1、汇编语言沈美明温冬婵课后答案 汇编语言程序设计(第二版) ( 清华大学 IBM-PC 汇编语言程序设计(第二版) 沈美明温冬婵 编著)第二章 1、答:直接由指令指定的I/O端口数为256个。 2、答: 3、答:字节单元:(30022H) = AB H,(30024H) = EF H 字单元: (30021H) = AB34 H,(30022H) = CDAB H。 4、答:3017:000A的存储单元的物理地址是3017AH, 3015:002A的存储单元的物理地址是3017AH, 3010:007A的存储单元的物理地址是3017AH。 5、答:该程序的第一个字的物理地址是0AAA40H。 6
2、、答:条件标志OF、SF、ZF、CF的值依次分别为0、0、0、0。 7、答:(1)AX、BX、CX、DX、AH、AL、BH、BL、CH、CL、DH、DL、 SP、BP、DI、SI(注意:学生尽量不要用SP参与加减运算) (2)CX (3)DX、AX、AH、AL (4)CS、DS、ES、SS (5)FLAGS (6)IP (7)SS、SP、BP 8、答:可以用来指示存储器地址的寄存器有BX、SP、BP、DI、SI、IP、CS、DS、 ES、SS。 9、答:唯一正确的是D。 第三章 2、答: (1) ADD DX, BX (2) ADD AL, BXSI (3) ADD BX+0B2H, CX (
3、4) ADD 0524H, 2A59H (5) ADD AL, 0B5H 3、答: (1)MOV BX, OFFSET BLOCK + 0AH MOV DX, BX (2)MOV BX, 0AH MOV DX, BLOCKBX (3)MOV BX, OFFSET BLOCK MOV SI, 0AH MOV DX, BXSI 4、答: (1)1200H (2)0100H (3)4C2AH (4)3412H (5)4C2AH (6)7856H (7)65B7H 6、答: MOV BX, 2000H LES DI, BX MOV AX, ES : DI 7、答: (1) 064DH (2) 0691
4、H (3) 05ECH 9、答: (1) MOV AX, BX+0CH MOV ZERO, AX (2) MOV AX, ARRAYBX MOV ZERO, AX 10、答: (1)(AX)= 1234H (2)(AX)= 0032H 11、答: (AX)= 1E00H 12、答: LEA BX, CSTRING MOV DL, BX MOV DH, BX+6 13、答: 14、答: LES BX, 2000 MOV AX, ES:BX 16、答: (1) 74D4H SF=0 ZF=0 CF=0 OF=0 (2) A5C1H SF=1 ZF=0 CF=0 OF=1 (3) 3240H SF=
5、0 ZF=0 CF=1 OF=0 (4) 0000H SF=0 ZF=1 CF=1 OF=0 17、答: (1) 0C754H SF=1 ZF=0 CF=1 OF=0 (2) 12B0H SF=0 ZF=0 CF=0 OF=0 (3) 45B0H SF=0 ZF=0 CF=0 OF=1 (4) 9F24H SF=1 ZF=0 CF=0 OF=0 21、答: (1) MOV AX, Z SUB AX, X ADD AX, W MOV Z, AX (2) MOV BX, X ADD BX, 6 MOV CX, R ADD CX, 9 MOV AX, W SUB AX, BX SUB AX, CX
6、MOV Z, AX (3) MOV AX, W IMUL X MOV BX, Y ADD BX, 6 IDIV BX MOV Z, AX MOV R, DX 22、答: NEG DX NEG AX SBB DX, 0 16、答: MOV AX, A MOV DX, A+2 TEST DX, 8000H JZ STORE ; 为正 NEG DX NEG AX SBB DX, 0 STORE : MOV B, AX MOV B+2, DX 17、答: (1) MOV AL, S SUB AL, 6 DAS ADD AL, V DAA MOV U, AL (2) MOV AL, Z SUB AL,
7、U DAS MOV U, AL MOV AL, X ADD AL, W DAA SUB AL, U DAS MOV U, AL 23、答: (1)(BX)= 9AH (2)(BX)= 61H (3)(BX)= 0FBH (4)(BX)= 1CH (5)(BX)= 0 (6)(BX)= 0E3H 26、答: 把(DX)(AX)中的双字左移四位(乘以16)。 20、答: MOV CL, 4 SHR AX, CL MOV BL, DL SHR DX, CL SHL BL, CL OR AH, BL 31、答: (1)CLD MOV CX, 132 MOV AL, 20H LEA DI, ARRAY
8、REP STOSB (2)CLD MOV CX, 9 MOV AL, * LEA DI, ADDR REPNE SCASB JNE L1 L2 : ; 找到 L1 : ; 未找到 (3)CLD MOV CX, 30 MOV AL, 20H LEA DI, NAME REPE SCASB JNE DO_NOT MOV CX, 30 MOV AL, $ LEA DI, NAME REP STOSB DO_NOT: (4)CLD MOV CX, 30 LEA SI, NAME LEA DI, ARRAY REP MOVSB STD MOV CX, 9 LEA SI, ADDR+8 LEA DI, A
9、RRAY+131 REP MOVSB 34、答: (1) 转L1 (2) 转L1 (3) 转L2 (4) 转L5 (5) 转L5 36、答: 2 p q 时,(AX) 2 2 p q 时,(AX) 1 38、答: (1)(AX)= 5 (BX)= 16 (CX)= 0 (DX)= 0 (2)(AX)= 2 (BX)= 4 (CX)= 3 (DX)= 1 (3)(AX)= 3 (BX)= 7 (CX)= 2 (DX)= 0 39、答: 第四章 1、答: (2) 源*作数和目的*作数同为存储器寻址方式。 (3) SI、DI同为变址寄存器。 (7) 目的*作数不能是代码段段寄存器CS。 (5) 缺少
10、 PTR 5、答: BYTE_VAR 42 59 54 45 0C EE 00 ? - 01 02 01 02 ? 00 ? 01 02 01 02 ? 00 ? 01 02 - 01 02 ? WORD_VAR 00 00 01 00 02 00 00 00 - 01 00 02 00 00 00 01 00 02 00 00 00 01 00 02 00 - 00 00 01 00 02 00 ? ? FB FF 59 42 45 54 56 02 - 8、答: PLENTH的值为22(16H)。 12、答: (1) 10025 (2) 25 (3) 2548 (4) 3 (5) 103
11、(6) 0FFFFH (7) 1 (8) 3 5假设数据段中数据定义如下: VAR DW 34 VAR1 DB 100, ABCD VAR2 DD 1 COUNT EQU $-VAR1 X DW 5 DUP (COUNT DUP (0) Y LABEL WORD Z DB 123456 V DW 2, $-VAR 执行下面程序段并回答问题。 MOV AX, COUNT ; (AX) = ? MOV BX, Z-X ; (BX) = ? MOV CX, V+2 ; (CX) = ? MOV DX, VAR ; (DX) = ? MOV Y+3, 2 MOV SI, Y+4 ; (SI) = ?
12、ADD Z+5, 1 MOV DI, WORD PTR Z+4 ; (DI) = ? 、答: (AX)= 9 (BX)= 90 (CX)= 109 (DX)= 3334H (SI)= 3600H (DI)= 3700H 14、答: (1) (AX)= 1 (2) (AX)= 2 (3) (CX)= 20 (4) (DX)= 40 (5) (CX)= 1 17、答: D_SEG SEGMENT D_WORD LABEL WORD AUGEND DD 99251 S_WORD LABEL WORD SUM DD ? D_SEG ENDS E_SEG SEGMENT E_WORD LABEL WOR
13、D ADDEND DD -15962 E_SEG ENDS C_SEG SEGMENT ASSUME CS:C_SEG, DS:D_SEG, ES:E_SEG MAIN PROC FAR START: PUSH DS MOV AX, 0 PUSH AX MOV AX, D_SEG MOV DS, AX MOV AX, E_SEG MOV ES, AX MOV AX, D_WORD MOV BX, D_WORD+2 ADD AX, ES:E_WORD ADC BX, ES:E_WORD+2 MOV S_WORD, AX MOV S_WORD+2, BX RET MAIN ENDP C_SEG E
14、NDS END START 16、答: DATASG SEGMENT AT 0E000H WORD_ARRAY LABEL WORD BYTE_ARRAY DB 100 DUP (?) DATASG ENDS STACKSG SEGMENT PARA STACK STACK DW 32 DUP (?) TOS LABEL WORD STACKSG ENDS CODESG SEGMENT ORG 1000H MAIN PROC FAR ASSUME CS:CODESG, DS:DATASG, ES:DATASG, SS:STACKSG START: MOV AX, STACKSG MOV SS,
15、 AX MOV SP, OFFSET TOS PUSH DS SUB AX, AX PUSH AX MOV AX, DATASG MOV DS, AX MOV ES, AX RET MAIN ENDP CODESG ENDS END START 9编写一个完整的程序,要求把含有23H,24H,25H,26H四个字符数据的数据区复制20次。 、答: DSEG SEGMENT VAR1 DB 23H,24H,25H,26H DSEG ENDS ESEG SEGMENT VAR2 DB 80 DUP (?) ESEG ENDS CSEG SEGMENT ASSUME CS:CSEG, DS:DSEG
16、, ES:ESEG MAIN PROC FAR START: PUSH DS MOV AX, 0 PUSH AX MOV AX, DSEG MOV DS, AX MOV AX, ESEG MOV ES, AX MOV DX, 20 CLD LEA DI, VAR2 AGAIN: LEA SI, VAR1 MOV CX, 4 REP MOVSB DEC DX JNZ AGAIN RET MAIN ENDP CSEG ENDS END START 第五章 1、答: mov cx, count lea si, string1 lea di, string2 again: mov al, si mov
17、 di, al inc si inc di loop again 2、答: code segment assume cs: code main proc far start: push ds mov ax, 0 push ax mov ah, 1 int 21h sub al, 30h cmp al, 0 jz exit mov cl, al mov ch, 0 again: mov ah, 2 mov dl, 7 int 21h loop again exit: ret main endp code ends end start 8、答: MOV CX,8 MOV DL,0 NEXT3: R
18、OR AX,1 JNC NEXT1 ROR AX,1 JNC NEXT2 INC DL NEXT2: LOOP NEXT3 ADD DL, 30H MOV AH, 2 INT 21H MOV AH, 4CH INT 21H NEXT1: ROR AX, 1 JMP NEXT2 12、答: mov cx, 100 lea di, mem mov ax, 0 cld comp: repne scasw jcxz exit push cx mov si, di sub di, 2 mov bx, di rep movsw mov word ptr di, 0 mov di, bx pop cx jm
19、p comp exit: 13、答: mov dx, 100 mov si, 0 repeat: mov al, stringsi cmp al, 30h jb goon cmp al, 39h ja goon or cl, 20h ; (cl)51 jmp exit goon: inc si dec dx jnz repeat and cl, 0dfh ; (cl)50 exit: 14、答: table dw 100h dup (?) mdata dw ? ; 存放出现次数最多的数 count dw 0 ; 存放出现次数 mov bx, 100h mov di, 0 ; di为数组TABL
20、E的指针 next: mov dx, 0 mov si, 0 mov ax, tabledi mov cx, 100h comp: cmp tablesi, ax jne addr inc dx addr: add si, 2 loop comp cmp dx, count jle chang mov count, dx mov mdata, ax chang: add di, 2 dec bx jnz next mov cx, count mov ax, mdata 19、答: a dw 15 dup (?) b dw 20 dup (?) c dw 15 dup (?) mov si, 0
21、 ; si为数组A的指针 mov bx, 0 ; bx为数组C的指针 mov cx, 15 loop1: mov di, 0 ; di为数组B的指针 push cx mov cx, 20 mov ax, asi loop2: cmp bdi, ax jne no mov cbx, ax add bx, 2 jmp next no: add di, 2 loop loop2 next: add si, 2 pop cx loop loop1 21、答: mov dx, 0 lea si, array mov ax, si mov bx, si+2 cmp ax, bx jne next1 inc
22、 dx next1: cmp si+4, ax jne next2 inc dx next2: cmp si+4, bx jne num inc dx num: cmp dx, 3 jl disp dec dx disp: mov ah, 2 add dl, 30h int 21h 第六章 2、答: 2、答: (1) NAME1 NAMELIST (2) MOV AX,DATA ;假设结构变量NAME1定义在数据段DATA中 MOV DS,AX MOV ES,AX ; MOV AH,10 LEA DX,NAME1 INT 21H ; MOV CL,NAME1.ACTLEN MOV CH,0 L
23、EA SI,NAME1.NAMEIN LEA DI,DISPFILE CLD REP MOVSB 6、答: SKIPLINES PROC NEAR PUSH CX PUSH DX MOV CX,AX NEXT: MOV AH,2 MOV DL,0AH INT 21H MOV AH,2 MOV DL,0DH INT 21H LOOP NEXT POP DX POP CX RET SKIPLINES ENDP 7、答: dseg segment num dw 76,69,84,90,73,88,99,63,100,80 n dw 10 s6 dw 0 s7 dw 0 s8 dw 0 s9 dw 0
24、 s10 dw 0 dseg ends code segment main proc far assume cs:code, ds:dseg start: push ds sub ax, ax push ax mov ax, dseg mov ds, ax call sub1 ret main endp sub1 proc near push ax push bx push cx push si mov si, 0 mov cx, n next: mov ax, numsi mov bx, 10 div bl mov bl, al cbw sub bx, 6 sal bx, 1 inc s6b
25、x add si,2 loop next pop si pop cx pop bx pop ax ret sub1 endp code ends end start 8、答: data segment maxlen db 40 n db ? table db 40 dup (?) char db a ; 查找字符a even addr dw 3 dup (?) data ends code segment assume cs:code, ds:data main proc far start: push ds mov ax, 0 push ax mov ax, data mov ds, ax
26、lea dx, maxlen mov ah, 0ah int 21h ; 从键盘接收字符串 mov addr, offset table mov addr+2, offset n mov addr+4, offset char mov bx, offset addr ; 通过地址表传送变量地址 call count ; 计算CHAR的出现次数 call display ; 显示 ret main endp count proc near ; count子程序 push si push di push ax push cx mov di, bx mov si, bx+2 mov cl, byte
27、 ptrsi mov ch, 0 mov si, bx+4 mov al, byte ptrsi mov bx, 0 again: cmp al, byte ptrdi jne l1 inc bx l1: inc di loop again pop cx pop ax pop di pop si ret count endp display proc near ; display子程序 call crlf ; 显示回车和换行 mov dl, char mov ah, 2 int 21h mov dl, 20h mov ah, 2 int 21h mov al, bl and al, 0fh add al, 30h cmp al, 3ah jl print add al, 7 print: mov
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