对于下列程序段: AGAIN:MOV ES:[DI],AL INC DI LOOP AGAIN 可用指令______完成相同的功能。( )
有一段程序如下:
MOV CX,100
LEA SI,XSI
MOV DI,OFFSET
CLD
REP MOVSW
有一段程序如下:
MOV CX,100
LEA SI,XSI
MOV DI,OFFSET
CLD
REP MOVSW
请指出以下各指令的源、目的操作数所使用的寻址方式。
(1)MOV SI,2100H
(2)SBB DISP[BX],7
(3)AND [DI],AX
(4)OR AX,[609EH]
(5)MOV [BX+DI+30H],CX
(6)PUSH ES:[BP]
(7)CALL DISP[DI]
分析下面程序段,
MOV AL,200
SAR AL,1
MOV BL,AL
MOV CL,2
SAR AL,CL
ADD AL,BL
试问程序段执行后
(BL)=?
(AL)=?
阅读下面的程序,回答问题
DATA SEGMENT
BUF DB '1234'
N=$-BUF
BCD DB N DUP(?)
DATA ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA
START:MOV AX,DATA
MOV DS,AX
LEA SI,BUF
MOV CX,N
LEA DI,BCD+N-1
LOOP1:MOV AL,[SI]
SUB AL,30H
MOV [DI],AL
INC SI
DEC DI
DEC CX
JNE LOOP1
MOV AH,4CH
INT 21H
CODE ENDS
END START
阅读下面的程序,回答问题
DATA SEGMENT
BUF DB '1234'
N=$-BUF
BCD DB N DUP(?)
DATA ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA
START:MOV AX,DATA
MOV DS,AX
LEA SI,BUF
MOV CX,N
LEA DI,BCD+N-1
LOOP1:MOV AL,[SI]
SUB AL,30H
MOV [DI],AL
INC SI
DEC DI
DEC CX
JNE LOOP1
MOV AH,4CH
INT 21H
CODE ENDS
END START
阅读下面的程序,回答问题
DATA SEGMENT
BUF DB '1234'
N=$-BUF
BCD DB N DUP(?)
DATA ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA
START:MOV AX,DATA
MOV DS,AX
LEA SI,BUF
MOV CX,N
LEA DI,BCD+N-1
LOOP1:MOV AL,[SI]
SUB AL,30H
MOV [DI],AL
INC SI
DEC DI
DEC CX
JNE LOOP1
MOV AH,4CH
INT 21H
CODE ENDS
END START
阅读下面的程序,回答问题
DATA SEGMENT
BUF DB '1234'
N=$-BUF
BCD DB N DUP(?)
DATA ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA
START:MOV AX,DATA
MOV DS,AX
LEA SI,BUF
MOV CX,N
LEA DI,BCD+N-1
LOOP1:MOV AL,[SI]
SUB AL,30H
MOV [DI],AL
INC SI
DEC DI
DEC CX
JNE LOOP1
MOV AH,4CH
INT 21H
CODE ENDS
END START
阅读下面的程序,回答问题
DATA SEGMENT
BUF DB '1234'
N=$-BUF
BCD DB N DUP(?)
DATA ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA
START:MOV AX,DATA
MOV DS,AX
LEA SI,BUF
MOV CX,N
LEA DI,BCD+N-1
LOOP1:MOV AL,[SI]
SUB AL,30H
MOV [DI],AL
INC SI
DEC DI
DEC CX
JNE LOOP1
MOV AH,4CH
INT 21H
CODE ENDS
END START
已知有程序段如下:
MOV AL,35H
MOV DL,AL
AND DL,0FH
AND AL,0F0H
MOV CL,4
SHR AL,CL
MOV BL,10
MUL BL
ADD AL,DL
执行之后,AL的值等于多少?该程序段完成了什么功能?
阅读下列程序,指出运行结果
MOV SI,2500H
MOV AX,1000H
MOV DS,AX
MOV CL,05H
NEXT:MOV[ SI] ,AL
INC AL
INC SI
DEC CL
JNZ NEX
TINT 3
程序运行后结果为:DS=()H SI=()HAX=()H CL=()H
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