习题答案ibm-pc-汇编语言程序设计(第二版,清)

⑴=1 0111 0001B=171H⑵=10 0111 0001 0000B=2710H⑶=1111 1111 1111B=0FFFH1.2⑴=2DH=45D⑶=0FFFFH=65535D1.3⑴=1111 1010B=250D⑵=0101 1011B=91D⑶=1111 1111 1111 1110B=65534D1.4⑴=0F1H⑶=0AACFH1.5⑴=0F7H1.6⑴=-40D or 216D⑵=-1D or 255D1.7⑴=+79D or “O”1.846H6FH…32H2EH2562.3(30022H)=0ABH (30024H)=0EFH (30021H)=0AB34H (30022H)=0CDABH 2.4PA1=3017AHPA2=3017AHPA3=3017AH2.5PA=0AAA40H2.62.71、AX,BX,CX,DX2、CX3、AX,BX,DX4、CS,DS,SS,ES5、PSW6、CS,IP7、SS.SP2.8CS,DS,SS,ES,BX,BP,SI,DI,SP,IP 2.91）M2）C3）D4）B5）A6）C7）K8）J9）F10）H11）G12） E13）I14）015）N第三章习题答案(IBM PC 第二版)3.11、无EA2、EA=7237H3、无EA4、EA=637DH5、EA=0D5B4H6、EA=8E18H7、EA=EA=(BX)+(SI)+D=1004FH；超过了段的边界，最高进位位丢失，因此EA=004FH。

3.21、ADD DX,BX2、ADD AL,[BX][SI]3、ADD [BX+0B2H],CX4、ADD [0524H],2A59H5、ADD AL,0B5H3.31、LEA BX,BLOCK+10MOV DX,[BX]2、LEA BX,BLOCKMOV DX,10[BX]3、LEA BX,BLOCKMOV DI,10MOV DX,[BX][DI]3.41、(AX)=1200H2、(AX)=0100H3、(AX)=4C2AH4、(AX)=3412H5、(AX)=4C2AH6、(AX)=7856H7、(AX)=65B7H3.51、（IP）=7CD9H2、（IP）=0600H3、（IP）=098AH3.6LDS SI,[2000]MOV AX,[SI]3.71、064DH2、0691H3、05ECH(0624H+02H-3AH)3.81、立即数寻址：无EA，PA不确定2、寄存器寻址：无PA3、直接寻址：PA=20100H4、直接寻址：PA＝20050H5、寄存器间接：PA＝20100H6、寄存器间接寻址：PA＝21100H7、寄存器间接寻址：PA＝15010H8、寄存器间接寻址：PA＝200A0H9、寄存器相对寻址：PA＝20110H10、寄存器相对寻址：PA＝20150H11、基址变址寻址：PA＝201A0H12、相对基址变址寻址：PA＝201F0H 3.91、MOV AX,12[BX]MOV ZERO,AX2、MOV AX,ARRAY[BX]MOV ZERO,AX3.10 （略）3.11(AX)=1E00H3.12MOV DL,CSTRING+0MOV DH,CSTRING+63.13 (略)3.14LDS SI,[2000H]MOV AX,[SI]3.152、SF=1 ZF=0 CF=0 OF=14、SF=0 ZF=1 CF=1 OF=03.16 （略）3.171、MOV AX,ZSUB AX,XADD AX,WMOV Z,AX2、ADD X,6ADD R,9MOV BX,XADD BX,RMOV AX,WSUB AX,BXMOV Z,AX3、ADD Y,6MOV AX,WIMUL XIDIV YMOV Z,AXMOV R,DX4、MOV AX,WSUB AX,XCWDIMUL BX,Y,5IDIV BXMOV BX,2IMUL AX,BXMOV Z ,AX3.181、(AX)=1234H 对CF、SF、ZF无影响2、(AX)=1234H 对CF、SF、ZF无影响3、(AX)=2341H CF＝1、对SF、ZF无影响4、(AX)=2340H 对CF无影响、SF＝0、ZF＝05、(AX)=2340H 对CF、SF、ZF无影响6、(AX)=8D00H CF＝0、对SF、ZF无定义最后(DX)=0000H,(AX)=8D00H3.19 （略）3.201、MOV AX,DATAXADD AX,DA TAYMOV DATAY,AX2、MOV AX,DATAXMOV DX,DATAX+2ADD AX,DA TAYADC DX,DATAY+2MOV DATAY,AXMOV DATAY,DX3、DA TX 和DATAY两个字数据之和加14、MOV AX,DATAXMUL DATAY5、MOV EAX,DA TAXMOV EBX,DA TAYMUL EBX6、MOV AX,DATAXMOV BL,23DIV BL7、MOV DX,DATAX+2MOV AX,DATAXDIV DATAY3.21NEG DXNEG AXSBB DX,0 ；带借位的减法3.22MOV AX,AMOV DX,A+2CMP DX,0JS FUSHUJMP CUNSHUFUSHU:NEG DXNEG AX；NEG对PSW的影响，AX为0时，CF＝0，否则CF＝1 SBB DX,0CUNSHU:MOV B,AXMOV B+2,DX3.231、(BX)=009AH2、(BX)=0061H3、(BX)=00FBH4、(BX)=001CH5、(BX)=0000H6、(BX)=00E3H3.24(BX)=00DAH3.25 （略）3.26(DX)(AX)中的双字左移4位3.271、(DX)=005CH3、(DX)=05C8H5、(DX)=2017H7、(DX)=00B9H9、(DX)=00DCH3.280C02DH3.291、LEA SI,CONAMELEA DI,PRLINEMOV CX,20CLDREP MOVSB2、LEA SI,CONAME+19LEA DI,PRLINE+19MOV CX,20STDREP MOVSB3、LEA SI,CONAME+2CLDLODSW4、LEA DI,PRLINE+5CLDSTOSB5、LEA DI,CONAMEMOV AL,20HMOV CX,20CLDREPNE SCASBJNZ NEXTDEC DIMOV BH,ES:[DI]3.30LEA DI,STRINGMOV AL,26HMOV CX,18CLDREPNE SCASBDEC DIMOV ES:[DI],20H3.311、LEA DI,PRINT_LINEMOV CX,132MOV AL,20HCLDREP STOSB2、LEA DI,STUDENT_ADDRMOV CX,9MOV AL,‟_‟CLDREPNE SCASB3、LEA DI,STUDENT_ADDR+8MOV CX,9MOV AL,‟_‟STDREPNE SCASB4、LEA DI,STUDENT_NAMEMOV CX,30MOV AL,20HCLDREPE SCASBJNZ NEXTDEC DIMOV CX,30MOV AL,‟*‟STDREP STOSB…NEXT:5、LEA SI,STUDENT_NAMELEA DI,PRINT_LINEMOV CX,30CLDREP MOVSBLEA SI,STUDENT_ADDR+8LEA DI,PRINT_LINE+131MOV CX,9STDREP MOVSB3.32LEA SI,OLDSLEA DI,NEWSMOV CX,5CLDREPE CMPSBJNZ NEW_LESS…NEW_LESS：3.331、CMP DX ,CXJA EXCEED…EXCEED:2、CMP BX ,AXJG EXCEED…EXCESS:3、CMP CX,0JCXZ ZERO…ZERO4、CMP BX,AXJO OVERFLOW…OVERFLOW5、CMP BX,AXJLE EQ_SMA…EQ_SMA6、CMP DX,CXJBE EQ_SMA…EQ_SMA3.341、L12、L13、L24、L3？5、L53.351、JNB/JBE/JNL/JLE2、JNB/JBE/JNL/JLE3、JNB/JNBE/JL/JLE4、JNB/JNBE/JNL/JNLE5、JNB/JNBE/JNL/JNLE6、JB/JBE/JL/JLE7、JB/JBE/JNL/JNLE8、JNB/JNBE/JL/JLE3.36如果2×P≤Q ，则2 →（AX）；否则1 →（AX）3.37TEST STATUS,15HJZ ROUTINE_4 ；全0JP ROUTINE_2 ；两个1TEST STATUS,1HJZ ROUTINE_3 ；一个1TEST STATUS,4HJZ ROUTINE_3 ；一个1TEST STATUS,10HJZ ROUTINE_3 ；一个1ROUTINE_1: ；三个1…ROUTINE_2:…ROUTINE_3:…ROUTINE4:…3.381、(AX)=0005H,(BX)=0010H,(CX)=0000H,(DX)=0000H2、(AX)=0002H,(BX)=0004H,(CX)=0003H,(DX)=0001H3、(AX)=0003H,(BX)=0007H,(CX)=0002H,(DX)=0000H3.39 （略）3.42 （略）3.48（BX）←(DX)×100H3.49方法1、MOV AL,BLCBWDIV CLSHL AL,1CBWMOV DX,AX 方法2、MOV AL,BLCBWDIV CLCBWMUL DX,AX,2 3.50（略）3.51MOV AX,0MOV BX,AXLEA SI,STRINGMOV CX,100H NEXT:CMP [SI],42HJA DAYJB XIAOYJMP DENGYU DAY:INC AXJMP DENGY XIAOY:INC BX DENGY:INC SILOOP NEXTMOV UP,AXMOV DOWN第三章习题答案(80X86) 3.231、MOV AL,SSUB AL,6DASMOV AL,VDAAMOV U,AL3.31MOV CL,4SHR AX,CLMOV BL,DLSHR DX,CLSHL BL,CLOR AH,BL3.43CMP AX,-1JLE XDCMPAX,1JL XYMOV AX,1JMP OUTXD:MOV AX,-1JMP OUTXY:MOV AX,0OUT:3.45MOV CX,100MOB BX,-1NEXT:INC BXADD ARRAY[BX],1LOOP NEXT第四章习题答案(IBM PC 第二版)4.1、(略)4.2、（略）4.3、（略）Array 4.4、（略）4.5、如又图所示4.6、DW 5150HDB 50H,51HDB 'PQ'4.7、DA TA SEGMENTFLD1B DB 'per~ter'FLD2B DB 32FLD3B DB 20HFLD4B DB 01011001BFLD5B DB '32654'FLD6B DB 10 DUP (0)FLD7B DB 'PART1',20,'PART2',50,'PART3',14FLD1W DW 0FFF0HFLD2W DW 01011001BFLD3W DW FLD7BFLD4W DW 5,6,7,8,9FLD5W DW 5 DUP (0)FLD6W DW FLD1W-FLD1B4.9、L的值为64.10、1、MOV AX,OFFSET LNAME2、MOV SI,WORD PTR LODE_LIST3、CODE_LENGTH EQU $-CODE_LIST4.11、TITLE QIUZUIDAZHI;********************************************** DA TA SEGMENTDB 5DA TA_LIST DW -1,0,2,5,4,5 DUP (?)MAX DW ?MIN DW ?DA TA ENDS;******************************************* CODE SEGMENTASSUME CS:CODE,DS:DATA MAIN PROC FARSTART: PUSH DSMOV AX,0PUSH AXMOV AX,DA TAMOV DS,AXLEA SI,DA TA_LISTMOV CX,4MOV AX,[SI]LOP1: ADD SI,2CMP AX,[SI]JGE BIGER1MOV AX,[SI]BIGER1: LOOP LOP1MOV MAX,AXLEA SI,DA TA_LISTMOV CX,4MOV AX,[SI]LOP2: ADD SI,2CMP AX,[SI]JLE BIGER2MOV AX,[SI]BIGER2: LOOP LOP2MOV MIN,AXRETMAIN ENDPCODE ENDSEND START4.12、(略)4.13、1、MOV AX,102、MOV BL,103、MOV CL,14.14、1、(AX)=12、(AX)=23、(AX)=204、(AX)=405、(AX)=14.15、（略）4.16、DA TA SEGMENT AT 0E000HA EQU THIS WORDB 100 DUP (?)DA TA ENDS;***********************************STACK SEGMENT STACK 'STACK'64 DUP (?)STACK ENDS;***********************************CDOE SEGMENTASSUME CS:CODE,DS:DATA,SS:STACKORG 1000HSTART:MOV AX,DA TAMOV DS,AX…CODE ENDSEND START4.17、D_SEG SEGMENTAUGEND DD 99251SUN DD ?D_SEG ENDS;*****************************E_SEG SEGMENTADDEND DD -15962E_SEG SEGMENT;*****************************C_SEG SEGMENTASSUME CS:C_SEG,DS:D_SEG,ES:E_SEG START:MOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXMOV AX,WORD PTR AUGENDADD AX,WORD PTR ADDENDMOV DX,WORD PTR AUGEND+2ADC DX,WORD PTR ADDEND+2MOV WORD PTR SUM,AXMOV WORD PTR SUM+2,DXMOV AH,4CHINT 21HC_SEG ENDSEND START4.18、1、结束汇编END START2、结束用户程序执行MOV AH,4CHINT 21H 4.19、（略）第四章习题答案(80X86)4.17、X_SEG SEGMENTX DW ?X_SEG ENDS;*********************Y_SEG SEGMENTY DW ?Y_SEG ENDS;*********************Z_SEG SEGMENTY DW ?Z_SEG ENDS;*********************CODE SEGMENTASSUME CS:CODE,DS:X_SEG,ES:Y_SEG,SS:Z_SEG START:MOV AX,X_SEGMOV DS,AXMOV AX,Y_SEGMOV ES,AXMOV AX,XADD AX,YADD AX,ZMOV X,AXMOV AH,4CHINT 21HCODE ENDSEND START第五章习题答案(IBM PC 第二版) 5.1、NEXT:MOV AH,07H ;不带回显INT 21HCMP AL,‟a‟JB STOPCMP AL,‟z‟JA STOPSUB AL,20HMOV DL,ALMOV AH,02HINT 21HJMP NEXTSTOP:MOV AH,4CHINT 21H5.2、NEXT:MOV AH,07H ;不带回显INT 21HCMP AL,‟a‟JB STOPCMP AL,‟z‟JA STOPDEC ALMOV DL,ALMOV CX,3DISPLAY:MOV AH,02HINT 21HINC DLLOOP DISPLAY STOP:MOV AH,4CHINT 21H5.3、DA TA SEGMENTSTORE DB 4 DUP(?) DA TA ENDS…NEXT:MOV CL,4MOV CH,4LEA BX,STOREA:MOV DX,AXAND DX,0FHMOV BYTE PTR [BX],DLINC BXSHR AX,CLDEC CHJNZ AB:MOV DL,STOREMOV CL,STORE+1MOV BL,STORE+2MOV AL,STORE+3MOV AH,4CHINT 21H5.4、DA TA SEGMENTSTRING1 DB …I am a student‟STRING2 DB …I am a student‟YES DB …MATCH‟,0DH,0AH,‟$‟NO DB …NO MA TH‟, 0DH,0AH,‟$‟DA TA ENDS;CODE SEGMENTASSUME CS:CODE,DS:DA TA,ES:DATA START:MOV AX,DA TAMOV DS,AXMOV ES,AXNEXT:LEA SI,STRING1LEA DI,STRING2MOV CX,STRING2-STRING1REPE CMPSBJNE DISPNOMOV AH,09HLEA DX,YESINT 21HJMP EXITDISPNO:MOV AH,09HLEA DX,NOINT 21HEXIT:MOV AH,4CHINT 21HCODE ENDSEND START5.5、START:MOV AH,07HINT 21HAND AL,0FHCBWMOV CX,AXBELL:MOV DL,07HMOV AH,02HINT 21HLOOP BELLEXIT:MOV AH,4CHINT 21H5.7、;********************************************DA TAS SEGMENTDA TA DW 100 DUP (?)DA TAS ENDS;--------------------------------CODES SEGMENTASSUME CS:CODES,DS:DA TASSTART:MOV AX,DATASMOV DS,AXBEGIN:MOV BX,0MOV CX,100COMPARE:MOV AX,DATA[BX] ;取数组的第一个偶数ADD BX,2TEST AX,01H ;是偶数吗？LOOPNZ COMPARE ;不是，比较下一个数JNZ STOP ;没有偶数，退出JCXZ STOP ;最后一个数是偶数，即为最小偶数，退出COMPARE1:MOV DX,DATA[BX] ;取数组的下一个偶数ADD BX,2TEST DX,01H ;是偶数吗？JNZ NEXT ;不是，比较下一个数CMP AX,DX ;(AX)<(DX)吗？JLE NEXTMOV AX,DX ;(AX)<(DX)，则置换(AX)为最小偶数NEXT:LOOP COMPARE1STOP:MOV AH,4CHINT 21HCODES ENDS;-------------------------------------------------END START;**************************************************** 5.9、MOV BX,0 ;用来存放4位16进制数MOV CH,4MOV CL,4INPUT:SHL BX,CLMOV AH,01HINT 21HCMP AL,39HJA AFAND AL,0FHJMP BINARYAF:AND AL,0FHADD AL,9BINARY:OR BL,ALDEC CHJNE INPUTDISPN:MOV CX,16DISP:MOV DL,0ROR BX,1RCL DL,1OR DL,30HMOV AH,02HINT 21HLOOP DISPEXIT:MOV AH,4CHINT 21H5.11、DA TA SEGMENTBUFF DB 50 DUP(?)COUNT DB 0DA TA ENDS…BIGIN:LEA BX,BUFFMOV COUNT,0INPUT:MOV AH,07HINT 21HMOV [BX],ALINC BXCMP AL,$JNZ INPUTLEA BX,BUFFNEXT:MUV CL,[BX]INC BXCMP CL,'$'JZ DISPCMP CL,30HJB NEXTCMP CL,39HJA NEXTCONT:INC COUNTJMP NEXTDISP:MOV DL, COUNTADD DL,30HMOV AH,02HINT 21H5.13、DA TA SEGMENTSTRING DB 100 DUP(?)DA TA ENDS;CODE SEGMENTASSUME CS:CODE,DS:DSTA START:…BEGIN:MOV CX,100MOV SI,0REPEAT:MOV AL,STRING[SI]CMP AL,30HJB GOONCMP AL,39HJA GOONOR CL,10HJMP EXITGOON:INC SILOOP REPEATAND CL,0DFHEXIT:MOV AH,4CHINT 21HCODE ENDSEND START5.15、DA TA SEGMENTM DW 4,3,2,1,0,-1,-2,-3,-4,-5 DA TA ENDS;CODE SEGMENTASSUME CS:CODE,DS:DATA START:MOV AX,DA TAMOV DS,AXMOV CX,9LEA DI,MMOV AX,[DI]MOV SI,DI ；SI指向最大数CMP AX,0JGE ANEG AXA:ADD DI,2MOV DX,[DI]CMP DX,0JGE BNEG DXB:CMP AX,DXJGE CXCHG AX,DXMOV SI,DIC:LOOP AMOV AX,[SI]MOV M+2*10,AXMOV M+2*(10+1)MOV AH,4CHINT 21HCODE ENDSEND START3.17、DA TA SEGMENTMEM DB 4 DUP(?)DA TA ENDS;CODE SEGMENTASSUME CS:CODE,DS:DATA START:MOV AX,DA TAMOV DS,AXMOV BX,0MOV CL,4MOV CH,4MOV DX,AXNEXT:AND AL,0FHADD AL,30HCMP AL,3AHJL S1ADD AL,07HS1:MOV MEM[BX],ALINC BXMOV AX,DXSHR AX,CLDEC CHJNZ NEXTMOV AH,4CHINT 21HCODE ENDSEND START5.19、DA TA SEGMENT ；双重循环A DW 15 DUP(?)B DW 20 DUP(?)C DW 20 DUP(?)DA TA ENDS;CODE SEGMENTASSUME CS:CODE,DS:DATA START:MOV AX,DA TAMOV DS,AXBEGIN:MOV SI,0 ；A的指针MOV BX,0 ；C的指针MOV CX,15LOP1:PUSH CXMOV DI,0 ；B的指针MOV CX,20MOV AX,A[SI]LOP2:CMP B[DI],AXJNE NOMOV C[BX],AXADD BX,2NO:ADD DI,2LOOP LOP2ADD SI,2POP CXLOOP LOP1MOV AH,4CHINT 21HCODE ENDSEND START5.21、;************************************************ DA TA SEGMENTARRAY DW 3 DUP(?)DA TA ENDS;******************************************** CODE SEGMENTASSUME CS:CODE,DS:DATASTART:MOV AX,DA TAMOV DS,AXMOV CX,3MOV BX,0NEXT:MOV AH,01HINT 21HMOV AH,0MOV ARRAY[BX],AXADD BX,2LOOP NEXTMOV DL,10MOV AH,02HINT 21HMOV DL,13MOV AH,02HINT 21HMOV AX, ARRAYCMP AX,ARRAY+2JZ S1MOV AX,ARRAY+2CMP AX,ARRAY+4JZ S2MOV AX,ARRAYCMP AX,ARRAY+4JZ S2JMP S3S1:MOV AX,ARRAY+2CMP AX,ARRAY+4JNZ S2MOV DL,'2'JMP DISPS2:MOV DL,'1'JMP DISPS3:MOV DL,'0'DISP:MOV AH,02HINT 21HMOV AH,4CHINT 21HCODE ENDS;*************************************************** END START5.23、DA TA SEGMENTA DW ?B DW ?DA TA ENDS;CODE SEGMENTASSUME CS:CODE,DS:DATASTART:MOV AX,DA TAMOV DS,AXBEGIN:MOV AX,AMOV BX,BXOR AX,BXTEST AX,0001HJZ CLASS ；两数均为奇数或偶数时转移TEST BX,0001HJZ EXITXCHG BX,AMOV B,BXJMP EXITCLASS:TEST BX,0001HJZ EXIT ；两数均为偶数时转移INC BINC AEXIT:MOV AH,4CHINT 21HCODE ENDSEND START5.25、（略）5.27、（略）第六章习题答案(IBM PC 第二版) 6.1、CRAY PROC FARPUSH DSXOR AX,AXPUSH AXADD AX,BXRETCRAY ENDP6.3、6.5、;*********************************TITLE 统计各分数段的学生人数;***********************************DA TA SEGMENTCHENGJ DB 76,69,84,90,73,88,99,63,100,80 S6 DB 0S7 DB 0S8 DB 0S9 DB 0S10 DB 0DA TA ENDS;-----------------------------------STACK SEGMENT STACKDW 20 DUP (?)STACK ENDS;-----------------------------------CODE SEGMENTASSUME CS:CODE,DS:DA TA,SS:STACK ;-----------------------------------MAIN PROC FARSTART:PUSH DSSUB AX,AXPUSH AXMOV AX,DATAMOV DS,AXMOV CX,10CALL COUNTCALL DISPRETMAIN ENDP;-----------------------------------COUNT PROC NEARPUSH SIPUSH CXPUSH AXPUSH BXMOV SI,0MOV CX,10NEXT1:MOV AL,[CHENGJ+SI]MOV AH,0MOV BL,10DIV BLMOV BH,0MOV BL,ALSUB BX,6INC [S6+BX]INC SILOOP NEXT1POP BXPOP AXPOP CXPOP SIRETCOUNT ENDP;-----------------------------------DISP PROC NEARPUSH BXPUSH CXPUSH DXPUSH AXMOV BX,0MOV CX,5NEXT2:MOV DL,[S6+BX]ADD DL,30HMOV AH,02HINT 21HINC BXLOOP NEXT2POP AXPOP DXPOP CXPOP BXRETDISP ENDPCODE ENDS;**********************************END START;**********************************6.9、（上机）;******************************************************DSEG SEGMENTINBUF DB 12 DUP (' ') ;输入缓冲区，初始值为空格OUTNAME DB 16 DUP (' ') ;姓名输出行，初始值为空格OUTPHONE DB 12 DUP (' '),0DH,0AH,'$' ;号码输出行，初始值为空格MESG1 DB 'INPUT NAME:','$'MESG2 DB 'INPUT A TELEPHONE NUMBER:','$'MESG3 DB 'NAME',12 DUP (' '),'TEL.',0DH,0AH,'$'CRLF DB 0DH,0AH,'$'DSEG ENDS ;以上定义数据段;----------------------------------------------------STACK SEGMENT STACKDW 100 DUP (?)TOS LABEL WORDSTACK ENDS ;以上定义堆栈段;----------------------------------------------------CSEG SEGMENTTELIST PROC FAR ;主程序TELISTASSUME CS:CSEG,DS:DSEG,ES:DSEG,SS:STACKSTART:MOV AX,STACKMOV SS,AX ;给SS赋值MOV SP,OFFSET TOS ;给SP赋值PUSH DS ;设置返回DOSSUB AX,AXPUSH AXMOV AX,DSEGMOV DS, AX ;给DS赋值MOV ES, AX ;给ES赋值BEGIN:LEA DX,MESG1MOV AH,09H ;显示字符串功能调用INT 21HCALL INPUT_NAME ;输入姓名LEA DX,MESG2MOV AH,09H ;显示字符串功能调用INT 21HCALL INPHONE ;输入电话号码CALL PRINTLINE ;显示姓名及电话号码RETTELIST ENDP;----------------------------------------------------INPUT_NAME PROC NEAR ;输入姓名子程序CALL GETCHAR ;调输入字符子程序输入姓名LEA SI,INBUF ;把INBUF中的姓名移入输出行OUTNAMELEA DI,OUTNAMEMOV CX,12CLDREP MOVSBRETINPUT_NAME ENDP ;INPUT_NAME子程序结束;----------------------------------------------------INPHONE PROC NEAR ;输入电话号码子程序CALL GETCHAR ;调输入字符子程序输入电话号码LEA SI,INBUF ;把INBUF中的电话号码移入输出行OUTPHONE LEA DI,OUTPHONEMOV CX, 12CLDREP MOVSBRETINPHONE ENDP ;INPHONE子程序结束;-----------------------------------------------GETCHAR PROC NEAR ;键盘输入子程序MOV AL,20H ;先将INBUF中填满空格字符MOV CX,12LEA DI,INBUFCLDREP STOSBMOV CX,12 ;向INBUF输入字符MOV DI,0INPUT:MOV AH,1 ;从键盘接收一个字符并回显的DOS功能调用INT 21HCMP AL,0DH ;输入回车符返回JZ QUITMOV [INBUF+DI], ALINC DILOOP INPUTQUIT:CALL DISP_CRLFRETGETCHAR ENDP ;GETCHAR子程序结束;----------------------------------------------------PRINTLINE PROC NEAR ;显示姓名及电话号码子程序LEA DX,MESG3MOV AH, 09H ;显示字符串功能调用INT 21HLEA DX,OUTNAME ;显示姓名及电话号码MOV AH,09H ;显示字符串功能调用INT 21HRETPRINTLINE ENDP ;PRINTLINE子程序结束;-----------------------------------------------DISP_CRLF PROC NEAR ;显示回车换行符子程序LEA DX,CRLFMOV AH,09HINT 21HRETDISP_CRLF ENDP ;DISP_CRLF子程序结束;----------------------------------------CSEG ENDS ;以上定义代码段;--------------------------------END START;***********************************6.11、1、CALL指令,CALL指令具有FAR属性2、根SUPBPRO的入口地址，入口地址为SUPBPRO本身6.13、（略）6.15、1、D_SEG SEGMENTX DW ?Y DW ?D_SEG ENDS;**********************E_SEG SEGMENTR DW ?E_SEG ENDS;***********************S_SEG SEGMENT STACKDW 20 DUP (0)S_SEG ENDS;*************************C_SEG SEGMENTASSUME CS:C_SEG,SS:S_SEG,ES:E_SEG,DS:D_SEG MAIN PROC FARSTART:PUSH DSXOR AX,AXPUSH AXMOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXCALL COMPUTERETMAIN ENDP;****************************COMPUTE PROC NEARMOV AX,XADD AX,YSUB AX,3MOV R,AXRETCOMPUTE ENDP;*****************************C_SEG ENDSEND MAIN ;或START2、D_SEG SEGMENTX DW ?Y DW ?D_SEG ENDS;**********************E_SEG SEGMENTR DW ?E_SEG ENDS;***********************S_SEG SEGMENT STACKDW 20 DUP (0)S_SEG ENDS;*************************C_SEG SEGMENTASSUME CS:C_SEG,SS:S_SEG,ES:E_SEG,DS:D_SEG MAIN PROC FARSTART:PUSH DSXOR AX,AXPUSH AXMOV AX,D_SEGMOV DS,AXMOV AX,E_SEGMOV ES,AXCALL FAR PTR COMPUTERETMAIN ENDP;****************************C_SEG ENDS;*************************************C_SEG1 SEGMENTCOMPUTE PROC FARMOV AX,XADD AX,YSUB AX,3MOV R,AXRETCOMPUTE ENDP;*****************************C_SEG1 ENDS;**************************END MAIN ;或START ;********************************** 3、（略）第六章习题答案(80X86)6.9、TITLE 显示不及格学生情况;*********************************** STUDENT STRUC ;定义结构BIANJ DB ' 'XINGM DB ' 'CHENGJ DB ?STUDENT ENDS;***********************************DA TA SEGMENTS1 STUDENT <'(1)',' ZHANG',20>S2 STUDENT <'(2)',' LI',30>S3 STUDENT <'(3)',' WANG',40>S4 STUDENT <'(4)',' CHEN',30>S5 STUDENT <'(5)',' YANG',20>DA TA ENDS;*********************************** STACK SEGMENT STACKDW 20 DUP (?)STACK ENDS;***********************************CODE SEGMENTASSUME CS:CODE,DS:DATA,SS:STACK MAIN PROC FARSTART: PUSH DSSUB AX,AXPUSH AXMOV AX,DA TAMOV DS,AXMOV CX,5MOV SI,OFFSET S1LOP: CMP [SI].CHENGJ,60JAE JIGEMOV BL,9MOV DI,SIBUJIGE: MOV DL,[DI]MOV AH,02HINT 21HINC DIDEC BLJNE BUJIGEJIGE: CALL CRLFADD SI,TYPE S1LOOP LOPMOV AH,4CHINT 21HMAIN ENDP;CRLF PROC NEARMOV DL,10MOV AH,02HINT 21HMOV DL,13MOV AH,02HINT 21HRETCRLF ENDP;CODE ENDSEND START ;*****************************第七章习题答案(IBM PC 第二版)7.1、CLRB MACRO DIZH CHANGDMOV CX,CHANGDMOV AL,' 'LEA DI,DIZHCLDREP STOSBENDM7.3、1、是2、是3、否4、否7.5、RESULT,OPERAND,COUNT应都为字节变量BIN_SUB MACRO RESULT,OPERAND,COUNTLOCAL NEXTPUSH CXPUSH BXPUSH AX？MOV CX,COUNTMOV AX,ALAE BX,OPERANDCLCNEXT:SBB AX,[BX]INC BXLOOP NEXTMOV RESULT,AXPOP AXPOP BXPOP CXENDM7.7、+ CNT DA TA,0+ CNT DA TA,1+ DATA0 DW ?+ DATA1 DW ?7.9、+ MOV TAB+0,0+ MOV TAB+1,1+ MOV TAB+2,2…+ MOV TAB+6,6第七章习题答案(80X86)TITLE 用宏指令定义显示字符串；******************************************* DINGY MACRO X,NAMEGOOD DB 'GOOD STUDENTS:CLASS&X &NAME' ENDM;******************************************** DA TA SEGMENTDINGY 1,ZHANGDB '0'DA TA ENDS;******************************************** STACK SEGMENT STACKDB 20 DUP(?)STACK ENDS;******************************************** CODE SEGMENTASSUME DS:DA TA,CS:CODE,SS:STACK MAIN PROC FARPUSH DSSUB AX,AXPUSH AXMOV AX,DA TAMOV DS,AXMOV DI,OFFSET GOODENXT:MOV DL,[DI]CMP DL,'0'JZ EXITMOV AH,02HINT 21HINC DIJMP NEXTEXIT:MAIN ENDP;****************************CODE ENDS;****************************END MAIN第八章习题答案(IBM PC 第二版)8.1、1、PA=00048H2、PA=00020H8.2、MOV DL,'T' MOV DL,'T'MOV AH,02H MOV AH,02HINT 21H CALL 86H×16D+84H 8.3、OUT 25H,1000H8.4、MOV DX,1000HIN AX,DX8.5、MOV DI,0MOV CX,80HNEXT:IN AL,51HTEST AL,02HJZ NEXTIN AL,50HMOV BUFF[DI],ALINC DIIN AL,51HTEST AL,00111000BJNZ ERR_ROUTLOOP NEXTJMP EXITERR_ROUT:…EXIT:…8.8、（SP）=00FAH（SS）=0300H（IP）=0040H（FLAGS）=0240H00A0H0900H0240H8.9、00050H,00051H,00052H,00053H四个字节8.10、PUSH DSMOV AX,SEG IN_ROUTMOV DS,AXMOV DX,OFFSET IN_ROUTMOV AL,09MOV AH,25H ;设置中断向量：DS:DX为中断向量；AL为中断类型号INT 21HPOP DS第九章习题答案(IBM PC 第二版)9.1、能否回显。