汇编的课后习题答案.pdf

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1、第二章1、答：直接由指令指定的I/O 端口数为256个。2、答：3、答：字节单元：(30022H)=AB H,(30024H)=EF H字单元：(30021H)=AB34 H,(30022H)=CDAB H.4、答：3017：000A的存储单元的物理地址是3017AH,3015：002A的存储单元的物理地址是3017AH,3010：007A的存储单元的物理地址是301 7AH。5、答：该程序的第一个字的物理地址是0AAA40H。6、答：条件标志OF、SF,ZF、CF的值依次分别为0、0,0,0。7、答：(1)AX,BX、CX、DX、AH、AL、BH,BL、CH,CL、DH、DL、SP、BP、D

2、l、SI(注意：学生尽量不要用SP参与加减运算)(2)CX(3)DX、AX、AH,AL(4)CS、DS、ES、SS(5)FLAGS(6)IPSS、SP.BP8、答：可以用来指示存储器地址的寄存器有BX、SP,BP、Dl、S l、IP、CS、DS、ES,SS.9、答：唯一正确的是D。第三章2、答：(1)ADD DX,BX(2)ADD AL,BXSI(3)ADD BX+0B2H,CX(4)ADD 0524H,2A59H(5)ADD AL,0B5H3、答：(1)MOV BX,OFFSET BLOCK+OAHMOV DX,BX(2)MOV BX,OAHMOV DX,BLOCKBX(3)MOV BX,O

3、FFSET BLOCKMOV SI,OAHMOV DX,BXSI4、答:(1)1200H(2)0100H(3)4C2AH(4)3412H(5)4C2AH(6)7856H(7)65B7H6、答：MOV BX,2000HLES DI,BXMOV AX,ES:DI7、答：(1)064DH(2)0691H(3)05ECH9、答：(1)MOV AX,BX+OCH MOV ZERO,AX(2)MOV AX,ARRAYBXMOV ZERO,AX10、答：(1)(AX)=1234H(2)(AX)=0032H11、答：(AX)=1E00H12、答：LEA BX,CSTRINGMOV DL,BXMOV DH,BX

4、+613、答:14、答：LES BX,2000MOV AX,ES:BX16、答：(1)74D4H SF=0 ZF=0 CF=0 OF=0(2)A5C1H SF=1 ZF=0 CF=0 OF=1(3)3240H SF=0 ZF=0 CF=1 OF=0(4)0000H SF=O ZF=1 CF=1 OF=017、答：(1)0C754H SF=1 ZF=O CF=1 OF=0(2)12B0H SF=O ZF=O CF=0 OF=0(3)45B0H SF=O ZF=O CF=0 OF=1(4)9F24H SF=1 ZF=O CF=0 OF=021、答：(1)MOV AX,ZSUB AX,XADD AX

5、,WMOV Z,AX(2)MOV BX,XADD BX,6MOV CX,RADD CX,9MOV AX,WSUB AX,BXSUB AX,CXMOV Z,AX(3)MOV AX,WIMUL XMOV BX,YADD BX,6IDIV BXMOV Z,AXMOV R,DX22、答：NEG DXNEG AXSBB DX,016、答：MOV AX,AMOV DX,A+2TEST DX,8000HJZ STORE;为正NEG DXNEG AXSBB DX,0STORE:MOV B,AXMOV B+2,DX17、答:(1)MOV AL,SSUB AL,6DASADD AL,VDAAMOV U,AL(2)

6、MOV AL,ZSUB AL,UDASMOV U,ALMOV AL,XADD AL,WDAASUB AL,UDASMOV U,AL23、答：(1)(BX)=9AH(2)(BX)=61H(3)(BX)=OFBH(4)(BX)=1CH(5)(BX)=0(6)(BX)=0E3H26、答：把(DX)(AX)中的双字左移四位(乘以16)。20、答：MOV CL,4SHR AX,CLMOV BL,DLSHR DX,CLSHL BL,CLOR AH,BL31、答：(1)CLDMOV CX,132MOV AL,20HLEA DI,ARRAYREP STOSB(2)CLDMOV CX,9MOV AL J*LEA

7、 DI,ADDRREPNE SCASBJNE L1L2:.;找到L1.i 未找到(3)CLDMOV CX,30MOV AL,20HLEA DI,NAMEREPE SCASBJNE DO_NOTMOV CX,30MOV AL,$LEA DI,NAMEREP STOSBDO_NOT:.(4)CLDMOV CX,30LEA SI,NAMELEA DI,ARRAYREP MOVSBSTDMOV CX,9LEA SI,ADDR+8LEA DI,ARRAY+131REP MOVSB34、答:(1)转L1(2)转L1(3)转L2(4)转L5(5)转L536、答：2 p q 时(AX)22 p q 时(AX)

8、138、答：(1)(AX)=5(BX)=16(CX)=0(DX)=0(2)(AX)=2(BX)=4(CX)=3(DX)=1(3)(AX)=3(BX)=7(CX)=2(DX)=039、答：第四章1、答：(2)源*作数和目的*作数同为存储器寻址方式。(3)SI、D I同为变址寄存器。(7)目的*作数不能是代码段段寄存器CSo(5)缺 少PTR5、答：BYTE_VAR 42 59 54 45 OC EE 00?-01 02 01 02?00?0102 01 02?00?01 02-01 02?WORD_VAR 00 00 01 00 02 00 00 00-01 00 02 00 00 00 01

9、0002 00 00 00 01 00 02 00-00 00 01 00 02 00?FB FF 59 42 45 54 56 02-8、答：PLENTH 的值为 22(16H)。12、答：(1)10025(2)25(3)2548(4)3(5)103(6)OFFFFH(7)1(8)35假设数据段中数据定义如下：VAR DW 34VAR1 DB 100,ABCDVAR2 DD 1COUNT EQU$-VAR1X DW 5 DUP(COUNT DUP(0)Y LABEL WORDZ DB,123456,V DW 2,$-VAR执行下面程序段并回答问题。MOV AX,COUNT;(AX)=?MOV

10、 BX,Z-X;(BX)=?MOV CX,V+2;(CX)=?MOV DX,VAR;(DX)=?MOV Y+3,2MOV SI,Y+4;(SI)=?ADD Z+5,1MOV DI,WORD PTR Z+4;(DI)=?、答：(AX)=9(BX)=90(CX)=109(DX)=3334H(SI)=3600H(D I)=3700H14、答：(1)(AX)=1(2)(AX)=2(3)(CX)=20(4)(DX)=40(5)(CX)=117、答：D_SEG SEGMENTD_WORD LABEL WORDAUGEND DD 99251S_WORD LABEL WORDSUM DD?D SEG ENDS

11、E_SEG SEGMENTE_WORD LABEL WORDADDEND DD-15962E_SEG ENDSC_SEG SEGMENTASSUME CS:C_SEG,DS:D_SEG,ES:E_SEGMAIN PROC FARSTART:PUSH DSMOV AX,0PUSH AXMOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXMOV AX,D_WORDMOV BX,D_WORD+2ADD AX,ES:E_WORDADC BX,ES:E_WORD+2MOV S_WORD,AXMOV S_WORD+2,BXRETMAIN ENDPC_SEG ENDSEND

12、START16、答：DATASG SEGMENT AT 0E000HWORD_ARRAY LABEL WORDBYTE_ARRAY DB 1 00 DUP(?)DATASG ENDSSTACKSG SEGMENT PARA STACK STACKDW 32 DUP(?)TOS LABEL WORDSTACKSG ENDSCODESG SEGMENTORG 1000HMAIN PROC FARASSUME CS:CODESG,DS:DATASG,ES:DATASG,SS:STACKSGSTART:MOV AX,STACKSGMOV SS,AXMOV SR OFFSET TOSPUSH DSSUB

13、 AX,AXPUSH AXMOV AX,DATASGMOV DS,AXMOV ES,AXRETMAIN ENDPCODESG ENDSEND START9 编写一个完整的程序，要求把含有23H,24H,25H,26H四个字符数据的数据区复制2 0 次。、答：DSEG SEGMENTVAR1 DB 23H,24H,25H,26HDSEG ENDSESEG SEGMENTVAR2 DB 80 DUP(?)ESEG ENDSCSEG SEGMENTASSUME CS:CSEG,DS:DSEG,ES:ESEGMAIN PROC FARSTART:PUSH DSMOV AX,0PUSH AXMOV AX

14、,DSEGMOV DS,AXMOV AX,ESEGMOV ES,AXMOV DX,20CLDLEA DI,VAR2AGAIN:LEA SI,VAR1MOV CX,4REP MOVSBDEC DXJNZ AGAINRETMAIN ENDPCSEG ENDSEND START第五章1、答：mov ex,countlea si,stringllea di,string2again:mov al,simov di,alinc siinc diloop again2、答：code segmentassume cs:codemain proc farstart:push dsmov ax,0push ax

15、mov ah,1int 21hsub al,30hemp al,0jz exitmov cl,almov ch,0again:mov ah,2mov dl,7int 21 hloop againexit:retmain endpcode endsend start8、答：MOV CX,8MOV DL,0NEXT3：ROR AX,1JNC NEXT1ROR AX,1JNC NEXT2INC DLNEXT2：LOOP NEXT3ADD DL,30HMOV AH,2INT21HMOV AH,4CHINT 21HNEXT1：ROR AX,1JMP NEXT212、答：mov ex,100lea di,

16、memmov ax,0cldcomp:repne scaswjexz exitpush exmov si,disub di,2mov bx,direp movswmov word ptr di,0mov di,bxpop exjmp compexit:13、答：mov dx,1 00mov si,0repeat:mov al,stringsiemp al,30hjb goonemp al,39hja goonor cl,20h;(cl)51jmp exitgoon:inc sidec dxjnz repeatand cl,Odfh;(cl)50exit:14、答：table dw 10Oh d

17、up(?)mdata dw?;存放出现次数最多的数count dw 0;存放出现次数mov bx,10Ohmov di,0;d i为数组TABLE的指针next:mov dx,0mov si,0mov ax,tabledimov ex,100hcomp:emp tablesi,axjne addrinc dxaddr:add si,2loop compcmp dx,countjle changmov count,dxmov mdata,axchang:add di,2dec bxjnz nextmov ex,countmov ax,mdata19、答：a dw 15 dup(?)b dw 20

18、 dup(?)c dw 15 dup(?)mov si,0;s i为数组A的指针mov bx,0;b x为数组C的指针mov ex,15loopl:mov di,0;d i为数组B的指针push exmov ex,20mov ax,asiIoop2:cmp bdi,axjne nomov cbx,axadd bx,2jmp nextno:add di,2loop Ioop2next:add si,2pop exloop loopl21、答:.mov dx,0lea si,arraymov ax,simov bx,si+2cmp ax,bxjne nextlinc dxn e xtl:cmp s

19、i+4,axjne next2inc dxnext2:cmp si+4,bxjne numinc dxnum:cmp dx,3jl dispdec dxdisp:mov ah,2add dl,30hint 21h2、答：第六章2、答：(1)NAME1 NAMELIST(2)MOV AX,DATA；假设结构变量NAME1定义在数据段DATA中MOV DS,AXMOV ES,AXMOV AH,10LEA DX,NAME1INT 21HMOV CL,NAME1.ACTLENMOV CH,0LEA SI,NAME1.NAMEINLEA DI,DISPFILECLDREP MOVSB6、答：SKIPLI

20、NES PROC NEARPUSH CXPUSH DXMOV CX,AXNEXT：MOV AH,2MOV DL,OAHINT 21HMOV AH,2MOV DL,ODHINT 21HLOOP NEXTPOP DXPOPCXRETSKIPLINES ENDP7、答：dseg segmentnum dw 76,69,84,90,73,88,99,63,1 00,80n dw 1 0s6 dw 0s7 dw 0s8 dw 0s9 dw 0s10 dw 0dseg endscode segmentmain proc farassume cs:code,ds:dsegstart:push dssub a

21、x,axpush axmov ax,dsegmov ds,axcall sub1retmain endpsub1 proc nearpush axpush bxpush expush simov si,0mov ex,nnext:mov ax,numsimov bx,10div blmov bl,alcbwsub bx,6sal bx,1inc s6bxadd si,2loop nextpop sipop expop bxpop axretsub1 endpcode endsend start8、答：data segmentmaxlen db 40n db?table db 40 dup(?)

22、char db a;查找字符aevenaddr dw 3 dup(?)data endscode segmentassume cs:code,ds:datamain proc farstart:push dsmov ax,0push axmov ax,datamov ds,axlea dx,maxlenmov ah,Oahint 21h;从键盘接收字符串mov ad dr,offset tablemov addr+2,offset nmov addr+4,offset charmov bx,offset addr;通过地址表传送变量地址call count;计算CHAR的出现次数call di

23、splay;显示retmain endpcount proc near;count 子程序push sipush dipush axpush exmov di,bxmov si,bx+2mov cl,byte ptrsimov ch,0mov si,bx+4mov al,byte ptrsimov bx,0again:emp al,byte ptrdijne 1 1inc bx11:inc diloop againpop expop axpop dipop siretcount endpdisplay proc near;display 子程序call crlf;显示回车和换行mov dl,c

24、harmov ah,2int 21 hmov dl,20hmov ah,2int 21 hmov al,bland al,Ofhadd al,30hcmp al,3ahjl printadd al,7print:mov dl,alint 21 hcall crlfretdisplay endpcrlf proc near;crlf 子程序mov dl,Odhmov ah,2int 21hmov dl,Oahmov ah,2int 21 hretcrlf endpcode endsend start第七章1、答：CLRB MACRO N,ARRAYCLDMOV CX,NMOV AL,20HLEA

25、 DI,ARRAYREP STOSBENDM2、答：WAGES MACRO RATE,HOURMOV AL,RATEMOV BL,HOURMUL BL；计算周工资ADD AX,AXADD AX,AX；一个月有4 周MOV WAG,AX；暂存中间结果ADD BL,BLADD BL,BL；计算月出工小时数MOV AL,BLMOV AH,0MOV BL,10DIV BLMOV BL,3MUL BL；计算奖金ADD WAG,AX；计算工资总数ENDM宏展开：1 MOV AL,R11 MOV BL,421 MUL BL1 ADD AX,AX1 ADD AX,AX1 MOV WAG,AX1 ADD BL,

26、BL1 ADD BL,BL1 MOV AL,BL1 MOV AH,01 MOV BL,101 DIV BL1 MOV BL,31 MUL BL1 ADD WAG,AX3、答：(1)调用 有效1 PUSH AX2 MOV AX,P12 SUB AX,P21 CMP AX,01 JGE?00001 NEG AX1?0000:MOV DISTANCE,AX1 POP AX(2)调用有效1 PUSH AX2 MOV AX,BX2 SUB AX,SI1 CMP AX,01 JGE?00011 NEG AX1?0001:MOV XDI,AX1 POP AX(3)调用无效，有错误指令1?0002:MOV 2

27、40H,AX(4)调用有效，但无意义宏展开略。4、答：SEND MACRO SCHARS,DCHARSLOCAL NEXT EXITPUSH AXPUSH SIMOV SI,0NEXT:MOV AL,SCHARSSIMOV DCHARSSI,ALCMP AL,24HJZ EXITINC SIJMP NEXTEXIT:POP SIPOP AXENDM5、答：BIN_SUB MACRO OPERAND,COUNT,RESULTLOCAL NEXTLEA BX,OPERANDMOV CX,COUNTDEC CXMOV AL,BXNEXT：INC BXSUB AL,BXLOOP NEXTMOV RES

28、ULT,ALENDM7、答：宏展开：2 DATAO DW?2 DATA1 DW?9、答：1 MOV TAB,01 MOV TAB+1,11 MOV TAB+2,21 MOV TAB+3,31 MOV TAB+4,41 MOV TAB+5,51 MOV TAB+6,610、答：STORE MACRO XI RPC X,0123456MOV TAB+&X,XENDMENDM12、答：FINSUM MACRO X,Y,SUMMOV AX,XMOV BX,Y.IF AX BXADD BX,BXADD AX,BXMOV SUM,AX.ELSEADD AX,AXADD AX,BXMOV SUM,AX.EN

29、DIFENDM14、答：DOS21 MACRO DOSFUNC,BUFFIFNB aBUFF?MOV AH.DOSFUNCMOV DX,OFFSET BUFFI NT 21HELSEMOV AH,DOSFUNCINT 21HENDIFENDM展开宏调用：1 MOV AH.011 INT21H1 MOV AH,0AH1 MOV DX,OFFSET IPFIELD1 I NT 21H第八章1、答：(1)00048H(2)00020H2、答：mov ah,02mov dl,Tmov bx,0mov ds,bxmov bx,21 h*4call far ptr bx3、答：mov al,12hout

30、25h,al4、答：mov dx,10OOhin ax,dx8、答：(SP)=OOFA(SS)=0300(IP)=0040(FLAGS)=00409、答：14H*4=50H,I.(00050H)=偏移地址，(00052H)=段地址10s 答：push dsmov ax,seg int_routmov ds,axmov dx,offset int_routmov al,09mov ah,25hint 21hpop ds12、答：D 3-D 2 fD 3 fD 4-D 5-D 1-D 3 -D513、答：D3-*D2-D4-D5-*D1-D3-D5第九章1、答：:者不同点是功能1输入字符可何显，功

31、能8输入字符不回显。2、答：dseg segmentscode db 7,5,9,1,3,6,8,0,2,4buffer db 10 dup(?)dseg endscseg segmentmain proc farmov ax,dsegmov ds,axmov si,0mov cx,10lea bx,scodeinput:mov ah,01;键盘输入int 21hand al,Ofh;ASCI I-数字xlat;换码mov buffersi,al;存入缓冲区inc si;修改地址指针loop input;输入下一个数字mov ax,4c00h;返回 DOSint 21hmain endpcse

32、g endsend main3、答：(24 行x80)+40 歹 ij)x2=3920=0F50H4、答：mov dh,Och;第 12 行mov dl,08h;第 8 歹 ijmov bh,0;0 页mov ah,2;置光标int 10h;BIOS 调用5、答：mov ah,7;屏幕初始化mov al,0;全屏为空格mov bh,07h;正常属性mov ch,12;左上行mov cl,0;左上列mov dh,22;右下行mov dl,39;右下列int 10h;Bl OS 调用6、答：(1)mov ah,3;读光标位置mov bh,0;页号int 10h;BIOS 调用(2)mov dh,1

33、8h;第25 行mov dl,0;第 0 歹 ijmov bh,0;0 页mov ah,2;置光标位置int 10h;BIOS 调用(3)mov ah,2;置光标位置mov bh,0;0 页mov dx,0;0 行 0 歹 Uint 10h;BIOS 调用mov ah,9;显示字符及属性mov al,M;字符mov bh,0;0 页mov bl,7;正常属性mov ex,1 ;显示次数int 10h;BIOS 调用7、答：message db Try again,you have*Count db ndb t starfighters le ft$.add cl,30h;二进制数ASCIImo

34、v count,cl;存入 COUNTlea dx,message;字符串地址mov ah,09;显示字符串int 21h;DOS 调用9、答：mess db What is the date(mm/dd/yy)07,$date db 10,0,10 dup()mov ah,09;显示字符串lea dx,mess;字符串地址int 21hmov ah,Oah;键盘输入lea dx,date;缓冲区地址int 21h11、答：MODE COM2:12,N,8,1第十章、1。答：(1)EXYRN SUBPRO：FAR(2)PUBLIC SUBPRO2、答：(1)PUBLIC QTY VALUE,P

35、RICE(2)EXTRN QTY:BYTE,VALUE:WORD,PRICE:WORD3、答：模块1:PUBLIC VAR1,VAR2,LAB3模块2:PUBLIC VAR3,LAB2EXTRN VAR1:DWORD,VAR2:BYTE,LAB1:NEAR,LAB3:FAR模块3:PUBLIC LAB3EXTRN VAR1:DWORD,VAR2:BYTE,LAB1:NEAR,LAB2:FAR4、答：EXTRN SUBMUL:FARSSEG SEGMENTDW 64 DUP(?)SSEG ENDSDSEG SEGMENTQTY DW?PRICE DW?DSEG ENDSCSEG1 SEGMENT

36、ASSUME CS:CSEG1,DS:DSEG,SS:SSEGCALLMUL PROC FARMOV AX,DSEGMOV DS,AXMOV AX,PRICEMOV BX,QTYCALL FAR PTR SUBMULCALLMUL ENDPCSEG1 ENDSEND CALLMULPUBLIC SUBMULCSEG2 SEGMENTASSUME CS:CSEG2SUBMUL PROC FARMUL BXRETSUBMUL ENDPCSEG2 ENDSEND5、答：1 D_SEG SEGMENT2 X DW 16 DUP(?)3 Y DW 16 DUP(?)4 D_SEG ENDS5 E_SEG

37、 SEGMENT6 R DW 16 DUP(?)7 E_SEG ENDS8 C_SEG SEGMENT9 ASSUME CS:C_SEG,DS:D_SEG,ES:E_SEG10 MAIN PROC FAR11 MOV AX,D_SEG12 MOV DS,AX13 MOV AX,E_SEG14 MOV ES,AX1 5 CALL COMPUTE16.17 MAIN ENDP18 COMPUTE PROC NEAR19 MOV CX,1620 MOV SI,021 MOV DI,022 MOV BX,023 NEXT:MOV AX,XSI24 ADD AX,YDI25 SUB AX,326 MOV

38、 ES:RBX,AX27 INC SI28 INC DI29 INC BX30 LOOP NEXT31 RET32 COMPUTE ENDP33 C_SEG ENDS34 END MAIN(1)D_SEG SEGMENTD_SEG ENDSE_SEG SEGMENTE SEG ENDSC_SEG1 SEGMENTASSUME CS:C_SEG,DS:D_SEG,ES:E_SEGMAIN PROC FARMOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXCALL FAR PTR COMPUTEMAIN ENDPC_SEG1 ENDSC_SEG2 SEGMENTA

39、SSUME CS:C_SEG2,DS:D_SEG,ES:E_SEGCOMPUTE PROC FARCOMPUTE ENDPC_SEG2 ENDSEND MAIN(2)D_SEG SEGMENTD_SEG ENDSE_SEG SEGMENTE_SEG ENDSEXTRN COMPUTE:FARPUBLIC X,Y RC_SEG1 SEGMENTASSUME CS:C_SEG,DS:D_SEG,ES:E_SEGMAIN PROC FARMOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXCALL FAR PTR COMPUTEMAIN ENDPC_SEG1 ENDS

40、END MAINEXTRN X:WORD,Y:WORD,R:WORDPUBLIC COMPUTEC_SEG2 SEGMENTASSUME CS:C_SEG2,DS:D_SEG,ES:E_SEGCOMPUTE PROC FARMOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXCOMPUTE ENDPC_SEG2 ENDSEND6、答：.model small.stack 10Oh.datapublicdatal dw 1228data2 dw 52400result dd?quotient dw?remainder dw?.codeextrn dmul:far,

41、ddiv:farstart:mov ax,datamov ds,axmov dx,datalmov ax,data2push axpush dxcall far ptr dmulcall far ptr ddivmov ah,4chint 21hend start.model smallextrn result:dword.codepublic dmuldmul proc far_push bpmov bp,spmov ax,bp+6mov bx,bp+8mul bxmov word ptr result,axmov word ptr result+2,dxpop bpretdmul endpend.model smallextrn quotient:word,remainder:word.codepublic ddivddiv proc far_push bpmov bp,spmov bx,bp+6mov ax,bp+8mov dx,0div bxmov quotient,axmov remainder,dxpop bpretdmul endpend