微机原理期末编程.docx

微机原理期末编程.docx

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微机原理期末编程

《微机原理》课程设计

课程号：

B0901270课堂号：

学号：

1209030313姓名：

万佳成绩：

设计内容：

1.从键盘输入一个十进制数，范围在-128~127，编程将其转换为二进制数，以十六进制数形式在显示器上显示。

2.已知字节变量buf1的内容是：

320,210,160,120,36,10,字节变量buf2的内容是：

813,260,180,93,22,11,2,编写程序将两个变量中的数据合并到字节变量buf3中，且要求buf3中的数据为降序排列。

3.从键盘输入十个任意正整数，范围在0~127之间，按降序排序后在屏幕输出。

编程实现该要求。

4.使用二重循环，在屏幕输出下述图形：

*****

***

*

第一题

stacksegmentstack'stack'

dw32dup（0）

stackends

datasegment

IBUFDB7,0,7DUP（0）

OBUFDB6DUP（0）

dataends

codesegment

beginprocfar

assumess:

stack,cs:

code,ds:

data

pushds

subax,ax

pushax

movax,data

movds,ax

MOVDX,OFFSETIBUF

MOVAH,10

INT21H

MOVDL,0AH

MOVAH,2

MOVCL,IBUF+1

MOVCH,0

MOVSI,OFFSETIBUF+2

CMPBYTEPTR[SI],'-'

PUSHF

JNESININC

INCSI

DECCX

SININC:

MOVAX,0

AGAIN:

MOVDX,10

MULDX

ANDBYTEPTR[SI],0FH

ADDAL,[SI]

ADCAH,0

INCSI

LOOPAGAIN

POPF

JNZNNEG

NEGAX

NNEG:

MOVCX,404H

MOVSI,4

MOVWORDPTROBUF[SI],'$H'

AG:

MOVDL,0FH

ANDDL,AL

ADDDL,30H

CMPDL,3AH

JCNOAD7

ADDDL,7

NOAD7:

DECSI

MOVOBUF[SI],DL

SHRAX,CL

DECCH

JNZAG

callcrlf;回车换行

MOVBX,OFFSETOBUF-1

COUNT:

INCBX

CMPBYTEPTR[BX],'0'

JECOUNT

MOVDX,BX

MOVDX,BX

MOVAH,9

INT21H

Ret

beginendp

crlfprocnear

movdl,0dh

movah,2

int21h

movdl,0ah

movah,2

int21h

ret

crlfendp

codeends

endbegin

第二题

datasegment

buf1dw320,210,160,120,36,10

buf2dw813,260,180,93,22,11,2

buf3db100dup（?

）,'$'

tmpdb100dup（?

）,'$'

dataends

codesegment

assumecs:

code,ds:

data

START:

movax,data

movds,ax

movdi,0

movsi,0

movbx,0

STEP:

cmpsi,14

jeSTEPA

cmpdi,12

jeSTEPB

movax,buf1[di]

cmpax,buf2[si]

jaSTEPA

movax,buf2[si]

cmpax,buf1[di]

jaSTEPB

STEPA:

cmpdi,12

jeexit

movax,0

movax,buf1[di]

movbp,0

chgA:

movcx,10

movdx,0

divcx

adddx,30h

movcx,dx

movtmp[bp],cl

incbp

cmpax,0

jeexitA

jmpchgA

exitA:

decbp

movcl,tmp[bp]

movbuf3[bx],cl

incbx

cmpbp,0

jelastA

jmpexitA

lastA:

movbuf3[bx],','

incbx

adddi,2

jmpSTEP

STEPB:

cmpsi,14

jeexit

movax,0

movax,buf2[si]

movbp,0

chgB:

movcx,10

movdx,0

divcx

adddx,30h

movcx,dx

movtmp[bp],cl

incbp

cmpax,0

jeexitB

jmpchgB

exitB:

decbp

movcl,tmp[bp]

movbuf3[bx],cl

incbx

cmpbp,0

jelastB

jmpexitB

lastB:

movbuf3[bx],','

incbx

addsi,2

jmpSTEP

exit:

leadx,buf3

movah,9

int21h

MOVAH,4CH

INT21H

codeends

ENDSTART

第三题

datasegment

arraydb10dup（?

）

Buf1db4,?

4dup（?

）

flagdb0

dataends

stacksegmentstack

dw32dup（?

）

stackends

codesegment

assumeds:

data,cs:

code,ss:

stack

BEGIN:

movax,data

movds,ax

movcx,10

leabx,array

Input:

leadx,Buf1

movah,0ah

int21h

movdl,0ah

movah,2

int21h

callbint

mov[bx],al

incbx

loopInput

;排序

movcx,9

OuterSort:

pushcx

leasi,array

InterSort:

moval,[si]

cmpal,[si+1]

jaeNext

xchgal,[si+1]

mov[si],al

Next:

incsi

loopInterSort

popcx

loopOuterSort

;输出

movcx,10

leabx,array

Output:

xorah,ah

moval,[bx]

callstart

incbx

loopOutput

movah,04ch

int21h

bintproc

movax,0

pushcx

pushbx

movcl,10

leasi,Buf1+2

Loop1:

movbl,[si]

cmpbl,0dh

jeCalcDone

subbl,'0'

mulcl

addal,bl

incsi

jmpLoop1

CalcDone:

popbx

popcx

ret

bintendp

startproc

pushcx

movflag,0

leasi,Buf1

movdl,100

movcl,10

Loop2:

xorah,ah

divdl

cmpal,0

jneNext2

cmpflag,0

jeNext3

Next2:

movflag,1

mov[si],al

addbyteptr[si],'0'

incsi

Next3:

pushax

movax,0

moval,dl

divcl

movdl,al

popax

moval,ah

cmpdl,0

jneLoop2

movbyteptr[si],''

movbyteptr[si+1],'$'

leadx,Buf1

movah,9

int21h

popcx

ret

startendp

codeends

endBEGIN

第四题

CODESEGMENT

ASSUMECS:

CODE

START:

MOVCX,3;控制外循环

MOVBX,5;控制*号个数

S:

PUSHBX

SYM:

MOVAH,02H

MOVDL,'*'

INT21H

DECBX

JNZSYM

POPBX

SUBBX,2;每行*号之间相差两个

CALLCRLF;回车换行

LOOPS

MOVAX,4C00H

INT21H

CRLF:

MOVAH,02H

MOVDL,0DH

INT21H

MOVDL,0AH

INT21H

RET

CODEENDS

ENDSTART