汇编实验报告

院系：计算机科学学院专业：计算机科学与技术年级： 2011级课程名称：80X86汇编语言程序设计学号：姓名：指导教师：2013年 06 月 10 日；*****************************************************************************；功能：计算成绩总分和平均分；作者：；日期：；****************************************************************************** .386.model flat；------------------------------------------------------------------------------ ExitProcess proto near32 stdcall,dwexitcode:dword;------------------------------------------------------------------------------ Include io.hCr equ 0dh ;回车Lf equ 0ah ;换行;------------------------------------------------------------------------------ .stack 4096 ;预留4096字节的堆栈;------------------------------------------------------------------------------ .DATA ;数据保留区prompt1 byte “please input the first score1:”,0prompt2 byte “please input the second score2:”,0prompt3 byte “please input the third score3:”,0prompt4 byte “please input the fouth score4:”,0out1 byte “the sum is:”sumstr byte 6 dup(?).cr.Lfout2 byte “the average is:”averstr byte 6 dup(?).cr.Lfsum word 0average word ?numstr byte 20 dup(?);------------------------------------------------------------------------------ .code_start:L1:output prompt1input numstr,10ayio numstrcmp ax,0JL L1cmp ax,100JG L1add sum,axL2:output prompt2input numstr,10ayio numstrcmp ax,0JL L2cmp ax,100JG L2add sum,axL3:output prompt4input numstr,10ayio numstrcmp ax,0JL L3cmp ax,100JG L3add sum,axL4:output prompt4 input numstr,10 ayio numstrcmp ax,0JL L4cmp ax,100JG L4add sum,axmov ax,sumitoa sumstr,ax mov bL,4idiv bLmov ah,0itoa avestr,ax output out1 output out2 public _start end；*****************************************************************************；功能：分类统计字符个数；作者：；日期：2013.05.12；****************************************************************************** .386.model flatExitProcess proto near32 stdcall, dwexitcode:dwordINCLUDE io.hcr EQU 0dhlf EQU 0ah.STACK 4096.DATAprompt1 BYTE cr,lf,lf,"Please enter the string(<80).",cr,lf,lfBYTE "The string is:",0stringIn BYTE 1 DUP (?)charLabel BYTE cr,lf,"The number of char is:"char BYTE 16 DUP (?)BYTE cr,lf,0numberLabel BYTE cr,lf,"The number of number is:"number BYTE 16 DUP (?)BYTE cr,lf,0otherLabel BYTE cr,lf,"The number of other is:"other BYTE 16 DUP (?)BYTE cr,lf,0.CODE_start: output prompt1 ；提示输入字符mov ax,0mov bx,0mov cx,0begin:input stringIn,1 ；输入字符cmp stringIn,0dhje forOut ；是回车则结束程序cmp stringIn,0dhjb forOther ；ASCII码值小于0dh则其他字符数加1cmp stringIn,2fhjbe forOther ；ASCII码值小于或等于2fh则其他字符数加1 cmp stringIn,39hjbe forNumber ；ASCII码值小于或等于39h则数字数加1 cmp stringIn,40hjbe forOther；ASCII码值小于或等于40h则其他字符数加1cmp stringIn,5ahjbe forChar ；ASCII码值小于或等于5ah则字母数加1 cmp stringIn,60hjbe forOther ；ASCII码值小于或等于60h则其他字符数加1cmp stringIn,7ahjbe forChar ；ASCII码值小于或等于7ah则字母数加1 jmp forOther ；其他字符数加1forOther: ；其他字符数加1，回到循环开始inc axjmp beginforNumber: ；数字数加1，回到循环开始inc bxjmp beginforChar: ；字母数加1，回到循环开始inc cxjmp beginforOut:itoa other,axoutput otherLabel ；输出其他字符数itoa number,bxoutput numberLabel ；输出数字数itoa char,cxoutput charLabel ；输出字母数INVOKE ExitProcess,0 ；退出程序，返回0 PUBLIC _startEND组内分工（可选）无实验结果分析及心得体会运行结果：Please enter the string(<80).The string is:fhhsj*/656（）+422=The number of char is:5The number of number is:6The number of other is:6心得体会：成绩评定教师签名：2013年月日流程图：代码：；*************************************************************************** ; 求毕达哥拉斯三角形的边; Author:; Date:；**************************************************************************** .386.MODEL FLATExitProcess PROTO NEAR32 stdcall, dwExitCode:DWORD；---------------------------------------------------------------------------- INCLUDE io.hcr EQU 0dh ；回车Lf EQU 0ah ；换行；---------------------------------------------------------------------------- .STACK 4096 ；预留4096字节的堆栈；---------------------------------------------------------------------------- .DATASum1 WORD ?Sum2 WORD ?flag BYTE ?num1 WORD ?num2 WORD ?prompt1 BYTE "please input a number: ",0prompt2 BYTE "The result is: ",cr,lfBYTE " A B C ",cr,lf,0prompt3 BYTE "Sorry, this triangle doesn't exist.",0count1 BYTE 6 DUP (? )count2 BYTE 6 DUP (? )count3 BYTE 6 DUP (? )BYTE cr, lf, 0.CODE_start:output prompt1 ；提示输入一个数input count3,6atoi count3 ；转换为整型mov bx,ax ；将输入的数拷贝到bx中mul ax ；求输入数的平方mov num1,ax ；将所求结果保存到num1中mov sum1,1 ；将1赋值给sum1mov flag,0 ；将0赋给flagfor1: inc sum1 ；sum1加1cmp sum1,bx ；比较sum1和输入的数的大小jae result ；如果大于等于则跳到resultmov ax,sum1 ；将sum1赋给axmul ax ；求出sum1的平方mov num2,axmov cx,bxsub cx,sum1 ；将cx减去sum1的值保存到cx中mov sum2,cx ；将cx赋给sum2for2: inc sum2 ；做加法运算，与1相加赋值给sum2 mov ax,sum2 ；将sum2赋给axcmp ax,bx ；比较ax和bx的大小jae for1 ；如果大于等于则跳到for1mul ax ；求出sum2的平方add ax,num2 ；求出sum1的平方加上sum2的平方赋给axcmp ax,num1 ；比较ax 和num1的大小jne for2 ；如果不等于则跳到 for2进行循环mov flag,1 ；如果等于，则将1赋给标记号flag itoa count3,bx ；将整数转换为ASCII码itoa count2,sum1itoa count1,sum2output count1 ；输出所得结果jmp for1 ；跳到for1进行循环result:cmp flag,1jne none ；如果不相等，则跳到nonejmp endfilenone:output prompt3endfile:INVOKE ExitProcess, 0 ；退出程序，并返回0PUBLIC _startEND；***************************************************************************** ；功能：数组排序；作者：；日期：2013.05.12；****************************************************************************** .386.model flat；------------------------------------------------------------------------------ ExitProcess proto near32 stdcall,dwexitcode:dword;------------------------------------------------------------------------------ Include io.h ;header file for input/outputCr equ 0dh ;carriage return characterLf equ 0ah ;line feed;------------------------------------------------------------------------------ .stack 4096 ;reserve 4096_byte stack;------------------------------------------------------------------------------ .DATA ;reserve storage for dataprompt1 BYTE "Please entry 10 numbers：",cr,Lf,0prompt2 BYTE "Number : ",0number DWORD 20 dup(?) ,0array DWORD 20 dup(?),0prompt3 BYTE "Now the range is : " ,0.CODE_start:output prompt1 ；提示输入10个数据lea ebx,arraymov ecx,numbers ；将元素拷贝到ecx寄存器inNum:output prompt2input number,20 ；输入数据元素atod number ；将数据转换为整型mov [ebx] ,eaxadd ebx, 4 ；进行加法运算，并将结果存储在ebx中loop inNum ；进行inNum循环output prompt3 ；输出结果mov ecx, numberscycle1:mov dx,0lea ebx,arraycycle2:inc dx ；做加法运算，源操作数为1 cmp dx,numbers ；比较dx与numbers的大小jnl cycle3 ；不小于则转移到cycle3mov ax,[ebx]cmp ax,[ebx+4]jg exchange ；大于则转移add ebx,4jmp cycle2exchange:xchg [ebx+4],axmov [ebx],axadd ebx,4jmp cycle2 ；跳转到cycle2cycle3:loop cycle1 ；循环cycle1mov eax,0lea ebx,arraymov ecx,numbersforout:mov eax,[ebx]dtoa number,eax ；将eax转换成ASCII码，并存入number中add ebx,4output [number+6]loop forout ；进行forout循环，输出结果INVOKE ExitProcess, 0 ；退出程序，并返回0 PUBLIC _startEND组内分工（可选）无实验结果分析及心得体会运行结果：Please entry 10 numbers：Number :-5Number :6Number :8Number :1Number :-11Number :3Number :2Number :8Number :-4Number :-6Now the range is :-11 -6 -5 -4 1 2 3 6 8 8心得体会：成绩评定教师签名：2013年月日_Start: mov ax,datamov ds,axmov es,axlea dx,msg1mov ah,09hint 21hlea dx,string1mov ah,0ahint 21hxor bx,bx ；进行bx与bx的异或运算 mov bl,string1+1lea dx,msg3mov ah,09hint 21hlea dx,msg2mov ah,09hint 21hlea dx,string2mov ah,0ahint 21hlea dx,msg3mov ah,09hint 21hcmp bl,string2+1 ；比较bl与string2+1大小ja lop1 ；若bl大于string2+1则转移到lop1 jb lop2 ；若bl小于string2+1则转移到lop2jmp lop3 ；跳转到lop3lop1: mov cl,string2+1jmp lop3lop2: mov cl,string1+1jmp lop3lop3: mov cl,blcldlea di,string2+2lea si,string1+2repz cmpsbja mat1jb mat2jz mat3mat1: mov result,“No Match”lea dx,resultmov ah,09hint 21hjmp exitmat2: mov result,“No Match” lea dx,resultmov ah,09hint 21hjmp exitmat3: mov result,“Match”lea dx,resultmov ah,09hint 21hjmp exitexit: mov ah,4chint 21hcode endsend start程序流程图：；***************************************************************************** ；功能：比较两个字符串；作者：；日期：；****************************************************************************** .386.model flat；------------------------------------------------------------------------------ ExitProcess proto near32 stdcall,dwexitcode:dword;------------------------------------------------------------------------------ Include io.h ;header file for input/outputCr equ 0dh ;carriage return characterLf equ 0ah ;line feed;------------------------------------------------------------------------------ .stack 4096 ;reserve 4096_byte stack;------------------------------------------------------------------------------ STACKSG segment stack 's'dw 64 dup('ST')STACKSG endsDATA segmentBUFFER DB 6,?,6 DUP(?) ;输入缓冲区，定义长度为6，预留实际输入字符个数的计数单元DECN DW ? ;十进制数，预留实际输入字符个数的计数单元OUTPUT DB 16 DUP('0'),'$' ;输出缓冲区，长度为16PRINT DB 'Input the integer: ','$' ;打印提示信息，必须以'$'结尾CRLF DB 0DH,0AH,'$';0DH为回车符，0AH为换行符DATA endsCODE segmentassume cs:CODE,ds:DATA,ss:STACKSG,ES:NOTHINGMAIN proc farmov ax, DATA ;DATA是立即数，不能直接送段寄存器dsmov ds, axmov ah, 9 ;功能号9表示输出字符串，打印提示信息lea dx, PRINTint 21hmov ah, 0AH ;输入十进制数串lea dx, BUFFERint 21hmov ah, 9 ;输出回车换行符lea dx, CRLFint 21hmov cl, BUFFER + 1 ;cl中为输入十进制整数串的实际长度,si指向字符串首字母xor ch, chlea si, BUFFER+2call EXCH_ASC ;将十进制ASC转换为十进制数并保存到ax中mov DECN, ax ;将所得的十进制数存入DECNlea di, OUTPUT+15 ;将OUTPUT 的最后一个字符的偏移量送至dicall DEC_ASC ;将十进制数装换为二进制数，并转换为ASClea dx, OUTPUT ;输出结果call DISPmov ax, 4C00Hint 21hMAIN endpEXCH_ASC procxor ax, ax ;清零ax寄存器jcxz M2 ;测试cx寄存器的内容，cx为十进制字串的长度，cx为0的时候转M2M1: mov bx, 10 ;设置进制为10mul bx ;ax乘以10mov bl, [si] ;将si指向的字符放入bl寄存器inc sisub bx, 30H ;将每一位的ASCII码值减去30H，变成十进制值add ax, bxloop M1 ;loop一次cx减少1M2: retEXCH_ASC endpDEC_ASC procmov bx, 2mov ax, DECNC2: cmp ax, 0jz c3mov dx, 0div bxor dl, 30H ;转换为ASCIImov [di], dldec dijmp short C2C3: retDEC_ASC endpDISP procmov ah, 9int 21hretDISP endpCODE endsend MAIN。