计算机等级考试三级PC技术上机17

1、2013年计算机等级考试三级pc技术上机（17）请编制程序，其功能是：从10个有符号字节数据中取出负数并计算其绝对值之和(字型)，然后存放在指定的内存区中，多余的空间填0。例如：内存中有： 80h,01h,02h,00h,ffh,7ch,feh,7bh,fdh,81h结果为 ： 80h,ffh,feh,fdh,81h,05h,01h,00h,00h,00h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

2、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:far,save:farn equ 10dseg segmentsource db n dup(?)result db n dup(0)name0 db input1.dat,0name1 db output1.dat,0dseg endsss

3、eg segment stackdb 256 dup(?)sseg endscseg segmentassume cs:cseg,ss:sseg,ds:dsegstart proc farpush dsxor ax,axpush axmov ax,dsegmov ds,axmov es,ax ;置附加段寄存器lea dx,source ;数据区起始地址lea si,name0 ;原始数据文件名mov cx,n ;字节数call load ;从input1.dat中读取数据;*begin*lea si,sourcelea di,resultmov dx,0mov cx,ncldcont: lod

4、sb_jge _mov di,alinc di_add dl,al_next: loop contmov di,dxadd di,2mov cx,20sub cx,dimov al,0_;*end*lea dx,result ;结果数据区首址lea si,name1 ;结果文件名mov cx,n ;字节数call save ;保存结果到文件retstart endpcseg endsend startqcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeau

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