微机原理及应用(第3版) - 吴宁 - 马旭东 - 部分习题 - 解答 下载本文

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+EA

3-5解:

(1)X (2)√ (3)X (4)X (5)X (6) √(9)X (10) √ (11) √ (12) √ (13) √ (14)X

3-6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=1FFEH

3-8解:

BX= 4154H, [2F246H]=6F30H

3-9解:

SI=0180H, DS=2000H

1

(7)X (8) √ (15) √ (16)X 3-10解: (1) CL=0F6H

(2) [1E46FH]=5678H

(3) BX=56H, AX=1E40H (4) SI=00F6H, [SI]=0024H

(5) AX=5678H, [09226H]=1234H

3-11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 BLOCK ADD BX,12 MOV CX,[BX]

(3) MOV BX, OFFSET BLOCK MOV CX,12[BX]

(4) MOV BX, OFFSET BLOCK MOV SI,7 MOV CX, [BX][SI]

3-14解: MOV BX,0A80H MOV AL,5 XLAT

3-16解: (1) LEA SI, NUM1 LEA DI, NUM2

MOV CX,2 CLC AGAIN: MOV AX, [SI] ADC AX, [DI] MOV [DI], AX INC SI INC SI INC DI INC DI LOOP AGAIN ADC AX,0

2

MOV (2) LEA MOV CLC MOV

AGAIN: INC

ADC ADC

LOOP MOV 3-17解: (1) MOV MOV MOV AGAIN: ADD DAA MOV MOV ADC DAA MOV MOV INC LOOP MOV (2) MOV SUB DAS MOV MOV SBB DAS MOV 3-18解: (1) MOV MUL MOV

[DI], AX SI, NUM2 CX,3 AL, [SI] SI

AL, [SI] AH, 0 AGAIN

[RES], AX

BX, OFFSET NUM2 CX, 4 AX, 0 AL, [BX]

DL,AL AL,AH AL,0

AH,AL AL,DL BX AGAIN [RES],AX AL, [NUM1] AL, [NUM2] [RES],AL

AL,[NUM1+1] AL, [NUM2+1] [RES+1],AL

AL, NUM1

BYTE PTR [NUM2] RES,AX

3

(2) MOV AX,NUM1 IMUL WORD PTR [NUM2] MOV [RES],AX MOV RES+2],DX (3) MOV AL, NUM1 MOV AH, 0 MOV BL, 46H DIV BL MOV RES, AX (4) MOV AX, NUM1 CWD MOV BX, NUM2 IDIV BX MOV [RES],AX MOV [RES+2],DX

3-23解: (1) AND BX, 0F7AFH (2) OR CX, 01H (3) XOR AX, 4020H (4) TEST DX, 0201H (5) XOR AL, 55H OR AL, 0AAH (6) MOV CL,4 SHL AX,CL MOV CL, 4 SHR AL,CL

3-26解: STRING DB 20 DUP(?) DESTIN DB 20 DUP(?) LEA SI, STRING LEA DI, DESTIN MOV CX, 20 CLD REP MOVSB

3-29解:

IP=009AH, CS=2000H, SP= 0F178H, [SP]=8FH, [SP+1]=3DH, [SP+2]=50H, [SP+3]=40H

4