汇编第六章答案.docx

汇编第六章答案.docx

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汇编第六章答案

第六章答案

=======================================

1.下面的程序段有错吗?

若有,请指出错误.

CRAYPROC

PUSHAX

ADDAX,BX

RET

ENDPCRAY

[解]:

当然有错误,ENDPCRAY写反了,应该将其改成CRAYENDP.

2.已知堆栈寄存器SS的内容是0F0A0H,堆栈指示器SP的内容是00B0H,先执行两条把8057H和0F79BH分别入栈的PUSH指令.然后再执行一POP指令.试画出示意图说明堆栈及SP内容的变化过程.

3.分析下面"6,3的程序",画出堆栈最满时各单元的地址及内容.

;             6.3题的程序

;===========================================

S_SEGSEGMENTAT1000H;DEFINESTACKSEGMENT

DW200DUP（?

）

TOSLABELWORD

S_SEGENDS

C_SEGSEGMENT;DEFINECODESEGMENT

ASSUMECS:

C_SEG,SS:

S_SEG

MOVAX,S_SEG

MOVSS,AX

MOVSP,OFFSETTOS

PUSHDS

MOVAX,0

PUSHAX

...

PUSHT_ADDR

PUSHAX

PUSHF

...

POPF

POPAX

POPT_ADDR

RET

C_SEGENDS;ENDOFCODESEGMENT

ENDC_SEG;ENDOFASSEMBLY

4.分析下面"6.4题的程序"的功能,写出堆栈最满时各单元的地址及内容.

;                 6.4题的程序

;====================================

STACKSEGMENTAT500H

DW128DUP（?

）

TOSLABELWORD

STACKENDS

CODESEGMENT;DEFINECODESEGMENT

MAINPROCFAR;MAINPARTOFPROGRAM

ASSUMECS:

CODE,SS:

STACK

START:

;STARTINGEXECUTIONADDRESS

MOVAX,STACK

MOVSS,AX

MOVSP,OFFSETTOS

PUSHDS

SUBAX,AX

PUSHAX

;MAINPARTOFPROGRAMGOESHERE

MOVAX,4321H

CALLHTOA

RET      ;RETURNTODOS

MAINENDP;ENDOFMAINPARTOFPROGRAM

HTOAPROCNEAR;DEFINESUBPROCEDUREHTOA

CMPAX,15

JLEB1

PUSHAX

PUSHBP

MOVBP,SP

MOVBX,[BP+2]

ANDBX,000FH

MOV[BP+2],BX

POPBP

MOVCL,4

SHRAX,CL

CALLHTOA

POPAX

B1:

ADDAL,30H

CMPAL,3AH

JLPRINTIT

ADDAL,07H

PRINTIT:

MOVDL,AL

MOVAH,2

INT21H

RET

HTOAENDP;ENDOFSUBPROCEDURE

CODEENDS;ENDOFCODESEGMENT

ENDSTART;ENDOFASSEMBLY

5.下面是6.5题的程序清单,请在清单中填入此程序执行过程中的堆栈变化.

0000      STACKSGSEGMENT

000020[.DW32DUP（?

）

?

?

?

?

0040        ]

          STACKSGENDS

0000      CODESGSEGMENTPARA'CODE'

0000      BEGINPROCFAR

          ASSUMECS:

CODESG,SS:

STACKSG

00001E   PUSHDS

00012BC0SUBAX,AX

000350   PUSHAX

0004E80008RCALLP10

0007CB        RET

0008      BEGINENDP

0008      B10PROC

0008E8000CRCALLC10

000BC3        RET

000C      B10ENDP

000C      C10PROC

000CC3        RET

000D      C10ENDP

000D      CODESGENDS

          ENDBEGIN

6.写一段子程序SKIPLINES,完成输出空行的功能.空行的行数在AX寄存器中.

[解]:

SKIPLINESPROCNEAR

PUSHCX

PUSHDX

MOVCX，AX

NEXT：

MOVAH，2

MOVDL，0AH

INT21H

MOVAH，2

MOVDL，0DH

INT21H

LOOPNEXT

POPDX

POPCX

RET

SKIPLINESENDP

7.设有10个学生的成绩分别是76,69,81,90,73,88,99,63,100和80分.试编制一个子程序统计60-69,70-79,80-89,90-99和100分的人数并分别存放到S6,S7,S8,S9和S10单元中.

DSEGSEGMENT

NUMDW76,69,84,90,73,88,99,63,100,80

NDW10

S6DW0

S7DW0

S8DW0

S9DW0

S10DW0

DSEGENDS

CODESEGMENT

MAINPROCFAR

ASSUMECS:

CODE,DS:

DSEG

START:

PUSHDS

SUBAX,AX

PUSHAX

MOVAX,DSEG

MOVDS,AX

CALLSUB1

RET

MAINENDP

SUB1PROCNEAR

PUSHAX

PUSHBX

PUSHCX

PUSHSI

MOVSI,0

MOVCX,N

NEXT:

MOVAX,NUM[SI]

MOVBX,10

DIVBL

MOVBL,AL

CBW

SUBBX,6

SALBX,1

INCS6[BX]

ADDSI,2

LOOPNEXT

POPSI

POPCX

POPBX

POPAX

RET

SUB1ENDP

CODEENDS

ENDSTART

（解法二）

datasgsegment

grade  db76,69,84,90,73,88,99,63,100,80

s6     db0

s7     db0

s8     db0

s9     db0

s10    db0

mess6db'60~69:

$'

mess7db'70~79:

$'

mess8db'80~89:

$'

mess9db'90~99:

$'

mess10db'100:

$'

datasgends

codesgsegment

main   procfar

       assumecs:

codesg,ds:

datasg

start:

       pushds

       subax,ax

       pushax

       movax,datasg

       movds,ax

       callsub1

       leadx,mess6

       calldispstr

       movdl,s6

       calldispscore

       callcrlf

       leadx,mess7

       calldispstr

       movdl,s7

       calldispscore    

       callcrlf

       leadx,mess8

       calldispstr

       movdl,s8

       calldispscore

       callcrlf

       leadx,mess9

       calldispstr

       movdl,s9

       calldispscore

       callcrlf

       leadx,mess10

       calldispstr

       movdl,s10

       calldispscore    

       callcrlf

       ret

main   endp

sub1   procnear

       movcx,10

       movsi,0

loop1:

moval,grade[si]

       cmpal,60

       jl  next5

       cmpal,70

       jgenext1

       incs6

       jmpshortnext5

next1:

cmpal,80

       jgenext2

       incs7

       jmpshortnext5

next2:

cmpal,90

       jgenext3

       incs8

       jmpshortnext5

next3:

cmpal,100

       jgnext5

       jenext4

       incs9

       jmpshortnext5

next4:

incs10

next5:

incsi

       looploop1

       ret

sub1   endp

dispstrprocnear

       movah,9

       int21h

dispstrendp

dispscoreprocnear

      adddl,30h

      movah,2

      int21h

dispscoreendp

crlfprocnear

   movdl,0dh

   movah,2

   int21h

   movdl,0ah

   movah,2

   int21h

   ret

crlfendp

codesgends

endstart

8.编写一个有主程序和子程序结构的程序模块.子程序的参数是一个N字节数组的首地址TABLE,数N及字符CHAR.要求在N字节数组中查找字符CHAR,并记录该字符的出现次数.;主程序则要求从键盘接收一串字符以建立字节数组TABLE,并逐个显示从键盘输入的每个字符CHAR以及它在TABLE数组中出现的次数.（为简化起见,假设出现次数<=15,可以用十六进制形式显示出来）

[解]:

DATASEGMENT

MAXLENDB40

NDB?

TABLEDB40DUP（?

）

CHARDB'A';查找字符'A'

EVEN

_ADDRDW3DUP（?

）

DATAENDS

CODESEGMENT

ASSUMECS:

CODE,DS:

DATA

MAINPROCFAR

START:

PUSHDS

MOVAX,0

PUSHAX

MOVAX,DATA

MOVDS,AX

LEADX,MAXLEN

MOVAH,0AH

INT21H  ;从键盘接收字符串

MOV_ADDR,OFFSETTABLE

MOV_ADDR+2,OFFSETN

MOV_ADDR+4,OFFSETCHAR

MOVBX,OFFSET_ADDR;通过地址表传送变量地址

CALLCOUNT  ;计算CHAR的出现次数

CALLDISPLAY;显示

RET

MAINENDP

COUNTPROCNEAR;COUNT子程序

PUSHSI

PUSHDI

PUSHAX

PUSHCX

MOVDI,[BX]

MOVSI,[BX+2]

MOVCL,BYTEPTR[SI]

MOVCH,0

MOVSI,[BX+4]

MOVAL,BYTEPTR[SI]

MOVBX,0

AGAIN:

CMPAL,BYTEPTR[DI]

JNEL1

INCBX

L1:

INCDI

LOOPAGAIN

POPCX

POPAX

POPDI

POPSI

RET

COUNTENDP

DISPLAYPROCNEAR;DISPLAY子程序

CALLCRLF  ;显示回车和换行

MOVDL,CHAR

MOVAH,2

INT21H

MOVDL,20H

MOVAH,2

INT21H

MOVAL,BL

ANDAL,0FH

ADDAL,30H

CMPAL,3AH

JLPRINT

ADDAL,7

PRINT:

MOVDL,AL

INT21H

CALLCRLF

RET

DISPLAYENDP

CRLFPROCNEAR;CRLF子程序

MOVDL,0DH

MOVAH,2

INT21H

MOVDL,0AH

MOVAH,2

INT21H

RET

CRLFENDP

CODEENDS

ENDSTART

9.编写一个子程序嵌套结构的程序模块,分别从键盘输入姓名及8个字符的电话号码,并以一定的格式显示出来.

主程序TELIST:

（1）显示提示符INPUTNAME:

;

（2）调用子程序INPUT_NAME输入姓名:

（3）显示提示符INPUTATELEPHONENUMBER:

;

（4）调用子程序INPHONE输入电话号码;

（5）调用子程序PRINTLINE显示姓名及电话号码;

子程序INPUT_NAME:

（1）调用键盘输入子程序GETCHAR,把输入的姓名存放在INBUF缓冲区中;

（2）把INBUF中的姓名移入输出行OUTNAME;

子程序INPHONE:

（1）调用键盘输入子程序GETCHAR,把输入的8位电话号码存放在INBUF缓冲区中;

（2）把INBUF中的号码移入输出行OUTPHONE.

子程序PRINTLINE:

显示姓名及电话号码,格式为:

NAME          TEL

****          ********

==========================================

;编写一个子程序嵌套结构的程序模块，分别从键盘输入姓名及8个字符的电话号码，并以一定的格式显示出来

（解法一）

datasegment

    No_of_namedb20

    No_of_phonedb8

    inbufdb20dup（?

）

    outnamedb20dup（?

）,'$'

    outphonedb8dup（?

）,'$'

    message1db'pleaseinputname:

','$'

    message2db'pleaseinputatelephonenumber:

','$'

    message3db'NAME',16dup（20h）,'TEL.',13,10,'$'

    errormessage1db'youshouldinput8numbers!

',13,10,'$'

    errormessage2db'youinputawrongnumber!

',13,10,'$'

    flagdb?

dataends

progsegment

mainprocfar

     assumecs:

prog,ds:

data

start:

      pushds

      subax,ax

      pushax

      movax,data

      movds,ax

      movflag,0

      leadx,message1

      movah,09

      int21h

      callinput_name

      callcrlf

      leadx,message2

      movah,09

      int21h

      callinphone

      callcrlf

      cmpflag,1

      je  exit

      callprintline

exit:

      ret

mainendp

input_nameprocnear

      pushax

      pushdx

      pushcx

      movcx,0

      movcl,No_of_name

      callgetchar

      movax,0

      moval,No_of_name

      subax,cx

      movcx,ax

      movsi,0

next1:

      moval,inbuf[si]

      movoutname[si],al

      incsi

      loopnext1

      popcx

      popdx

      popax

      ret

input_nameendp

inphoneprocnear

      pushax

      pushdx

      movcx,0

      movcl,No_of_phone

      callgetchar

      cmpcx,0

      jnzerror1

      movcl,No_of_phone  

      movsi,0

next2:

      moval,inbuf[si]

      cmpal,30h

      jl  error2

      cmpal,39h

      ja  error2

      movoutphone[si],al

      incsi

      loopnext2

      jmpexit2

error1:

      callcrlf

      leadx,errormessage1

      movah,09

      int21h

      movflag,1

      jmpexit2

error2:

      callcrlf

      leadx,errormessage2

      movah,09

      int21h

      movflag,1

      jmpexit2

exit2:

      popdx

      popax

       ret

inphoneendp

getcharprocnear

       pushax

       pushdx

       movdi,0

rotate:

       movah,01

       int21h

       cmpal,0dh

       je  exit1

       movinbuf[di],al

       incdi

       looprotate

exit1:

       popdx

       popax

       ret

getcharendp

crlfprocnear

        pushax

        pushdx

        movdl,0dh

        movah,2

        int21h

        movdl,0ah

        movah,2

        int21h

        popdx

        popax

        ret

crlfendp

printlineprocnear

        pushax

        pushdx

        leadx,message3

        movah,09

        int21h

leadx,outname

movah,09

int21h

leadx,outphone

movah,09

int21h

        popdx

        popax

        ret

printlineendp

progends

endmain

==========================================

;从键盘输入姓名及8个字符的电话号码，并以一定的格式显示出来

datareasegment

inbuf   label  byte                        ;nameparameterlist:

maxnlen   db