微机原理与接口技术试题库(含答案)汇总

四、完成下列练习

1、1000H,5000H,8000H 2、答案:

(1) CMP CX,DX

JNB HIEQU : :

HIEQU:

(2) CMP AX,BX

JG GREAT : :

GREAT:

(3) CMP CX,0

JZ ZERO : :

ZERO:

(4) CMP AX,BX

JLE SMAEQU : :

SMAEQU:

(5) CMP CX,DX

JBE LOWEQU : :

LOWEQU:

(6) CMP AX,0

JS NEGAT : :

NEGAT:

(7) TEST AX,AX

JP IMAGE : :

IMAGE:

3、答案:

(1)MOV DS,SEG SOURCE MOV ES,SEGT DEST

MOV SI,OFFSET SOURCE MOV DI,OFFEST DEST MOV CX,N

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AGAIN: MOV AL,[SI]

MOV ES:[DI],AL INC SI INC DI

LOOP AGAIN

(2)MOV DS,SEG SOURCE

MOV SI,OFFSET SOURCE

MOV CX,N AGAIN: MOV AL,[SI] INC SI

LOOP AGAIN

(3) MOV ES,SEGT DEST

MOV DI,OFFEST DEST MOV CX,N

AGAIN: MOV ES:[DI],AL INC DI

LOOP AGAIN

(4) MOV ES,SEGT DEST

MOV DI,OFFEST DEST

MOV CX,N

AGAIN: CMP AL,ES:[DI] INC DI

LOOP AGAIN

五、是非判断题

(1)A(2)B(3)B(4)A(5)B(6)A(7)A(8)A(9)A(10)A 六、请阅读下列程序,分析其功能或结果

1、读取键盘输入,并显示在显示器上。 2、 在打印机一输出字符‘B’,并回车。

3、屏幕初始化,左上角行号为2,左上角列号为5,右下角行号为22,右下角列号为38,使用整个空白窗口。

4、在屏幕上输出PRESS ANY KEY并换行回车。 5、在屏幕上输出1 七、编写程序

1、答案: abc: mov ah,1 int 21h cmp al,’a’ jb stop cmp al,’z’ ja stop sub al,20h mov dl,al mov ah,2 int 21h

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jmp abc stop: ret 2、答案:

datarea segment string1 db ‘asfioa’ string2 db ‘xcviyoaf’ mess1 db ‘MATCH’,’$’ mess2 db ‘NO MATCH’,’$’ datarea ends prognam segment main proc far assume cs:prognam,ds:datarea start: push ds sub ax,ax push ax mov ax,datarea mov ds,ax mov es,ax begin: mov cx, string2-string1 mov bx, mess1-string2 cmp bx,cx jnz dispno lea dx,addr lea si,string1 lea di,string2 repe cmpsb jne dispno mov ah,9 lea dx,mess1 int 21h ret dispno: mov ah, 9 lea dx, mess2 int 21h ret main endp prognam ends end start 3、答案: data segment array dw 3 dup(?)

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data ends code segment main proc far assume cs:code,ds:data start: push ds sub ax,ax push ax mov ax,data mov ds,ax mov cx,3 lea si,array begin: push cx mov cl,4 mov di,4 mov dl, ‘ ‘ mov ah,02 int 21h mov dx,0 input: mov ah,01 int 21h and al,0fh shl dx,cl or dl,al dec di jne input mov [si],dx add si,2 pop cx loop begin comp: lea si,array mov dl,0 mov ax,[si] mov bx,[si+2] cmp ax,bx jne next1 add dl,2 next1: cmp [si+4],ax jne next2 add dx,2

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next2: cmp [si+4],bx jne num add dl,2 num: cmp dx,3 jl disp mov dl,3 disp: mov ah,2 add dl,30h int 21h ret

main endp code ends end start 4、答案: dseg segment a dw ? b dw ? dseg ends cseg segment mainproc far

assume cs:cseg,ds:dseg start:push ds sub ax,ax push ax mov ax,dseg mov ds,ax begin: mov ax,a mov bx,b xor ax,bx test ax,0001 jz class test bx,0001 jz exit xchg bx,a mov b,bx jmp exit class: test bx,0001 jz exit inc b inc a exit: ret

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