MCS_8086习题4_答案.doc

习题4答案：1. 从屏幕上输入大写字母，转换为小写字母并输出（生成.COM文件）要求：程序具有可读性、容错性code segment paraassume cs:code, ss:ss_seg, ds:dataorg0100Hmain proc farmain1:leadx, str1;显示提示信息movah, 9int21hmovah, 1;输入小写字母int21hcmpal, A;判断输入字符是否为azjberrorcmpal, Zjaerrorjmpnexterror:leadx, str2;显示错误提示信息movah, 9int21hjmpmain1;跳转，重新输入next:subal, 20hpushaxleadx, str3;显示输出提示信息movah, 9int21hpopaxmovdl, al;显示转换后的大写字母movah, 2int21hmovax, 4c00h;返回操作系统int21hmain endpstr1db0dh, 0ah, Please input(a-z): $str2db0dh, 0ah, The input is error!$str3db0dh, 0ah, The result is: $code endsend main2. 编写一子程序asc2bin，将ASCII转换为二进制数要求：输入参数：AL中存放需要转换的ASCII输出参数：AL中存放转换后的二进制数并返回;功 能：将ASCII转换为二进制数;输入参数：AL中存放需要转换的ASCII;输出参数：AL中存放转换后得到的二进制数asc2binprocsubal, 30hcmpal, 9jbeasc2bin_1subal, 7asc2bin_1: retasc2binendp3. 内存中存放8个16位有符号数，求8个数值之和，并将结果存放在内存变量SUM中注：程序中应用到字扩展为双字的指令CWDdata segment parabufdw-1, 30000, 35000, 36000, 37000, 20000, 10000, -2sumdd0data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axleabx, buf;bx指向buf首地址movcx, 8main1:movax, bxcwd;有符号数字扩展为双字addword ptr sum, ax;32位数相加adcword ptr sum + 2, dxincbxincbxloopmain1;循环8次movax, 4c00hint21hmain endpcode endsend main4. 内存中存放8个8位有符号数，请按从大到小顺序排列data segment parabufdb-1, 30, 35, 36, 37, 20, 100, -2data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axmovcx, 8main1:movbx, 0;外循环movdi, 0pushcxmain2:moval, bufbx;内循环cmpal, bufbx+1jgenextxchgal, bufbx+1;不符合规则, 则交换数据movbufbx, almovdi, 1next:incbxloopmain2popcxcmpdi, 0;判断内循环是否发生数据交换jzexitloopmain1 exit:movax, 4c00hint21hmain endpcode endsend main5. 内存中有8个16位数，请编写程序将8个数倒序排放例：定义内存中8个数bufdw100, 3, 1, 20, 40, -2, 7, 10程序运行后，buf开始应为：buf dw 10, 7, -2, 40, 20, 1, 3, 100data segment parabufdw-1, 30000, 35000, 36000, 37000, 20000, 10000, -2data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axmovbx, 0movcx, 8main1:pushbufbxincbxincbxloopmain1;8个数依次入堆栈leabx, bufmovcx, 8main2:popbxincbxincbxloopmain2;8个数依次出栈movax, 4c00hint21hmain endpcode endsend main6. 从键盘输入4位十进制数，然后以16进制形式显示在屏幕上.例：键盘输入：1024屏幕上应显示：0400H要求：键盘输入和显示结果时均应有提示data segment parabufdb4 dup(0)vardw0str1db0dh, 0ah, Please input four numbers(0-9): $str2db0dh, 0ah, The input is error, please try again. $str2db0dh, 0ah, The hex result is: $data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axmain1:leadx, str1;显示数据输入提示信息movah, 9int21hmovbx, 0movcx, 4;循环输入4个数main2:movah, 1;DOS功能调用，输入数据int21hcmpal, 0;判断输入字符是否为0-9jberrorcmpal, 9jaerror;若不为0-9，则进行错误处理subal, 30h;ASCII转换为二进制，得到0-9movbufbx, al;存入buf开始的缓冲区incbxloopmain2jmpnext;数据输入正确后,跳转到后续处理error:leadx, str2;显示错误提示信息movah, 9int21hjmpmain1;跳转到重新输入next:movax, 0;(0*10+B3)*10+B2)*10+B1)*10+B0movdx, 0;得到输入的4位十进制数movsi, 10movbx, 0movcx, 4main3:mulsi;相乘后dx仍然保持0movdl, bufbxmovdh, 0addax, dxincbxloopmain3;循环4次乘10movvar, ax;得到的4位十进制数存放到var中leadx, str3movah, 9int21h;显示输出提示符movch, 4;以16进制显示输入的数据, 循环4次movcl, 4main4:rolvar, cl;循环左移4位movax, varandax, 000fhcallbin2asc;二进制转换为ASCIIcallpchar;显示16进制decchjnzmain4moval, Hcallpcharmovax, 4c00hint21hmain endp;功能:将一个二进制数转换为ASCII;输入参数:AL中存放要转换的二进制数;输出参数:AL中存放转换后的ASCIIbin2ascprocandal, 0fhaddal, 30hcmpal, 39hjbebin2asc_1addal, 07hbin2asc_1:retbin2ascendp;功能：显示字符;输入参数：AL中存放要显示字符的ASCII;输出参数：无pcharprocmovdl, almovah, 2int21hretpcharendpcode endsend main7. 数据段从100H开始存放字符串str1，从200H开始存放str2，二者均以NULL字符为结束符，编写程序将str2拷贝到str1末尾，形成一个完整字符串例：ORG 100Hstr1db0dh, 0ah, Hello , 0ORG200Hstr2dbAutomation!, 0程序运行后结果应为：str1db0dh, 0ah, Hello Automation!, 0data segment paraORG100Hstr1db0dh, 0ah, Hello , 0ORG200Hstr2dbAutomation!, 0data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axmoves, axleadi, str1;es:di指向str1首地址moval, 0repnzscasb;查找str1结束符NULLdecdi;找到str1结束符，并让es:di指向该位置leasi, str2;ds:si指向str2首地址cld;DF = 0main1:lodsb;AL (ds:si), si = si + 1stosb;(es:di) AL, di = di + 1cmpal, 0;判断是否到str2的结束符jnzmain1leasi, str1;ds:si指向拷贝后的str1首地址main2:lodsb;显示拷贝后的str1字符串cmpal, 0jzexitcallpcharjmpmain2exit:movax, 4c00hint21hmain endp;功能：显示字符;输入参数：AL中存放要显示字符的ASCII;输出参数：无pcharprocmovdl, almovah, 2int21hretpcharendpcode endsend main8. 以10进制形式显示内存中一有符号字节数据例：vardb0ffH屏幕应显示：The result is: -1data segment paravardb7fhstr1db0dh, 0ah, The result is: bufdb4 dup( )db$data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axmovbuf, +cmpvar, 0;判断var是正数,还是负数jgenextmovbuf, -negvar;若var为负，则得到其相反数next:moval, varmovcx, 3movdl, 10leabx, buf+3main1:movah, 0divdl; (AX / 10)商 aladdah, 30h; (AX / 10)余数 ahmovbx, ahdecbxloopmain1;循环3次,分别得到百、十、个位leadx, str1;显示10进制数movah, 9int21hexit:movax, 4c00hint21hmain endpcode endsend main9. 将一个16位的无符号数var, 转换为非压缩格式BCD码，存放在内存中buf开始的单元中。（按高位在前、低位在后的顺序存放）data segment paravardw800Hbufdb5 dup(0)data endsss_seg segment stackdw100 dup(0)ss_seg endscode segment paraassume cs:code, ds:data, ss:ss_segmain proc farmovax, datamovds, axmovcx, 16;(0*2 + B15)*2 + B14)*2 + + )*2 + B0main1:shlvar, 1;得到var的Bi位leabx, 4pushcxmovcx, 5main2:moval, bufbx;执行buf*2 + Bi操作adcal, alAAA非压缩格式BCD码调整movbufbx, aldecbxloopmain2;内循环为5次popcxloopmain1;外循环为16次exit:movax, 4c00hint21hmain endpcode endsend main选作题：10. 内存中从str开始存放一字符串，结束符为NULL字符，请编写程序统计该字符串中单词的个数例：str1db0dh, 0ah, Hello world, welcome to DUT. CPU is central processing unit!, 0h统计.中的单词个数，结果为10datasegmentparastr1dblong long ago, there lived a king, the king asked the artist to dbdraw a horse for him ,0wordsdw0dataendsss_segsegmentstackdb256dup(?)ss_segendscodesegmentassumecs:code,ds:data,ss:ss_segmainprocfarpushdsxorax, axpushaxmovax, datamovds, axmovcx, 0;用cx存放单词数leasi, str1movbl, ;bl总保存当前字符的前一个字符cldmain_3:lodsbandal, al;判断al是否为结束符0jzmain_1cmpal, ;比较是否是空格jnzmain_2main_4:cmpbl, ;比较前一个字符是否为空格，不是则此空格;不能算一个单词jzmain_5cmpbl, ,;比较前一个字符是否为逗号，如果是则此空;格也不能算一个单词jzmain_5inccx;只有当前字符为 或,而且前一个字符为有;效字符时，才对单词数加1jmpmain_5main_2:cmpal, ,;比较是否是,号jzmain_4main_5:movbl, al;进入这里表明此时al中内容不是0, , ,三种;符号,保存al到bljmpmain_3main_1:cmpbl, ;判断结束符前面是否为单词，还是为一个 或,jzmain_6cmpbl, ,jzmain_6inccx;若结束符前是一个有效字符，那么单词数应该加1main_6:movwords, cxretmainendpcodeendsendmain11 编写程序用键盘17键，模拟7个音符，进行键盘演奏！注：需要对8253的定时器2和8255的PB口进行初始化datasegmentparaTable_Ldw131,147,165,175,196,220,247,262 ;低音1-7的频率Table_Mdw262,294,330,349,392,440,494,523 ;高音1-7的频率shift_1equ21hshift_2equ40hshift_3equ23hshift_4equ24hshift_5equ25hshift_6equ5Ehshift_7equ26hshift_8equ2Ahdataendsss_segsegmentstackdb256dup(?)ss_segendscodesegmentassumecs:code,ds:data,ss:ss_segmainprocfarpushds;将psp+0h处段地址和偏移量压入堆栈以便返回OSxorax, axpushaxmovax, data;装载数据段寄存器ds,使之指向当前数据段movds, axmain_1:movah, 0;利用bios的16号中断，扫描键盘按键int16h ;AL中返回按键ASCII码，AH中返回按键扫描码cmpal, 0dh;判断是否为回车键，是回车键就退出jzmain_exitcmpal, 1jbmain_2cmpal, 8jamain_2andax, 000fh;说明按下的是低音1-7，与操作将ASCII;码转换到数字1-8decal;将数值转换成以0开始的数字，以便从表中查找频率值shlal, 1;将数值乘以2，因每个频率值是字变量leabx, Table_Lmain_12:movsi, axmovdi, bx+si;从表中取频率值movbx, 5000;发声时间callGenSoundjmpmain_1main_2:cmpal, shift_1;判断是否按下shift+1键jnzmain_3movax, 0jmpmain_11main_3:cmpal, shift_2;判断是否按下shift+2键jnzmain_4movax, 2jmpmain_11main_4:cmpal, shift_3;判断是否按下shift+3键jnzmain_5movax, 4jmpmain_11main_5:cmpal, shift_4;判断是否按下shift+4键jnzmain_6