计算机等级考试三级PC技术上机（7）

1、2013年计算机等级考试三级pc技术上机（7）请编制程序，其功能是：内存中共有30个字节型数据，找出其中的两个ascii字符串并进行校验。欲寻找ascii字符串由13个字符组成：#(23h)，7个ascii字符，*(2ah)，2个ascii字符，回车符(0dh)，换行符(0ah)。校验方法为：对字符#及字符*之间的7个ascii字符进行异或操作，若异或操作结果的asccii字符表示(异或操作结果高4位的ascii字符表示在前，低4位的ascii字符表示在后)与原字符串中字符*之后的两个ascii字符相同，则将原字符串原样保存;反之，则将原字符串中的所有字符(共13个)均用字符!(21h)代替。

2、例如：内存中有 33h,35h,23h(#,第一个字符串开始),46h,41h,30h,2eh,34h,3dh,31h,2ah(*),32h,31h,0dh,0ah,46h,23h(#,第二个字符串开始),46h,41h,30h,2eh,34h,3dh,30h,2ah(*),32h,31h,0dh,0ah,55h结果为 23h,46h,41h,30h,2eh,34h,3dh,31h,2ah(*),32h,31h(校验正确，该字符串原样保持),0dh,0ah,21h,21h,21h,21h,21h,21h,21h,21h,21h,21h,21h,21h,21h(校验错，整个字符串用字符!代替)。

3、部分程序已给出, 其中原始数据由过程load从文件input1.dat中读入source开始的内存单元中。运算结果要求从result开始存放,由过程save保存到文件output1.dat中。填空begin和end之间已给出的一段源程序使其完整(空白已用横线标出,每行空白一般只需一条指令, 但采用功能相当的多条指令亦可),或删除begin和end之间原有的代码并自行编程来完成要求的功能。对程序必须进行汇编,并与io.obj链接产生prog1.exe执行文件,最终运行程序产生结果(无结果或结果不正确者均不得分)。调试中若发现整个程序中存在错误之处, 请加以修改。试题程序：extrn load:f

4、ar,save:farn equ 26stac segment stackdb 128 dup(?)stac endsdata segmentsource db n+4 dup(?)result db n dup(0)hh db 2 dup(0)name0 db input1.dat,0name1 db output1.dat,0data endscode segmentassume cs:code,ds:data,ss:stacstart proc farpush dsxor ax,axpush axmov ax,datamov ds,axlea dx,source ;数据区起始地址lea

5、si,name0 ;原始数据文件名mov cx,n+4 ;字节数call load ;从input1.dat中读取数据;*begin*mov si,0mov di,0mov bx,2 ;两个ascii字符串repeat: mov ah,0search: mov al,sourcesiinc sicmp al,#_ searchmov resultdi,alinc disearch1:mov al,sourcesiinc sicmp al,_je asciimov resultdi,alinc dixor ah,aljmp _ascii: mov resultdi,alinc dipush di

6、mov di,0mov dh,2mov dl,ah ;异或结果暂存在dl中mov cl,4 ;先将异或结果高4位转换成ascii字符shr ah,cl ;本行开始的4行语句将一个十六进制数转换为ascii码change: cmp ah,10jl add_0add ah,a-0-10add_0: add ah,0mov hhdi,ahinc didec dhjz extmov ah,dl ;再将异或结果低4位转换成ascii字符and ah,0fhjmp changeext: pop dimov al,sourcesimov resultdi,alinc siinc dimov ah,sourc

7、esimov resultdi,ahinc siinc di待添加的隐藏文字内容3mov dl,hhmov dh,hh+1cmp ax,dx_ errmov al,0dh ;校验正确mov resultdi,alinc dimov al,0ahmov resultdi,alinc dijmp lperr: sub di,11 ;校验错误mov al,!mov cx,_cover: mov resultdi,alinc diloop coverlp: _jz exitjmp repeatexit: nop;*end*lea dx,result ;结果数据区首址lea si,name1 ;结果文件

8、名mov cx,n ;字节数call save ;保存结果到文件retstart endpcode endsend start29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9d1b6c8795ad9c91a

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10、6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af5d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afad9c91fe4ced1b6c8795ad29c91afb6c8795ad29c91afb6c8795ad29c91afd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91d1b6c87

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17、5ad9c91afed1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8d1b6c8795ad9c91afe4ced195ad9c91afe4ced1b6c8795ad91afe4ced1b6cd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af8795ad9c91fe4ced1d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced

18、1b6c8795ad9c91fe4ced1b6c8795ad29c91af5d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afad9c91fe4ced1b6c8795ad29c91afb6c8795ad29c91afb6c8795ad29c91afd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91d1b6c8795ad9c91afe4ced1b

19、6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91affe4ced1b6c8795ad29c91afc8795ad9c91fe4ced1b6c8795ad29c91afd1b6c8795ad9c91afed1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af795ad9c91fe4ced1b6c8795add1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af29c91af4ced1b6c8795ad1b6c8795ad9c91afe4ced1b6c8795ad91afd1b6c8795ad9c91afe4ced1b6c8795