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发布时间 : 星期四 文章微机原理(王忠民版 课后答案)更新完毕开始阅读014816da80eb6294dd886c66

IP DI SI 0000H 000AH 0008H 8094H 1403H 1 AX BP CF 2050AH 2050BH 2050CH 37H C5H 2FH 94H 不变 不变

(1)MOV DX,[BX+2] (2)PUSH CX (3)MOV CX,BX (4)TEST AX,01 (5)MOV AL,[SI] (6)ADC DAA (7)INC (8)DEC

AL,[DI] SI DI

;(DX)=0006H,(BX)=0004H ;(SP)=0FFEH

;(CX)=0004H,(BX)=0004H ;(AX)=8094H,(CF)=0 ;(AL)=87H

;(AL)=0CCH,(CF)=0 ;(AL)=32H ;(SI)=0009H ;(DI)=0009H

(9)MOV [DI],AL ;((DI))=94H

(10)XCHG AX,DX ;(AX)=17C6H,(DX)=8094H (11)XOR AH,BL

;(AH)=84H,(BL)=04H

(12)JMP DX ;(IP)=17C6H 18、(DS)=2000H,(BX)=1256H,(SI)=528FH,偏移量=20A1H,(232F7H)=3280H,(264E5H)=2450H,

试求执行下述指令后的结果。

(1) JMP BX

;(IP)=1256 H ;(IP)=3280 H

(2) JMP TABLE[BX]

(3) JMP [BX][SI] ;(IP)=2450 H

19、设(IP)=3D8FH,(CS)=4050H,(SP)=0F17H,当执行CALL 2000H:0094H后,试指出(IP)、(CS)、

(SP)、((SP))、((SP)+1)、((SP)+2)和((SP)+3)的内容。

CALL指令是5字节指令,下一条指令地址为4050H:3D94H 所以执行后 (IP)=0094H,(CS)=2000H、(SP)=0F13H ((SP))=94H,((SP)+1)=00H,((SP)+2)=00H,((SP)+3)=20H

第五章 汇编语言程序设计

2.PLENTH的值为0022,它表示当前已分配单元空间; 3.L的值为6;

5.(AX)=000AH;(BL)=0AH;(CL)=01H; 10.

MOV AX, 4A82H

MOV DL,AH AND DL,0F0H MOV CL,4

SHR DL,CL PUSH AX

AND AH,0FH MOV BH,AH AND AL,0F0H MOV BL,AL MOV CL,4 SHR BL,CL MOV CL,BH POP AX

AND AL,0FH

11. data segment

string1 db 'I am a student.' string2 db 'I am a student .' yes db 'match',0dh,0ah,'$' no db 'no match',0dh,0ah,'$'

data ends

code segment

assume cs:code,ds:data,es:data start: push ds

sub ax,ax push ax mov ax,data mov ds,ax mov es,ax lea si,string1 lea di,string2 mov cx,string2-string1 cld

repe cmpsb jnz dispno

mov ah,9 lea dx,yes int 21h jmp exit

dispno: mov ah,9 lea dx,no int 21h

exit: MOV AH,4CH INT 21H code ends

13. DATA SEGMENT

M DB 11H,22H,33H,44H,64H,87H,34,29,67H,88H,0F6H DB 43H,0B7H,96H,0A3H,233,56H,23H,56H,89H

C P

EQU 20 DB 20 DUP(?)

N DB 20 DUP(?)

PLUS DB 'PLUS',0DH,0AH,'$' MINUS DB 'MINUS',0DH,0AH,'$' J DB 2 DUP(?) DATA ENDS

CODE SEGMENT ASSUME CS:CODE, DS:DATA START: MOV AX,DATA MOV DS,AX

LEA SI,M LEA DI,P LEA BX,N MOV CX,C MOV DX,0

LOOP1: LODSB

TEST AL,80H

JNZ MINUS1 ;负数转移 MOV [DI],AL

INC DI

INC DH ;存正数个数 JMP AGAIN

MINUS1: MOV [BX],AL

INC BX

INC DL

;存负数个数

AGAIN: DEC CX

JNZ LOOP1

MOV WORD PTR J,DX ;存结果

MOV DX,OFFSET MINUS MOV AH,9

INT 21H ;显示提示信息 MOV BL,J MOV CH,2 ROTATE: MOV CL,4 ROL BL,CL MOV AL,BL AND AL,0FH ADD AL,30H CMP AL,3AH JL POR ADD AL,7 POR: MOV DL,AL MOV AH,2 INT 21H

DEC CH JNZ ROTATE ;十六进制形式输出负数个数 MOV AH,2 MOV DL,0DH INT 21H MOV DL,0AH

INT 21H

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

MOV DH,J+1 MOV CH,2 ROTATE1: MOV CL,4 ROL DH,CL MOV AL,DH AND AL,0FH ADD AL,30H CMP AL,3AH JL POR1 ADD AL,7