单片机习题答案.pdf

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1、230 何宏单片机原理与接口技术习题解答第 4章习题解答4-1 简述以下基 本概念。答指 令：CPU根据人的 意图来执行某种操作的命令。指令 系统：一台电脑所能执行的全部指令集合。机器 语言：用二进制编码表示，电脑能直接识别和执行的语言。汇编 语言：用助记符、符号和数字来表示指令的程序语言。高级 语言：独立 于 机器的，在编 程时不需 要对机器结构及其指令系统有深入了解的通用性语言。4-2 什么是电脑 的指令 和指 令系 统?答见 1题。4-3 简述 80C51汇编 指令 格式。答操作 码 目的操纵 数 ，源 操作数 4-4 简述 80C51的寻 址方 式和 所能 涉及的寻址空间。答立 即数寻

2、址：程序 存储器ROM。直接 寻址：片内 RAM 低 128B 和 特殊功能寄存器。寄存 器寻址：R0 R7，A，AB，Cy，DPTR。寄存 器间接 寻址：片内 RAM 低 128B，片外RAM。变址 寻址：程序存储器 64KB。相对 寻址：程序存储器 256B范围。位寻 址：片 内 RAM 的 20H 2FH字节地址，部分特殊功能寄存器。4-5 要访问特殊 功能寄 存器 和片 外数据 存储器，应 采用 哪些寻址方式?答 SFR：直接寻址，位 寻址，寄存器寻址；片外 RAM：寄存器间接寻址。4-6 在 80C51 片 内 RAM 中，已知(30H)=38H，(38H)=40H，(40H)=48

3、H，(48H)=90H。请 分析 下面 各是什么指令，说 明源 操作数 的寻址方式 及按 顺序执 行后的结果。答 MOV A，40H 直接 寻址 MOV R0，A 寄存器寻址 MOV P1，#0F0H 立即数寻址 MOV R0，30H 直接寻址 MOV DPTR，#3848H 立即数 寻址 MOV 40H，38H 直接 寻址 MOV R0，30H 直接寻址 MOV P0，R0 寄存器寻址 MOV 18H，#30H 立即数寻址 MOV A，R0 寄存器间接寻址 MOV P2，P1 直接寻址231 均为 数据传 送指令，结果(参见以下图)为RAM 30H 38H 38H 40H 40H 48H 4

4、8H 90H (18H)=30H，(30H)=38H，(38H)=40H(40H)=40H，(48H)=90H R0=38H，A=40H，P0 38H，P1 FOH，P2=FOH，DPTR=3848H 4-7 对 80C51 片 内 RAM 高 128B 的地 址空 间寻址要注意 什么?答用 直接寻址，寄存 器寻址，位寻址。4-8 指出以下指 令的本 质区 别。答 MOV A，data 直接 寻址 MOV A，#data 立即数寻址 MOV data1，data2 直接寻址 MOV 74H，#78H 立即数寻址4-9 设 R0的 内 容 为 32H，A的 内 容 为 48H，片 内 RAM 的

5、 32H内 容 为 80H，40H的 内容 为 08H。请 指出 在执行以下 程序 段后各单元内容的 变化。MOV A，R0；(R0)=80H A MOV R0，40H；(40H)=08H (R0)MOV 40H，A；(A)=80H 40H MOV R0，#35H；35H R0 解 (R0)=35H(A)80H(32H)08H (40H)=80H 4-10 如何 访问 SFR，可 使用 哪些 寻址方 式?答访问 SFR：直 接寻 址，位寻址，寄存器寻址。4-11 如何 访问 片外 RAM 单元，可使用哪 些寻址方式?答只能 采用寄存器间接寻址(用 MOVX 指令)。4-12 如何 访问 片内

6、RAM 单元，可使用哪 些寻址方式?答低 128B：直 接寻址，位寻址，寄存器间接寻址，寄存器寻址(R0 R7)。高 128B：直 接寻址，位寻址，寄存器寻址。4-13 如何 访问 片内外程序存 储器，采用哪 些寻 址方 式?答采用 变址寻址(用 MOVC 指令)。4-14 说明 十进 制调整的原因 和方 法。答压缩 BCD码在 进行 加法运算时本应逢十进一，而电 脑只 将其当作 十六进制数处理，此时得到的结果不正确。用 DAA指令 调整(加 06H，60H，66H)。4-15 说明 80C51 的布尔 处理 机功 能。答用来进行位操 作。文档编码:CB10G1K4G4O7 HJ7A2M1O1

7、C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X

8、3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K

9、4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1

10、O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q

11、6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G

12、1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2

13、M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3232 4-16 已 知(A)=83H，(R0)=17H，(17H)=34H，请 指 出 在 执 行 以 下 程 序 段 后 A的 内容。解 ANL A，#17H ；83H 17H=03H A ORL 17H，A；34H 03H=37H 17H XRL A，R0；03H异 或 37H=34H CPL A；34H求 反

14、等于 CBH 所以(A)=CBH 4-17 使用 位操 作指令实现以 下逻 辑操作。要求不得 改变 未涉及位的内容。SETB ACC.0 或 SETB EOH (2)清 除累加器高 4位4-18 编写 程序，将片内RAM R0 R7的 内容 传送 到 20H 27H单 元。解 MOV 27H，R7 MOV 23H，R3 MOV 26H，R6 MOV 22H，R2 MOV 25H，R5 MOV 21H，R1 MOV 24H，R4 MOV 20H，R0 4-19 编 写 程 序，将 片 内 RAM 的 20H，21H，22H三 个 连 续 单 元 的 内 容 依 次 存 入2FH，2EH，2DH中

15、。解 MOV 2FH，20H MOV 2EH，21H MOV 2DH，22H 4-20 编 写程序，进行两个 16位数 的减 法：6F5DH一 13B4H，结果存入片 内 RAM的 30H和 31H单元，30H存 差的 低 8位。解 CLR C MOV A，#5DH；被 减数低 8位 A MOV R2，#B4H；减 数低 8位 R2 SUBB A，R2；被 减数减去减数，差 A MOV 30H，A；低 8位结果 30H MOV A，#6FH；被 减数高 8位 A MOV R2，#13H；减 数高 8位 R2 SUBB A，R2；被 减数减去减数，差 A MOV 31H，A；高 8位结果 31H

16、 文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K

17、4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1

18、O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q

19、6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G

20、1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2

21、M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I

22、9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3233 4-21 编 写程 序，假设累加器 A的内 容分 别满足以下 条件 时，则程 序转 至 LABEL存 储单 元。设 A中 有的是无符 号数。解 (1)A 10 CJNE A，#10，L1 ；(A)与 10比较，不等转 L1 L2：LJMP LABEL；相

23、等转LABEL L1：JNC L2 ；(A)大 于 10，转 LABEL (2)A10 CJNE A，#10，L1 ；(A)与 10比较，不等转 L1 SJMP L3 ；相等转L3 L1：JNC L2；(A)大 于 10，转 L2 SJMP L4 ；(A)小 于 10，转 L4 L2：JMP LABEL ；无条件转 LABEL (3)A 10 CJNE A，#10，L1；(A)与 10 比 较，不等 转 L1 L2：LJMP LABEL；相 等转 LABEL L1：JC，L2；(A)小 于 10，转 LABEL 4-22 已知 SP=25H，PC=2345H，(24H)=12H，(25H)=3

24、4H，(26H)=56H。问 此时执 行“RET”指令 后，SP=?PC=?解 SP 23H，PC=3412H 4-23 已 知 SP=25H，PC=2345H，标 号 LABEL所 在 的 地 址 为 3456H。问 执 行 长 调用 指令“LCALL LABEL”后，堆 栈指针和堆 栈内容发生 什么 变化?PC的值 等于 什么?解 SP 27H，(26H)=48H，(27H)23H，PC 3456H 4-24 上题 中 LCALL能否 直接 换成 ACALL指令，为什 么?如果 使用 ACALL指 令，则 可调 用的 地址范围是多 少?解不能。ACALL是短 转指令，可调用的地址范围是2

25、KB。4-25 阅读 以下 程序，要求：(1)说 明程序功能；(2)试 修改程序，使片内 RAM 的内容成为 如下图的结 果。文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ

26、1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码

27、:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7

28、 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2

29、ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档

30、编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4

31、O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3234 MOV R2，#0AH 50H 00H MOV R0，#50H 51H 01H CLR A 52H 02H LOOP：MOV R0

32、，A 53H 03H INC R0 54H 04H DJNZ R2，LOOP 55H 05H DONE：56H 06H 57H 07H 58H 08H 59H 09H 解(1)功能是将片内RAM 中 50H 59H单 元清零。(2)7A(OA)(7850)DA(FC)(3)在 INC R0 后 添一句INC A。4-26 设(R0)=7EH，(DPTR)=10FEH，片 内 RAM 中 7EH单 元 的 内 容 为 0FFH，7F单元 的内 容为 38H，试为以下程 序注 释其运行结果。解 INC R0(7EH)00H INC R0 (R0)7FH INC R0 (7FH)39H INC DP

33、TR (DPTR)10FFH INC DPTR (DPTR)1100H INC DPTR (DPTR)1101H 4-27 以 下程序 段经 汇编 后，从 1000H开始的各有 关存 储单 元的内容将是 什么?ORG 1000H TAB1 EQU 1234H TAB2 EQU 3000H DB“START”DW TAB1，TAB2，70H 解(1000H)=53H (1001H)=54H (1002H)=41H(1003H)=52H (1004H)=54H (1005H)=12H(1006H)=34H (1007H)=30H (1008H)=00H(1009H)00H(100AH)=70H 4

34、-28 阅读 以下 程序，并要求：(1)说 明程序功能；(2)写 出涉及的寄 存器及 片内 RAM 单 元(如下图)的最 后结果。40H 98H AFH 文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7

35、HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 Z

36、J1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编

37、码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O

38、7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2

39、 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文

40、档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3235 MOV R0，#40H；40H R0 MOV A，R0；98H A INC R0；41H R0 ADD

41、 A，R0；98+(A)=47H A INC R0 MOV R0，A；结果存入 42H单元 CLR A；清 A ADDC A，#0；进位位存入 A INC R0 MOV R0，A；进位位存入43H 解功 能：将 40H，41H单元 中 的 内容 相 加结 果 放在 42H单 元，进 位 放 在 43H单元，(R0)=43H，(A)1，(40H)=98H，(41H)AFH，(42H)47H，(43H)=01H。4-29 同上 题要 求，程序如下：61H F2H CCH MOV A，61H；F2H A MOV B，#02H；02H B MUL AB；F2H 02H E4H A ADD A，62H；

42、积 的低 8位 加上 CCH A MOV 63H，A；结 果送 63H CLR A；清 A ADDC A，B ；积 的高 8位 加进位位 A MOV 64H，A；结 果送 64H 解功能：将 61H单元 的内容乘2，低 8位再 加 上 62H单 元的内容放人63H，将 结果的高8位放在64H单元。(A)02H (B)01H(61H)F2H (62H)CCH (63H)BOH (64H)02H 4-30 编写 程序，采 用“与”运算，判 断 8位 二进 制数 是奇 数个 1还是偶数个 1。解 MOV A，#XXH ；待判断的数A ANL A，#0FFH；与 0FFH相 与 JB P，REL；是

43、奇数转REL；是 偶数程序顺序执 行 REL：4-31 编写 程序，采用“或”运算，使任意 8位 二进 制数 的符号位必为1。解 MOV A，XXH；取数据 A ORL A，#80H；使该数符号位为 1 MOV XXH，A；保存该数据文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 H

44、J7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ

45、1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码

46、:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7

47、 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2

48、ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档

49、编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3文档编码:CB10G1K4G4O7 HJ7A2M1O1C2 ZJ1U3I9Q6X3236 4-32 请 思 考：

50、采 用“异 或”运 算 怎 样 可 使 一 带 符 号 数 的 符 号 位 改 变，数 据位 不变?怎样 可使 该数必然为 零?解 (1)符号位改变，数据位不变：MOV A，XXH；取 数据A XRL A，#80H；异或80H A (2)使 该数为零：MOV A，XXH；该数A MOV R0，A ；该数R0 XRL A，R0 ；该数自身相异或第 5章习题解答5-1 编 写 程 序，查 找 在 片 内 RAM 中 的 20H 50H单 元 中 是 否 有 0AAH这 一 数 据。假 设有，则 51H单 元置为01H；假设未找 到，则51H单元置为 00H。解 MOV R2，#31H ；数据块长度