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信息学院自动化大三上微机原理试卷 部分大题JPG[/fly] 下面是课后答案!~ 答案(4)=101111.00101B
=2F.28H(2)=BC.EH=188.875D
(3)=0.D8H=0.84375D(1)n=8位
11011111B(DFH)
n=16位 ***1B(FFDFH)
(3)n=8位
10000000B(80H)
n=16位 ***0B(FF80H)
(4)n=8位
01111111B(7FH)
n=16位 007FH(***1B)(1)-32768
(2)-347(6)1,01100000 溢出
(7)11111110 不溢出(1)X>Y(2)XY(1)无符号数 136(88H)
原码-8 反码-77H(-119D)
补码-78H(-120D)
8421码 88(2)C9H(201)
-49H(-73D)
-36H(-54D)
-37H(-55D)
不是8421码 第四章作业答案p158
2.(3)(5)(6)(8)(10)对,其余错
4.(1)32000H;(2)33100H;(3)321000H;(4)42000H 5.E0000H~EFFFFH
6.(SP)=1FFEH ;(AX)=5000H ;(BX)=5000H
7.SUB指令:(AL)=0FFH ;(CF)=1;(OF)=0;(ZF)=0 CMP指令:(AL)=08H;(CF)=1;(OF)=0;(ZF)=0
8.(1)AL=7AH;(CF)=0;(SF)=0;(OF)=0;(PF)=0;(AF)=0;(ZF)=0(3)AL=23H;(CF)=1;(SF)=0;(OF)=0;(PF)=0;(AF)=1;(ZF)=0 9.(1)AND AH, 0FH(2)XOR AL, 0F0H(3)MOV CL, 4
SHR AL,CL(4)MOV CL, 4
SHL AL,CL
10.(AX)=4004H;(BX)=01B4H 11.(AX)=3520H
13.(BX)=4154H;(2F246H)=6F30H 14.(BX)=139EH
15.(SI)=0180H ;(DS)=2000H 16.书中题目有误
(CL)=0F6H(1E4F6H)=5678H(BX)=0056H(AX)=1E40H((SI))=0024H
(AX)=5678H;(09226H)=1234H 17.(1)(DX)=0006H;(BX)=0004H
(2)(SP)=0FFEH;((SP))=0FFFFH
(3)(CX)=0004H;(BX)=0004H
(4)(AX)=8094H;(CF)=0
(5)(AL)=87H
(6)(AL)=0CCH;(CF)=0
(AL)=32H
(7)(SI)=0009H;
(8)(DI)=0009H
(9)((DI))=94H
(10)(AX)=17C6H;(DX)=8094H
(11)(AH)=84H;(BL)=04H
(12)(IP)=17C6H
教材10题 code segment
aume cs:code start:
mov ax,1234h;测试数据
mov cl,4
mov dx,ax
and dl,0fh
and dh,0fh
mov bl,dh
mov bh,al
ror bh,cl
and bh,0fh
mov ch,bh
ror ah,cl
and ah,0fh
mov al,ah
mov cl,ch
mov ah,4ch
int 21h code ends
end start
教材11题 DATA
SEGMENT STRING1 DB 'HELLO.' C1
EQU $-STRING1 STRING2 DB 'HELLO!' C2
EQU $-STRING2 S1
DB 'MATCH$' S2
DB 'NOMATCH$' DATA
ENDS CODE
SEGMENT
ASSUME CS:CODE,DS:DATA START:
MOV AX,DATA
MOV DS,AX
MOV SI,C1
CMP SI,C2
JNZ N_M
MOV BX,0 L1:
MOV AL,STRING1[BX]
CMP AL,STRING2[BX]
JNZ N_M
INC BX
DEC SI
JNZ L1
MOV DX,OFFSET S1
JMP EXIT N_M:
MOV DX,OFFSET S2 EXIT:
MOV AH,9
INT 21H
MOV AH,4CH
INT 21H CODE
ENDS
END START
教材12题 CODE SEGMENT
ASSUME CS:CODE START:;MOV DL,7;MOV AH,2;INT 21H MOV AH,1
INT 21H
CMP AL,30H
JBE EXIT
CMP AL,39H
JA EXIT
SUB AL,30H
MOV CL,AL L1:
MOV DL,07H
MOV AH,2
INT 21H
MOV BX,1FFFH;延时 L3:
MOV DX,0FFFFH L2:
DEC DX
JNZ L2
DEC BX
JNZ L3
DEC CL
JNZ L1 EXIT:
MOV AH,4CH
INT 21H CODE ENDS
END START
教材13题 DATA SEGMENT M
DB 1,2,3,4,5,6,7,8,9,89,-1,-2 DB-3,-4,-5,-6,-7,-8,-9,0 P
DB 20 DUP(?)N
DB 20 DUP(?)DATA ENDS CODE SEGMENT
ASSUME CS:CODE,DS:DATA START:
MOV AX,DATA
MOV DS,AX
MOV CX,20
MOV SI,0
MOV DI,0
MOV BX,OFFSET M L:
MOV AL,[BX]
CMP AL,0
JGE POS
MOV N[DI],AL
INC DI
JMP L1 POS: JZ L1
MOV P[SI],AL
INC SI L1:
INC BX
LOOP L
MOV BX,SI
CALL DISP
MOV DL,0DH
MOV AH,2
INT 21H
MOV DL,0AH
MOV AH,2
INT 21H
MOV BX,DI
CALL DISP
MOV AH,4CH
INT 21H DISP PROC
MOV AX,BX
MOV CL,10
DIV CL
PUSH AX
MOV DL,AL
ADD DL,30H
MOV AH,2
INT 21H
POP AX
MOV DL,AH
ADD DL,30H
MOV AH,2
INT 21H
RET DISP ENDP CODE ENDS
END START 教材16题 DATA SEGMENT M
DW 1,2,3,-4,-55,-6,7,-8 COUNT EQU $-M
DW 2 DUP(?)DATA ENDS CODE SEGMENT
ASSUME CS:CODE,DS:DATA START: MOV AX,DATA
MOV DS,AX
MOV SI,0
MOV DI,0
MOV CX,COUNT
SHR CX,1
MOV BX,0 L3:
MOV AX,M[SI]
CMP AX,0
JGE L1
NEG AX L1:
CMP BX,AX
JG L2
XCHG BX,AX
MOV DI,SI L2:
INC SI
INC SI
LOOP L3
MOV AX,M[DI]
MOV M[COUNT],AX
MOV M[COUNT+2],DI
MOV AH,4CH
INT 21H CODE ENDS
END START