微机原理、汇编与接口技术习题答案 第6章习题答案.doc

微机原理、汇编与接口技术第六章习题答案 6.16.2 6.3 6.4 stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax, axpush axCALL I6116；调用写入6116子程序MOV AH, 2；回车换行MOV DL, 0DHINT 21HMOV DL, 0AHINT 21HCALL O6116；调用读出6116子程序retstartendp；写入6116：将键入字符的ASCII码写入6116I6116 PROCMOV BL, 0；定义页地址码（16个地址为一页）MOV CX, 128；定义页数LOP1: PUSH CXMOV DX, 380H；将页地址从74LS273输出MOV AL, BLOUT DX, ALMOV CX, 16；定义页内单元数MOV DX, 390H；将键入数写入16个单元LOP2: MOV AH, 1INT 21HOUT DX, ALINC DXLOOP LOP2INC BL；换页POP CXLOOP LOP1；页数未满换页重新写入16个单元RETI6116 ENDP；读出6116：将写入6116的数据读出送显示O6116 PROCMOV BL, 0MOV CX, 128LOP3 PUSH CXMOV DX, 390HMOV AL, BLOUT DX, ALMOV CX, 16MOV DX 380H ；将16个单元的数据送显示LOP4: IN AL, DXPUSH DXMOV DL, AL；将6116中读出的ASCII码送DLMOV AH, 2；2号功能调用显示DL中的字符INT 21HPOP DXINC DXLOOP LOP4INC BLPOP CXLOOP LOP3 RETO6116 ENDPcodeends end start6.5 stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax, axpush axMOV AH, 2；用2号功能调用输出字符，调用号送AHMOV CX, 100；字符个数送计数器MOV DX, 380HWAIT0: IN AL, DX；从244读入TEST AL, 80H；测试状态位是否为0JNZ WAIT0；不为0，等待继续测试状态WAIT1: IN AL, DX；状态为0，继续读入TEST AL, 80H；测试状态是否为1JZ WAIT1；不为1，等待继续测试状态IN AL, DX；状态为1，读入数据即ASCII码AND AL, 7FH；去掉状态位MOV DL, AL；ASCII码送DL，输出字符INT 21HLOOP WAIT0-3；跳到给DX赋值380H的MOV指令，该指令为3字节retstartendpcodeendsend start6.6 stack segment stack stackdw 32 dup(0)stack endsdatasegmentOBUF DB 4 DUP(0)dataendscodesegmentbegin proc farassume ss: stack, cs: code, ds: datapush dssub ax,axpush axmov ax,datamov ds,axMOV BX,0FFSET OBUF+3 ；建立指针MOV BYTE PTR BX,$ ；存字符串结束符$MOV DX,380H ；读入二进制数IN AL,DXMOV CH,10AG: MOV AH,0 ；无符号数扩展为16位DIV CHADD AH,30H ；转换为ASCII码DEC BXMOV BX,AH ；存入输出数据区中OR AL,ALJNZ AGMOV DX,BXMOV AH,9INT 21Hretbegin endpcodeendsend begin 6.7 用除2取余法将BCD数转换为二进制数的程序如下：stack segment stack stackdw 32 dup(0)stack endsdata segmentIBUF DB 4, 0, 4 dup(0)data endscode segmentstart proc farassume ss: stack, cs: code, ds: datapush ds sub ax,axpush axmov ax,datamov ds,axMOV DX,OFFSET IBUF ；键入十进制数MOV AH,10INT 21HMOV DX,0 ；将键入数以压缩BCD数形式放AX中MOV CH,IBUF+1 ；将键入数的个数送计数器CH中MOV CL,4MOV BX,2AGAIN1: AND IBUFBX,0FH ；将数据区中的ASCII码转换为BCD数SHL AX,CL ；将BCD数左移1位OR DL,IBUFBX ；将数据区中的1位BCD放入DX的最低4位INC BXDEC CHJNZ AGAIN1MOV CX,8 ；用除2取余法将DX中的压缩BCD数转换为二进制数AGAIN2: SHR DX,1RCR AL,1MOV AH,DLAND AH,88HSHR AH,1SHR AH,1SUB DL,AHSHR AH,1SUB DL,AHLOOP AGAIN2MOV DX,380H ；二进制数从74LS273输出OUT DX,ALretstart endpcode endsend start用（百位10十位）10+个位的方法将BCD数转换为二进制数的程序段如下：MOV CL,IBUF+1 ；将键入数的个数送计数器CX中MOV CH,0MOV SI,2MOV AL, 0 ；开始将十进制数转换为二进制数AGAIN: MOV AH, 10 ；(010+a2)10+)10+a0MUL AHAND BYTE PTR SI, 0FH ；将十进制数的ASCII码转换为BCD数ADD AL，SIINC SILOOP AGAIN6.8 stack segment stack stackdw 32 dup(0)stack endsdata segmentSEGPT DB 3FH,6,5BH,4FH,66H,7,7FH,6FH,6FH,76H,77H,7CH,39H,5EH,79H,71Hdata endscode segmentstart proc farassume ss: stack, cs: code, ds: datapush dssub ax,axpush axmov ax,datamov ds,axMOV DX,380HIN AL,DXSUB AL,30HCMP AL,0AHJB NS7SUB AL,7MOV BL,ALMOV BH,0MOV AL,SEGPTBXOUT DX,ALretstartendpcodeendsend start 6.9 stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax, axpush axMOV DX, 380HLOP1: IN AL, DX ；读入设备1和设置2的状态AND AL, 6 ；测试2台设备的状态，同时使AL的D0位为0JZ LOP2 ；都正常，即AL的D1位和D2位都为0，AL为0MOV AL, 1 ；有一台异常AL不为0，AL置1，即AL的D0位为1LOP2: OUT DX, ALMOV AH, 11 ；11号功能调用，键入任一键退出程序INT 21HCMP AL, 0 ；无键入AL=0，有键入AL=FFHJE LOP1 ；无键入，去LOP1继续监视2台设备retstartendpcodeendsend start6.10 stack segment stack stackdw 32 dup(0)stack endscodesegmentbegin proc farassume ss: stack, cs: codepush dssub ax, axpush axMOV DX, 380HIN AL, DX ；读入原码TEST AL, 80H ；判原码数的符号JZ NONEG ；正数的原码和补码相同，不用求补AND AL, 7FH ；负数，求其绝对值NEG AL ；对绝对值求补得该负数的补码NONEG: OUT DX, ALretbegin endpcodeendsend begin6.11 在显示器上显示：行*列的程序如下：stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax, axpush axMOV DX, 380H；检测全键盘MOV AL, 0；键盘的8行全为低电平OUT DX, ALLOP1: IN AL, DX；读入列值CMP AL, 0FFH；列值是FFH吗？JE LOP1；是，全键盘无键闭合，继续检测MOV AH, 7FH ；键盘的行检测的输出值，被检测行为0，其余行为1LOP2: MOV AL, AH；检测键盘的一行OUT DX, AL；输出行检测值IN AL, DX；读入列值，列值不为FFH有键闭合CMP AL, 0FFH；被检测行有键闭合吗？JNE LOP3；有键闭合，结束逐行扫描去LOP3ROR AH, 1；该行无键闭合，改变行检测的输出值检测另一行JMP LOP2LOP3: MOV DL, 0；求行值07，即AH中的那个0的位置AS1: SHR AH ,1；0在最低位DL=0，0在最高位DL=7JNC AS2INC DLJMP AS1AS2: ADD DL, 30H；将行值转换为ASCII码MOV AH, 2；输出行值INT 21HMOV DL, *；输出乘号：*INT 21HMOV DL, 0；求列值07，即AL中的那个0的位置AS3: SHR AL, 1JNC AS4INC DLJMP AS3AS4: ADD DL, 30H；将列值转换为ASCII码INT 21H；输出列值retstartendpcodeendsend start6.12 stack segment stack stackdw 32 dup(0)stack endsdatasegmentSEGPTDB 77H,7CH,39H,5EH,79H,71H,6FH,76Hdataendscodesegmentstartproc farassume ss: stack, cs: code, ds: datapush dssub ax, axpush axMOV DX, 380HLOP: IN AL, DXMOV BX,0LOP1: SHR AL,1JNC LOP2INC BLJMP LOP1LOP2: MOV AL,SEGPTBXOUT DX, ALMOV AH,11INT 21HCMP AL,0JE LOPretstartendpcodeendsend start 6.13 stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax, axpush axMOV DX 260HIN AL, DXMOV AH, ALINC DXIN AL DXCMP AL, AHJNC NXCHGXCHG AL, AHNXCHG: SUB AL, AHDASINC DXINC DXOUT DX, ALretstartendpcodeendsend start6.14 stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax, axpush axMOV BL, 240 MOV DX, 260HLOP: ADD BL, 16MOV AL, BL OUT DX, ALLOOP $；延时MOV AH, 11INT 21HCMP AL, 0JE LOPretstartendpcodeendsend start6.15 输出周期性的方波、锯齿波、三角波、正弦波的程序如下：stack segment stack stackdw 32 dup (0)stack endsdatasegmentOBF DB 80H, , 0FFH, , 80H, , 00H, ；正弦波的数据（一周期）N EQU $ OBFdataendscodesegmentstartproc farassume ss: stack, cs: code, ds: datapush dssub ax, axpush axmov ax, datamov ds, axMOV DX, 380HAGAIN: MOV AH, 1；键入一字符，应为大写！INT 21HCMP AL, FJE FBCMP AL, JJE JCBCMP AL, SJE SJBCMP AL, ZJE ZXB；按下F、J、B和Zret；4个键以外的键退出该程序FB: MOV AL, 0；方波输出OUT DX, ALCALL DELAY1NOT ALOUT DX, ALCALL DELAY1 MOV AH, 11INT 21HCMP AL, 0JE FBJMP AGAINJCB:INC AL；锯齿波输出OUT DX, ALCALL DELAY2MOV BL, AL；保护ALMOV AH, 11CMP AL, 0MOV AL, BL；恢复ALJE JCBJMP AGAINSJB:INC AL；三角波输出OUT DX, ALCALL DELAY3CMP AL, 1JNE SJBSJB1: DEC ALOUT DX, ALCALL DELAY3CMP AL, 0JNE SJB1MOV AH, 11INT 21HCMP AL, 0JE SJBJMP AGAINZXB: MOV BX, OFFSET OFB；正弦波输出MOV CX, NZXB1: MOV AL, BXOUT DX, ALCALL DELAY4INC BXLOOP ZXB1MOV AH, 11INT 21HCMP AL, 0JE ZXBJMP AGAINstartendpDELAY1: PROCPUSH CXMOV CX, 100DL11: MOV SI, 0DL12: DEC SIJNZ DL12LOOP DL11POP CXRETDELAY1 ENDPDELAY2；延时程序结构一样DELAY3；仅CX取值不同；故略DELAY4codeends end start6.1 stack segment stack stackdw 32 dup(0)stack endscodesegmentstartproc farassume ss: stack, cs: codepush dssub ax,axpush axmov ax,datamov ds,axMOV CL,4LOP1: MOV AX,0LOP2: PUSH AXMOV DX, 381HROR AX,CL ；高8位数据移入ALOUT DX, ALMOV DX, 380HMOV AL,AH ；低4位数据移入AL的低4位SHR AL,CLOUT DX, ALOUT DX, AL；再写一遍MOV AH,11 ；检查键盘有无键入INT 21HCMP AL,0JNE BAK ；键盘有键入，退出POP AXINC AX CMP AX,1000HJNE LOP2JMP LOP1BAK: POP AXretstart endpcode ends end start6.17stack segment stack stack dw 32 dup(0)stack endsdata segmentBUF DB 16 DUP(0)data endscode segmentstart proc