做同样操作，但不保存结果Programmingin16-bitassemblylanguage1)AddthedatainAL,BLandCL,lettheresultintheDX2)AddthedatainALandaBYTEdatainmemorywhichsymboladdressisNUMusingBase-Plus-Indexaddressing.TheresultinthenextaddressofNUM.RightAnsweris:1）MOVDX,0ADDDL,ALADCDH,0ADDDL,BLADCDH,0ADDDL,CLADCDH,02)MOVBX,OFFSETNUM(或LEABX,NUM)MOVSI,0ADDAL,[BX][SI]INCSIMOV[BX][SI],ALWeek6HowmanywayyouknowtocleartheAXwith0?Pleasegivetheway.RightAnsweris:1）MOVAX,02)ANDAX,03)SUBAX,AX4)XORAX,AX5)MOVBL,0MULBLProgrammingin16-bitassemblylanguage1)UseshiftinstructiontocalculateDX=3×AX＋7×BX，supposethedataareassignedandnocarry.2)UselogicinstructiontoimplementtheconversionfromASCIIof0~9tounpackedBCD,andfromunpackedBCDtoASCII.3)ImplementDX.AXshiftright4bits.RightAnsweris:1)MOVCX,AXSHLAX,1;AX*2ADDAX,CX;AX*3MOVDX,BXMOVCL,3SHLDX,CL;BX*8SUBDX,BX;BX*7ADDDX,AX2)MOVAL,’0’LOP:ANDAL,0FHINCALCMPAL,0AHJNELOPMOVAL,0LOP1:ORAL,30HINCALCMPAL,‘:’JNELOP13)MOVCL，4SHRAX，CLMOVBL，DLSHRDX，CLSHLBL，CLORAH，BLSupposeDS=2000H.BX=1256H,SI=528FH,displacementTABLE=20A1H，[232F7H]=3280H,[264E5H]=2450H,afterrunfollownearindirectjumpinstructions,IP=?i.JMPBXii.JMPTABLE[BX]iii.JMP[BX][SI]RightAnsweris:i.1256Hii.3280Hiii.2450HDeterminethejumpconditionsofthefollowingprogram.（1）XORAX,1E1EHJEEQUAL（2）TESTAL,10000001BJNZTHERE（3）CMPCX,64HJBTHERERightAnsweris:(1)AX=1E1EH(2)AL的最高位和/或最低位为1(3)(CX)64HIfthetargetaddressinshortjumpbeyondrangeof-128~127,howtohandle?Giveanexample.RightAnsweris:可以把跳转条件做一些调整，利用无条件跳转的跳转范围大，来达到目的，如：CMPAX,33JELONG……;大于127字节LONG:改为：CMPAX,33JNESHOETJMPLONGSHORT:……;大于127字节……LONG:WhatisA10andA20inthefollowshortjumpinstruction?（inhexadecimal）（1）0110HEB