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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 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 (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) = ? 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 WORD 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 ENDS 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, 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, 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
1、答: ……
mov cx, count lea si, string1 lea di, string2 again:
mov al, [si] mov [di], al inc si inc di
第五章
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: ROR 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