搜索
《汇编语言程序设计》试卷模拟试卷一20120后的括号内。1ABCD2ABPBIPCSPDPSW3SS=0805HSP=40HA08090HB08089HC0808EHD08450H4AMOVAXOFFSETABMOVAXACMOVAXA+1DMOVAXABX5AMOV[DI][SI]BMOV[DX+DI]ALCMOVALBXDMOVWORDPTR[BX]0100H6CS=2300HDS=2400HCS和DS值是PUSHCSPOPDSACS=0BCS=2400HCCS=2400HDCS=2300HDS=2300HDS=2300HDS=2400HDS=2300H7AX=MOVAL95HCBWA0095HB0F95HC0FF95HD9095H8ADDAXBXAX的内容为2BA0HPF1的叙述正确的是A1的个数为偶数B1的个数为奇数CD1的个数为偶数9AX=1000HNEGAXNOTAXAAX=1001HBAX=0FFFHCAX=1000HDAX=0111H10ALCFAMOVAL,00HBANDAL,00HCXORAL,ALDSUBAL,AL11CX寄存器低4位清零的正确指令是AANDCX0F0HBANDCX0FFF0HCANDCX1110HDMOVCX012BX=0的同时也使CF=0OF=0的指令是AXORBXBXBORBXBXCANDBXBXDCMPBXBX13REPEACX0且ZF0BCX0且ZF1CCX0或ZF0DCX0或ZF114JBEACF=0ORZF=0BCF=0ORZF=1CCF=1ORZF=0DCF=1ORZF=115BX和SIBX≥SI转向HIGH的正确指令是AJAEHIGHBJBEHIGHCJEGHIGHDJLEHIGH16Loop规定循环次数必定在寄存器ADX中BDL中CCX中DCL中17DECWORDPTRBXABCD18DAB中的内容是DAWDW2A05HDABDB0FAH:MOVALBYTEPRTDAWSUBDABALA0DAHB0FAHC0F5HD0D0H19使用汇编语言的伪操作命令定义VALDB2DUP(1,2,3DUP(3),2DUP(1,0))则在VAL存储区内前十个字节单元的数据是A1,2,3,3,2,1,0,1,2,3B1,2,3,3,3,3,2,1,0,1C2,1,2,3,3,2,1,0,2,1D1,2,3,3,3,1,0,1,0,120NUM=60HNUMLT60HAND30ORNUMGE60HAND40A0B30HC40D0FFFFH51201INTEL8088管理的存储器空间最大容量为______________I/O端口地址空间最大容量为_____________。28086/8088______。38086/8088________、________、________、________。4AX=2000HBX=1200HDS=3000HDI=0002H(31200H)=50H(31201H)=02H(31202H)=40HCF0。请写出下列各条指令独立执行完后有关寄存器及存储单元的内位ZF、CF的值。①ADDAX1200H问AX=________HZF=________②SUBAXBX问AX=________HZF=________③MOVAXBX问AX=________HCF=________④NEGWORDPTR1200H问(31200H)=________HCF=________5DS=2200HBX=1000HSI=0100HD=0A2B1H各种寻址方式下①使用D的直接寻址________②使用BX的寄存器间接寻址________③使用BX和D的寄存器相对寻址________④使用BX、SI和D的相对基址变址寻址________⑤使用BX、SI的基址变址寻址________8SP=0100HSS=0300HPSW=0240H0900H及偏移地址为00A0H的单元中有一条中断指令INT11HINT11HSP、CS、IP、PSWPC00000040B3188ACC4DF800F0--41F800F0C5188ACC0000005039E700F0A0198ACC--2EE800F0D2EF00F08086CPU中PSW1514131211109876543210OFDFIFTFSFZFAFPFCF32016NUMDB?MOVAH1INT21HCMPAL39HJBENEXTSUBAL7NEXTSUBAL30HMOVNUMAL1FNUM=_____28NUM=_____3_____。2阅读下面的程序段6MOVCX16MOVBX0MOVDX1AGTESTAXDXJZNEXTINCBXNEXTSHLDX1LOOPAG12(AX)=1234H完后(BX)=38BUF1DB-12-34-56-78-9BUF2DB9DUP(0)┇XORBLBLMOVSI0MOVCX9LOPMOVALBUF1[SI]TESTAL80HJZKINCBLNEGALKXORAL01HMOVBUF2[SI]ALINCSILOOPLOP??1BUF2中的各字节数据2BL=______________26121AX置全0AX置非0。试在空白处各VAR1DW××××VAR2DW××××┇MOVAXVAR1MOVBXVAR2XORAXBX①TESTBX8000H②MOVAX0NEXT??2DAT单元存放某一数N-6≦N≦6N的平方值SQRTABLEDB0,1,4,9,16,25DATDBNSQRDB?LEABX,TABLEMOVAL,DAT①JGENEXT②NEXTXLATHLT2201N1和N2分别有10H这些较小的数据存入N3数据区中。DATASEGMENTN1DB??N2DB??N3DB10HDUP(0)DATAENDS2SUMai=a1+a2+...+a20a1......a20依次存放在以BUF为首址的SUMDATASEGMENTBUFDW??SUMDW?DATAENDS模拟试卷二20120后的括号内。1序翻译成机器码程序的实用程序是ABCD2ADFOFSFBDFIFTFCOFCFPFDAFOFSF38088/8086A64K个字B32K个字节C1M个字节D64K个字节4DS=12A0HES=22A0HSS=33A0HBP=0174HMOVAXDSBP中源操作数的物理地址是A12A074HB22B74HC33B74HD12B74H5AMOVSS:[BX+DI],1000HBMOVDX,1000HCMOVWORDPTR[BX],1000HDMOVDS,2000H6SP=2110HPOPAXSP寄存器的值是A2111HB2112HC210FHD210EH7AX=MOVAL85HCBWA0085HB0F85HC0FF85HD8085H8MOVAL81HADDALALADCALALAAL=05HBAL=204HCAL=04HDAL=205H9NEGCF。设置CFACF置0BCF置0C0CF置0DCF置010CL寄存器的内容乘以4的正确操作是AROLCL1BMUL4CSHLCL1DMOVCL2ROLCL1SHLCL1SHLCLCL11AX=1200HCF置“1”的指令是AORAX,AXBNEGAXCNOTAXDDECAL12BHBH中AADDBH01HBORBH01HCXORBH01HDTESTBH01H13REPNEACX0且ZF0BCX0且ZF1CCX0或ZF0DCX0或ZF114JNBEACF=0ANDZF=0BCF=0ANDZF=1CCF=1ANDZF=0DCF=1ANDZF=115AL≥BL时分支去LOP1“CMPALBL”指令后应跟的分支指令是AJNCLOP1BJALOP1CJCLOP1DJGELOP116LoopNZ控制循环继续执行的条件是ACX≠0且ZF=1BCX≠0且ZF=0CCX≠0或ZF=1DCX≠0或ZF=0178086APROC和ENDPBNAME和ENDCSEGMENT和ENDSDSEGMENT和ASSUME18ARRAYDW69$+410H1$+4┇MOVAXARRAY+4设变量ARRAY的偏移量是0084HAX中的内容是A0009B008CHC0090HD0010H19BUF1DB3DUP02DUP123COUNTEQU$-BUF1符号COUNTA6B8C16D1820MOVCL55HXOR0F0H执行后CLA05HB50HC0A5HD0F5H61201n______________≤N______________。21M________的物理地址。3______________时由______________处理义体插入到______________处。子程序是在______________时由______________处理的。子程序执行速度比宏指令____________________________。4(SS)=2000HMOVSP,1234HPUSHAXSP=______________AL中的数据在内存地址______________AH中的数据在内存地址______________中。5DS=0F800HDI=180AH(0F980AH)=0064HMOVCL5SAR[DI]CL0F980A=______________CF=______________。6(AX)=0122HCF、SF、ZF、OF的初始状态为0“SUBAX0FFFHAX=________CF=_______SF=_______ZF=_______OF=_______。2481DS=2000HBX=0100H(20100H)=30H(20101H)=10H(1)执行MOVDXBXDX=_____执行LEADXBXDX=_____(2)简述两条指令区别2AL中第0、2位置13201lab14CODESGSEGMENTASSUMECS:CODESGBEGIN:MOVAX,1MOVBX,2MOVDX,3MOVCX,4L20:INCAXADDBX,AXSHRDX,1LOOPEL20lab1:MOVAH,4CHINT21HCODESGENDSENDBEGIN28STRINGDB‘ABCDEFGHIJ'':MOVAH,01;从键盘输入字符1~9INT21HANDAL0FHDECALXORAHAHMOVBXOFFSETSTRINGADDBXAXMOVBL[BX]MOVAH02HINT21H:1243DD2DB0l23456789┇LEASIDD2LEADIDD2+1MOVCX5LOPMOVAL[SI]XCHGAL[DI]MOV[SI]ALADDSI2ADDDI2DECCXJNZLOP上述程8①该程序段完成什么功能?②DD2开始的l0个字节数应是什么值?五、程序填空题(本大题共2612分)1AL和BLAL中的数据变负并送到AH中0→AH①____________JNSKMOVAH,0JMPEND0K②____________MOVAH,ALEND0??2FLDALFLDDW10,-20,30,-60,-71,80,79,56①_______MOVCX,8XORAX,AXR1:②_______ADDSI,2LOOPR1MOVCL,8IDIVCLHLT3共201BUF放于BUF+1和BUF+252AX、BXAX-BXAX中。请编写程序段。53XX字符用INT21HDL=字符ASCIIAH=2。(10分)DATASEGMENTXDW(?)DATAENDS模拟试卷三20120后的括号内。1END语句的叙述正确的是AEND语句是一可执行语句BEND语句表示程序执行到此结束CEND语句表示源程序到此结束DEND语句在汇编后要产生机器码241F100H4个字的物理地址是A1F105HB1F106HC1F107HD1F108H3DS=1200HCS=1400HA2K字节B4K字节C8K字节D16K字节4MOVBYTEPTRBX+SI0”中目的操作数的寻址方式是ABCD5AXCHGAX[BX]BXCHGAXDSCXCHGAX