微机原理：顺序、分支、循环、子程序设计

[公司名称]实验二顺序、分支、循环、子程序设计微机原理上机实验[作者姓名][日期]预习报告零、相关知识1，LEA指令和MOV指令的区别一、顺序结构计算m*n-w，最终结果保存在AX其中m=12, n=12, w=4最终结果为AX=008CH=140Dname "sequence strcture"org 100hMOV AX, 1000HMOV DS, AXMOV SI, 1000HMOV [SI], 12MOV [SI+1], 12MOV [SI+2], 4MOV AL, [SI]MOV BL, [SI+1]MUL BLMOV BX, 0MOV BL, [SI+2]SUB AX, BXHLT二、分支结构在DS数据段偏移地址为DATA开始的顺序80个单元中，存放着某班80个同学的微机原理考试成绩。

现欲编程序统计>=90、89~70、69~60和<60分的人数，并将统计的结果存放在当前数据段偏移地址为BUFFER的顺序单元中name “branch structure”START:MOV DX, 0000HMOV BX, 0000HMOV CX, 80LEA SI, DATALEA DI, BUFFERGOON:MOV AL, [SI]CMP AL, 90JC NEXT3INC DHJMP STORNEXT3:CMP AL, 70JC NEXT5INC DLJMP STORNEXT5:CMP AL, 60JC NEXT7INC BHJMP STORNEXT7:INC BLSTOR:INC SILOOP GOONMOV [DI], DHMOV [DI+1], DLMOV [DI+2], BHMOV [DI+3], BLHLT三、循环结构在DS所决定的数据段，从偏移地址BUFFER开始顺序存放100个无符号16位数。

现欲编程序将这100个数按大小顺序排序（以下程序使用的是快速排序算法，也可以使用冒泡排序，嵌套两层循环）name “circle structure”LEA DI, BUFFERMOV BL, 99NEXT0:MOV SI, DIMOV CL, BLNEXT3:MOV AX, [SI]ADD, SI, 2CMP AX, [SI]JNC NEXT 5MOV DX, [SI]MOV [SI-2], DXMOV [SI], AXNEXT5:DEC CLJNZ NEXT3DEC BLJNZ NEXT0HLT四、子程序编写程序，完成将一组BCD数转换成16位二进制数。

（BCD数转换成16位二进制数用子程序实现）下面提供三种方案，分别用寄存器、储存器和堆栈传递参数①使用寄存器传递参数name "text subprogram using register"STACK SEGMENT STACKDB 256 DUP(?)STACK ENDSDATA SEGMENTBCD1 DB 07H, 06H, 07H, 02H, 03HLENG1 DB 5RESULT DW ?DATA ENDSCODEM SEGMENTASSUME CS:CODEM, DS:DATA, SS:STACKSTART:MOV AX, DATAMOV DS, AXLEA SI, BCD1MOV CX, LENG1CALL FAR PTR BCD-16BMOV RESULT, DXMOV AH, 4CHINT 21HCODEM ENDSCODES SEGMENTASSUME CS:CODES;the part of subprogramBCD-16B PROC FARPUSH BXPUSH AXADD SI, CXMOV DX, 0BCDL:DEC SIPUSH CXMOV AL, [SI]AND AL, 0FHCBWMOV BX, AXMOV AX, DXMOV CX, 10MUL CXMOV DX, AXADD DX, BXPOP CXLOOP BCDLPOP AXPOP BXRETBCD-16B ENDPCODES ENDSEND START②使用储存器传递参数name "text subprogram using store" STACK SEGMENT STACKDB 256 DUP(?)STACK ENDSDATA SEGMENTBCD1 DB 07H, 06H, 07H, 02H, 03HADSEG DW ?ADOFST DW ?LENG1 DW ?RESULT DW ?DATA ENDSCODEM SEGMENTASSUME CS:CODEM, DS:DATA, SS:STACK START:MOV AX, DATAMOV DS, AX;entrance parameterMOV AX, SEG BCD1MOV ADSEG, AXMOV AX, OFFSET BCD1MOV ADOFST, AXMOV LENG1, 5CALL FAR PTR BCD-16BMOV AH, 4CHINT 21HCODEM ENDSCODES SEGMENTASSUME CS:CODES;the part of subprogramBCD-16B PROC FARPUSH BXPUSH AXPUSH SIPUSH DSMOV DS, ADSEGMOV SI, ADOFSTMOV CX, LENG1XOR DX, DXADD SI, CXBCDL:DEC SIPUSH CXMOV AL, [SI]AND AL, 0FHCBWMOV BX, AXMOV AX, DXMOV CX, 10MUL CXMOV DX, AXADD DX, BXPOP CXLOOP BCDLPOP AXPOP BXPOP DSPOP SIRETBCD-16B ENDPCODES ENDSEND START③使用堆栈传递参数name "text subprogram using stack"STACK SEGMENT STACKDB 256 DUP(?)STACK ENDSDATA SEGMENTBCD1 DB 07H, 06H, 07H, 02H, 03HLENG1 DW ?RESULT DW ?DATA ENDSCODEM SEGMENTASSUME CS:CODEM, DS:DATA, SS:STACK START:MOV AX, DATAMOV DS, AX;pushPUSH DSLEA SI, BCD1PUSH SIMOV CX, LENG1PUSH CXCALL FAR PTR BCD-16BPOP DXMOV RESULT, DXMOV AH, 4CHINT 21HCODEM ENDSCODES SEGMENTASSUME CS:CODES;the part of subprogramBCD-16B PROC FARPUSH BPMOV BP, SPPUSH DSPUSH SIPUSH CXPUSH BXPUSH AXMOV CX, [BP]+6MOV SI, [BP]+8MOV DS, [BP]+10ADD SI, CXMOV DX, 0BCDL:DEC SIPUSH CXMOV AL, [SI]AND AL, 0FHCBWMOV BX, AXMOV AX, DXMOV CX, 10MUL CXMOV DX, AXADD DX, BXPOP CXLOOP BCDLPOP AXPOP BXPOP CXPOP DSPOP SIPOP BPRET ABCD-16B ENDP CODES ENDSEND START正式实验报告一、实验要求1．教材P74例1。

m=6,n=4,w=7。

结果Q也放在内存中。

2．教材P75例3。

3．从键盘上输入1个数，判断其奇偶性，如果是奇数，屏幕上显示”It is odd”，否则显示”It is even”。

4．从键盘上输入N个字符（N<16），求这N个字符中’A’的个数，并将’A’的个数显示在屏幕上。

5. 已知BUF1中有N1个按从小到大的顺序排列的互不相等的无符号数，BUF2中有N2个从小到大的顺序排列的互不相等的无符号数。

编写程序将BUF1和BUF2中的数合并到BUF3中，使在BUF3中存放的数互不相等且按从小到大的顺序排列。

6. 需要预习报告。

二、实验目的学习顺序、分支、循环结构以及子程序的编程设计。

三、实验过程（调试分析）3.1 textbook page74-1最后的计算结果应该为AX=6*4-7=17D=11Horg 100hMOV AX, 1000HMOV DS, AXMOV SI, 1000HMOV [SI], 6MOV [SI+1], 4MOV [SI+2], 7MOV AL, [SI]MOV BL, [SI+1]MUL BL; 6*4-->ALMOV BX, 0MOV BL, [SI+2]SUB AX, BXHLT3.2 textbook page75-3用"0","1","C","D","E","O","P","Q","R","S","$"的ASCII值代表10位学生的分数，分别为48/49/67/68/69/79/80/81/82/83分name "textbook page75-3"data segmentSCORES DB "0","1","C","D","E","O","P","Q","R","S","$";48/49/67/68/69/79/80/81/82/83endscode segmentstart:ASSUME DS:DATA, CS:CODEMOV AX, DATAMOV DS, AXMOV CX, 10MOV SI, OFFSET SCORES;<--> MOV SI, OFFSET SCORES MOV DI, 0000HGOON:MOV AL, [SI]CMP AL, 90JC NEXT3INC DHJMP STORNEXT3:CMP AL, 70JC NEXT5INC DLJMP STORNEXT5:CMP AL, 60JC NEXT7INC BHJMP STORNEXT7:INC BLSTOR:INC SILOOP GOONMOV [DI], DHMOV [DI+1], DLMOV [DI+2], BHMOV [DI+3], BLendsend start由结果可知：DH=0，DL=5，BH=3，BL=2说明90以上0人，89~70有5人，69~60有3人，小于60有2人3.3 experiment 2-3name "experiment 2-3"data segmentHEX1 DB 0AH,0DH,"It is even$"HEX2 DB 0AH,0DH,"It is odd$";0AH,0DH换行+回车endscode segmentstart:ASSUME DS:DATAMOV AX, DATAMOV DS, AX;read from the keyboardMOV CX, 16DO:MOV AH, 01HINT 21HCMP AL, 13JZ NEXT0MOV BL, ALLOOP DONEXT0:SUB BL, 48MOV AL, BLMOV BL, 2DIV BLCMP AH, 1JNZ NEXT1JZ NEXT2NEXT1:LEA DX, HEX1MOV AH, 09HINT 21HHLTNEXT2:LEA DX, HEX2MOV AH, 09HINT 21H endsend start若输入的为偶数若输入的为奇数3.4 experiment 2-4 name "experiment 2-4" data segmentHEX DB 16 DUP(0) endscode segmentstart:ASSUME DS:DATAMOV AX, DATAMOV DS, AXMOV CX, 16;read from keyboard read:CMP CX, 1JZ STOPMOV AH, 01HINT 21HCMP AL, 65JZ countCMP AL, 13JZ STOPLOOP read;count 'A' characterscount:INC BLLOOP readSTOP:MOV AH, 02HMOV DL, 0AHINT 21H; 打印换行MOV DL, 0DHINT 21H;打印回车ADD BL, 48MOV DL, BLINT 21Hendsend start3.5 experiment 2-5①冒泡排序，没有在BUF3中除去重复的字符此题设两个字符串为BUF1 = ”ACEGIK”BUF2 = ”BDFJLHMN”name "experiment 2-5";此程序不需要压入堆栈保护的数据data segmentBUF1 DB "ACEGIK$"BUF2 DB "BDFJLHMN$"BUF3 DB 14 DUP(0)endsstack segmentDB 256 DUP(0)endscode segmentstart:ASSUME DS:DATA, CS:CODEL, SS:STACKMOV AX, DATAMOV DS, AXLEA BX, BUF1LEA SI, BUF2LEA DI, BUF3CALL removeCALL reorder;段内转移，若段间转移，测试一下格式;print on the screenLEA DX, BUF3MOV AH, 09HINT 21HHLTremove proc nearMOV CX, 6DO1:MOV AX, [BX]MOV [DI], AXINC BXINC DILOOP DO1MOV CX, 8DO2:MOV AX, [SI]MOV [DI], AXINC SIINC DILOOP DO2RETremove endpreorder proc nearMOV CX, 13;注意是13，思考清楚JUDGE1:LEA DI, BUF3PUSH CXMOV CX, 13JUDGE2:MOV AL, [DI]MOV BL, [DI+1]CMP AL, BLJC NEXT1MOV [DI+1], ALMOV [DI], BLNEXT1:INC DILOOP JUDGE2POP CXLOOP JUDGE1RETreorder endpend start②采用串的合并的算法，可以除去BUF3中重复的字符name "experiment 2-5plus"data segmentBUF1 DB "ACEGIK$"BUF2 DB "BCFHILMN$"BUF3 DB 14 DUP(0)endsstack segmentDB 256 DUP(0)endscode segmentstart:ASSUME DS:DATA, CS:CODEL, SS:STACKMOV AX, DATAMOV DS, AXLEA BX, BUF1LEA SI, BUF2LEA DI, BUF3NEXT0:MOV AX, [BX]MOV DX, [SI]CMP AL, DL; 写成CMP AX, DX会出错JC NEXT1JZ NEXT1_5JA NEXT2NEXT1:MOV [DI], AXINC DIINC BXJMP NEXT3NEXT1_5:MOV [DI], AXINC BXINC SIINC DIJMP NEXT3NEXT2:MOV [DI], DXINC DIINC SIJMP NEXT3NEXT3:CMP [BX+1], 36JZ NEXT5CMP [SI+1], 36JZ NEXT6LOOP NEXT0NEXT5:INC SICMP [SI], 36JZ PRINTMOV DX, [SI]MOV [DI], DXINC DILOOP NEXT5NEXT6:INC BXCMP [BX], 36JZ PRINTMOV AX, [BX]MOV [DI], AXINC DILOOP NEXT6。