微机原理与接口技术 顾晖 习题参考答案

INC DI LOOP AGAIN

EXIT: MOV AH,4CH INT 21H CODE ENDS END START 11

18. 答： 19. 答： CODE SEGMENT ASSUME CS:CODE START:

MOV AX,5678H MOV DX,1234H NOT AX NOT DX ADD AX,1 ADC DX,0 EXIT:

MOV AH,4CH INT 21H CODE ENDS END START

;本程序未考虑溢出的情况。 DATA SEGMENT A1 DW 5050H

A2 DW ? ;存A1的反码 A3 DW ? ;存A1的补码 DATA ENDS CODE SEGMENT

ASSUME CS:CODE,DS:DATA START:

MOV AX,DATA MOV DS,AX MOV AX,A1 NOT AX MOV A2,AX INC AX MOV A3,AX EXIT:

MOV AH,4CH INT 21H CODE ENDS END START

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20. 答： 21. 答：

DATA SEGMENT ;AT 5000H ORG 3481H DAT DB 12H DB ?,?,? DATA ENDS CODE SEGMENT

ASSUME CS:CODE,DS:DATA START:

MOV AX,DATA MOV DS,AX MOV AL,DAT NEG AL

MOV DAT+1,AL MOV AL,DAT

XOR AL,00001111B MOV DAT+2,AL MOV AL,DAT

OR AL,11110000B MOV DAT+3,AL EXIT:

MOV AH,4CH INT 21H CODE ENDS END START COUNT=1000 DATA SEGMENT ORG 1000H

DAT DB 10 DUP (12H,-5,-3,0,-128,56H,98H,4,128,200) ORG 2000H MINDAT DB ? DATA ENDS CODE SEGMENT

ASSUME CS:CODE,DS:DATA START:

MOV AX,DATA MOV DS,AX LEA SI,DAT MOV CX,COUNT DEC CX

MOV AL,[SI] NEXT: INC SI

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CMP AL,[SI] JLE ISMIN MOV AL,[SI] ISMIN: LOOP NEXT MOV MINDAT,AL EXIT:

MOV AH,4CH INT 21H CODE ENDS END START 13

22. 答： DATA SEGMENT

STRING1 DB 'hELLO!' COUNT1=$-STRING1 STRING2 DB 'hEL1O!' COUNT2=$-STRING2 IM DB 'MATCH$' NM DB 'NOT MATCH$' DATA ENDS CODE SEGMENT

ASSUME CS:CODE,DS:DATA START:

MOV AX,DATA MOV DS,AX

LEA SI,STRING1 LEA DI,STRING2 MOV CX,COUNT1 MOV BX,COUNT2 CMP CX,BX

JNZ DISPNOTMATCH NEXT:

MOV AL,[SI] MOV AH,[DI] CMP AL,AH

JNZ DISPNOTMATCH INC SI INC DI LOOP NEXT ISMATCH:

MOV DX,OFFSET IM MOV AH,9 INT 21H

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JMP EXIT

DISPNOTMATCH: MOV DX,OFFSET NM MOV AH,9 INT 21H EXIT:

MOV AH,4CH INT 21H CODE ENDS END START 14

23. 答： 24. 答： DSEG SEGMENT DATA DB 5,6,7,8 DW ?

DATA2 DB 1,10,100,20 DSEG ENDS CODE SEGMENT

ASSUME CS:CODE,DS:DSEG START:

MOV AX,DSEG MOV DS,AX MOV DX,0 MOV CX,4 LEA SI,DATA LEA DI,DATA2 NEXT:

MOV AL,[DI] MOV BL,[SI] CALL DOMUL ADD DX,AX INC DI INC SI LOOP NEXT

MOV WORD PTR DATA+4,DX EXIT:

MOV AH,4CH INT 21H DOMUL PROC MUL BL RET

DOMUL ENDP CODE ENDS

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