微机作业答案(第三版教材).doc

微机 作业答案
CH1
P19
2. 运算器、控制器、存储器、输入设备、输出设备
4.
(1) 29.6875 D
(2) 10010011B,93H
(3) 940D
(4) 01100000B or 96D
(5) FFH, -5BH
(6) -0EH, -72H, -71H, 8EH
(7) (a) 45H or 69D ,no overflow
(b) -6EH or -110D ,no overflow
(c) -46H or -70D , no overflow
(d) -76H , overflow

;补充作业
1.已知数据段12345H单元内容55H,X单元内容为66H，连续执行下列指令后，AL=?，X单元内容=?
MOV AX,1000H
MOV DS,AX
MOV BX,2345H
MOV AL,[BX]
MOV SI,X单元偏移地址
MOV [SI],BL
2. 已知DS=3000H,ES=4000H,BX=1000H,
内存单元地址 内容
31000H 34H
31001H 12H
41000H 78H
41001H 56H
请问执行 1. MOV AX,[BX]; AX=?
2.MOV AX,ES:[BX]; AX=?
3. MOV AH,ES:[BX];AH=?
3. 一个有16个字的数据区，它的起始地址为70A0H:DDF6H,请写出这个数据区首末字单元的物理地址。
4. BUF DB 'THE QUICK BROWN FOX'
LL EQU $-BUF
S1 DB 'HELLO'
LLL EQU $-S1
LLLL EQU $-BUF
请分别写出 LL=?, LLL=?,LLLL=?
(1) AL=55H, X= 45H
(2) AX = 1234H , AX = 5678H , AH = 78H
(3) START ADDRESS: 70A0:DDF6 -> START HARDWARE ADDRESS: 7E7F6H
END HARDWARE ADDRESS: 7E814H
(4) 19,5,24

CH3
1． 写出下列用逻辑地址表示的存储单元的物理地址。
（1）1234H:5678H (179B8H)
2．列表写出下列指令中目标操作数、源操作数的寻址方式，如果有非法的内存操作数请改正，并写出CPU所寻址的逻辑段。
（1）MOV BX, 50 寄存器寻址，立即寻址，
（2）CMP [BX], 100 间接寻址，立即寻址，CMP WORD PTR [BX], 100
CMP BYTE PTR [BX], 100; 数据段
（3）ADD [SI], 1000 间接寻址，立即寻址， ADD WORD PTR [SI], 1000，数据段
（4）MOV BP, SP 寄存器寻址，寄存器寻址，
（5）MOV BX, [BP + 4] 寄存器寻址，基址寻址，堆栈段
（6）MOV AX, [BX + DI + 5] 寄存器寻址，基址加变址寻址，数据段
3．以216为模，将C678H分别和下列各数相加，列表写出十六进制和数，以及A、C、O、P、S、Z 六种状态标志的值。
（1）CF23H (959BH,0,1,0,0,1,0)

P130-131
5(1)
-------------------------
.486
DATA SEGMENT USE16
FIRST DB 12H,34H
SECOND DB 56H,78H
DATA ENDS

CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG:MOV AX,DATA
MOV DS,AX
MOV AL,FIRST
XCHG SECOND,AL
MOV FIRST,AL
MOV AL,FIRST+1
XCHG SECOND+1,AL
MOV FIRST+1,AL
MOV AH,4CH
INT 21H
CODE ENDS
END BEG

5 (2)
-----------------------
.486
DATA SEGMENT USE16
FIRST DB 12H,34H
SECOND DB 56H,78H
DATA ENDS
; Not define the stack segment,use default stack instead.

CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG:MOV AX,DATA
MOV DS,AX

PUSH WORD PTR FIRST
PUSH WORD PTR SECOND

POP WORD PTR FIRST
POP WORD PTR SECOND

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

6.
-------------------------------
TABLE DW 158,258,358,458

contents in the memory as:
9EH,00H,02H,01H,66H,01H,CAH,01H

The number in the AX register is 6601H (26113D)

7.
--------------------------------

1193182 -> 1234DE H

............
MOV EAX, 1193182
MOV EDX, 0
MOV ECX, 433
DIV ECX
MOV XX, AX

8
-------------------------
.486
CODE SEGMENT USE16
ASSUEM CS:CODE
BEG:SAL AL,4
SAL AX,4
MOV AH,4CH
INT 21H
CODE ENDS
END BEG

9.
--------------------------
.486
CODE SEGMENT USE16
ASSUME CS:CODE
BEG: MOV CX,8
MOV AL,3 ; supposed that the number in AL register is 03H
CIR: SHR AL,1
RCL BL,1
LOOP CIR
MOV AL,BL
MOV AH,4CH
INT 21H
CODE ENDS
END BEG

10(1)
----------------------------
.486
DATA SEGMENT USE16
BUF DB 50 DUP(?)
PLUS DB 50 DUP(?)
MINUS DB 50 DUP(?)
DATA ENDS
CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG: MOV AX,DATA
MOV DS,AX
MOV BX,OFFSET BUF
MOV SI,OFFSET PLUS
MOV DI,OFFSET MINUS
MOV CX,50
CIR: MOV AL,[BX]
CMP BYTE PTR [BX],0
JZ NEXT
JG TOPLUS
MOV [DI],AL
INC DI
JMP NEXT
TOPLUS: MOV [SI],AL
INC SI
NEXT: INC BX
LOOP CIR
EXIT: MOV AH,4CH
INT 21H
CODE ENDS
END BEG

10(2)
------------------------------
.486
DATA SEGMENT USE16
BUF DB 50 DUP(?)
NOTO DB 50 DUP(?)
DATA ENDS
CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG: MOV AX,DATA
MOV DS,AX
MOV BX,OFFSET BUF
MOV SI,OFFSET NOTO
MOV CX,50
CIR: CMP BYTE PTR [BX],0
JZ NEXT
MOV AL,[BX]
MOV [SI],AL
INC SI
NEXT: INC BX
LOOP CIR
EXIT: MOV AH,4CH
INT 21H
CODE ENDS
END BEG

10 (3)
------------------------------------
.486
DATA SEGMENT USE16
BUF DB 50 DUP(?)
DATA ENDS
CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG: MOV AX,DATA
MOV DS,AX
MOV BX,OFFSET BUF
MOV CX,50
CIR: CMP BYTE PTR [BX],0
JGE NEXT
NEG BYTE PTR [BX]
NEXT: INC BX
LOOP CIR
EXIT: MOV AH,4CH
INT 21H
CODE ENDS
END BEG

10(4)
------------------------------------
.486
DATA SEGMENT USE16
BUF DB 50 DUP(?)
DATA ENDS
CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG: MOV AX,DATA
MOV DS,AX
MOV BX,OFFSET BUF
BT DWORD PTR [BX],31
JNC EXIT
NEG DWORD PTR [BX]
EXIT: MOV AH,4CH
INT 21H
CODE ENDS
END BEG

-------------------------------------------------------------------------

CH4
1.
.486
DATA SEGMENT USE16
NUMBER DB ?
FLAG DB -1
DATA ENDS
CODE SEGMENT USE16
ASSUME CS:CODE,DS:DATA
BEG: MOV AX,DATA
MOV DS,AX
CMP NUMBER,5
JNA EXIT
CMP NUMBER,24
JA EXIT
MOV FLAG,0
EXIT: MOV AH,4CH
INT 21H
CODE ENDS
END BEG

Assignment 4#
SUPPLEMENT:

SS= 2250H, SP= 0136H
SS= 2250H, SP= 013AH