微机原理习题_解答
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微机原理习题3-3解：（1）立即数寻址（2）基址寻址，EA=BX+DISP，PA=DS*16+EA（3）寄存器寻址（4）基址加变址寻址，EA=BX+SI，PA=DS*16+EA（5）基址寻址，EA=BP，PA=SS*16+EA （6）基址寻址，EA=BX+10H，PA=DS*16+EA （7）基址寻址，EA=BX，PA=ES*16+EA（8）基址加变址寻址，EA=BX+SI+20H，PA=DS*16+EA3-5解：(1)X （2）√（3）X （4）√（5）X (6) √(7)X (8) √(9)X (10) √(11) √(12) √(13) √(14)X (15) √(16)X3-6解：(1) X BX和BP不能同时使用(2) X 源操作数和目的操作数不能同时为存储器(3) X 不能将立即数赋值给段寄存器(4) X 不能给CS赋值(5) X 立即数不能为目的操作数(6) √(7) X 段寄存器不能互相赋值(8) X 移位指令中的移位次数只能是1或者为CL(9) X NOT指令是单操作数指令(10) √(11) X 不可以把立即数入栈(12) 直接端口地址必须小于等于0FFH(13) √(14) 不能用减法(15) 不能用减法(16) √3-7解：（1）AX=3355H，SP=1FFEH（2）AX=3355H，DX=4466H, SP=1FFEH3-8解：BX= 4154H, [2F246H]=6F30H3-9解：SI=0180H, DS=2000H3-10解：(1) CL=0F6H(2) [1E46FH]=5678H(3) BX=56H, AX=1E40H(4) SI=00F6H, [SI]=0024H(5) AX=5678H, [09226H]=1234H3-11解：MOV AX,[2C0H]MOV AX,100[DI]MOV AX,[BP]MOV AX,80H[DI][BX]3-13解：(1) MOV CX, [BLOCK+12](2) MOV BX, OFFSET BLOCKADD BX,12MOV CX,[BX](3) MOV BX, OFFSET BLOCKMOV CX,12[BX](4) MOV BX, OFFSET BLOCKMOV SI,7MOV CX, [BX][SI]3-14解：MOV BX,0A80HMOV AL,5XLAT3-16解：(1) LEA SI, NUM1LEA DI, NUM2MOV CX,2CLCAGAIN:MOV AX, [SI]ADC AX, [DI]MOV [DI], AXINC SIINC SIINC DIINC DILOOP AGAINADC AX,0MOV [DI], AX(2) LEA SI, NUM2MOV CX,3CLCMOV AL, [SI]AGAIN:INC SIADC AL, [SI]ADC AH, 0LOOP AGAINMOV [RES], AX3-17解：(1) MOV BX, OFFSET NUM2MOV CX, 4MOV AL, 0AGAIN:ADD AL, [BX]DAAMOV DL,ALMOV AL,AHADC AL,0DAAMOV AH,ALMOV AL,DLINC BXLOOP AGAINMOV [RES],AX(2) MOV AL, [NUM1]SUB AL, [NUM2]DASMOV [RES],ALMOV AL,[NUM1+1]SBB AL, [NUM2+1]DASMOV [RES+1],AL3-18解：(1) MOV AL, NUM1MUL BYTE PTR [NUM2]MOV RES,AX(2) MOV AX,NUM1IMUL WORD PTR [NUM2]MOV [RES],AXMOV RES+2],DX(3) MOV AL, NUM1MOV AH, 0MOV BL, 46HDIV BLMOV RES, AX(4) MOV AX, NUM1CWDMOV BX, NUM2IDIV BXMOV [RES],AXMOV [RES+2],DX3-23解：(1) AND BX, 0F7AFH(2) OR CX, 01H(3) XOR AX, 4020H(4) TEST DX, 0201H(5) XOR AL, 55HOR AL, 0AAH(6) MOV CL,4SHL AX,CLMOV CL, 4SHR AL,CL3-26解：STRING DB 20 DUP(?)DESTIN DB 20 DUP(?)LEA SI, STRINGLEA DI, DESTINMOV CX, 20CLDREP MOVSB3-29解：IP=009AH, CS=2000H, SP= 0F178H, [SP]=8FH, [SP+1]=3DH, [SP+2]=50H, [SP+3]=40HX DW ?Y DW ?Z DW ?S DW ?MOV AX, XCMP AX, YJNZ XNEYCMP AX, ZJNZ SEQ1MOV S,2JMP EXIT XNEY:CMP AL,ZJZ SEQ1MOV S,0JMP EXITSEQ1:MOV S,1EXIT:3-37解：MOV AX, 0B800HMOV DS, AXMOV BX, 0MOV CX,100MOV DX,0MOV AX,0 AGAIN:ADD AX, [BX]ADC DX, 0INC BXINC BXLOOP AGAINV AR2V AR3 V AR4（1）V AR1的偏移地址是30HV AR2的偏移地址是3AHV AR3的偏移地址是40H(2)DA TA1=50H, CNT= 6(3)[V AR2+2] =02H4-10解：DA TA SEGMENTORG 100HV ARW DW 1234H,5678HV ARD DD 12345678HBUFF DB 10 DUP(?)MESS DB ’HELLO’BEGIN:MOV AX, OFFSET MESS ;AX=112HMOV AX, TYPE BUFF+TYPE MESS+TYPE V ARD ;AX=1+1+4=6MOV AX, SIZE V ARW+SIZE BUFF +SIZE MESS ;AX=2+10+1=13MOV AX, LENGTH V ARW+LENGTH V ARD ;AX=1+1=2MOV AX, LENGTH BUFF+SIZE V ARW ;AX=10+2=12MOV AX, TYPE BEGIN ;AX=FFFFHMOV AX, OFFSET BEGIN ;AX=04-12解：DA TA SEGMENT WORDDABUF DB 100 DUP(?)DA TA ENDSSTACK SEGMENT PARA STACKDB 100 DUP(?)STACK ENDSCODE SEGMENT ‘CODE’ASSUME CS:CODE,DS:DA TA,ES:DATA,SS:STACKSTART:MOV AX, DA TAMOV DS,AXMOV ES,AXMOV AX,STACKMOV SS,AXMOV CX,100CLDMOV BX,OFFSET DABUFMOV AL,64HREP SCASBMOV AH,4CHINT 21HCODE ENDSEND START4-16解：DA TA SEGMENTX DB 4 DUP(?)Y DB 4 DUP(?)Z DB 4 DUP(?)DA TA ENDSCODE SEGMENTSTART:MOV AX,DA TAMOV DS,AXMOV SI, OFFSET XMOV DI, OFFSET YMOV BX, OFFSET ZMOV CX,4CLCAGAIN:MOV AL,[SI]ADC AL,[DI]MOV [BX],ALINC SIINC DIINC BXLOOP AGAIN4-20解：DA TA SEGMENTFIRST DB 99 DUP(？),’$’DA TA ENDSCODE SEGMENTSTART:MOV AX,DA TAMOV DS,AXMOV BX,OFFSET FIRSTMOV CX,100MOV DX,0 AGAIN:MOV AL,[BX]CMP AL.’$’JZ EXITCMP AL,’A’JNZ NEXTINC DX NEXT:INC BXLOOP AGAIN EXIT:MOV AH,4CHINT 21HCODE ENDSEND START。
