匯編第六章答案

本文格式為Word版，下載可任意編輯——匯編第六章答案第六章答案

=======================================1.下面的程序段有錯嗎?若有,請指出錯誤.CRAYPROCPUSHAXADDAX,BXRET

ENDPCRAY

[解]:當然有錯誤,ENDPCRAY寫反了,應當將其改成CRAYENDP.

2.已知堆棧寄放器SS的內(nèi)容是0F0A0H,堆棧指示器SP的內(nèi)容是00B0H,先執(zhí)行兩條把8057H和0F79BH分別入棧的PUSH指令.然后再執(zhí)行一POP指令.試畫出示意圖說明堆棧及SP內(nèi)容的變化過程.

3.分析下面\的程序\畫出堆棧最滿時各單元的地址及內(nèi)容.;6.3題的程序

;===========================================S_SEGSEGMENTAT1000H;DEFINESTACKSEGMENTDW200DUP(?)TOSLABELWORD

S_SEGENDS

C_SEGSEGMENT;DEFINECODESEGMENTASSUMECS:C_SEG,SS:S_SEGMOVAX,S_SEGMOVSS,AX

MOVSP,OFFSETTOSPUSHDSMOVAX,0PUSHAX...

PUSHT_ADDRPUSHAXPUSHF...POPFPOPAX

POPT_ADDRRET

C_SEGENDS;ENDOFCODESEGMENTENDC_SEG;ENDOFASSEMBLY

4.分析下面\題的程序\的功能,寫出堆棧最滿時各單元的地址及內(nèi)容.;6.4題的程序

;====================================STACKSEGMENTAT500HDW128DUP(?)TOSLABELWORDSTACKENDS

CODESEGMENT;DEFINECODESEGMENTMAINPROCFAR;MAINPARTOFPROGRAM

ASSUMECS:CODE,SS:STACK

START:;STARTINGEXECUTIONADDRESSMOVAX,STACKMOVSS,AX

MOVSP,OFFSETTOSPUSHDSSUBAX,AXPUSHAX

;MAINPARTOFPROGRAMGOESHEREMOVAX,4321HCALLHTOA

RET;RETURNTODOS

MAINENDP;ENDOFMAINPARTOFPROGRAMHTOAPROCNEAR;DEFINESUBPROCEDUREHTOACMPAX,15JLEB1PUSHAXPUSHBPMOVBP,SP

MOVBX,[BP+2]ANDBX,000FHMOV[BP+2],BXPOPBPMOVCL,4SHRAX,CLCALLHTOAPOPAXB1:

ADDAL,30HCMPAL,3AHJLPRINTITADDAL,07HPRINTIT:MOVDL,ALMOVAH,2INT21HRET

HTOAENDP;ENDOFSUBPROCEDURECODEENDS;ENDOFCODESEGMENTENDSTART;ENDOFASSEMBLY

5.下面是6.5題的程序清單,請在清單中填入此程序執(zhí)行過程中的堆棧變化.0000STACKSGSEGMENT000020[.DW32DUP(?)????

0040]

STACKSGENDS

0000CODESGSEGMENTPARA'CODE'0000BEGINPROCFAR

ASSUMECS:CODESG,SS:STACKSG00001EPUSHDS00012BC0SUBAX,AX000350PUSHAX

0004E80008RCALLP100007CBRET0008BEGINENDP0008B10PROC0008E8000CRCALLC10000BC3RET000CB10ENDP000CC10PROC000CC3RET000DC10ENDP000DCODESGENDSENDBEGIN

6.寫一段子程序SKIPLINES,完成輸出空行的功能.空行的行數(shù)在AX寄放器中.[解]:

SKIPLINESPROCNEARPUSHCXPUSHDXMOVCX，AX

NEXT：MOVAH，2MOVDL，0AHINT21HMOVAH，2MOVDL，0DHINT21HLOOPNEXTPOPDXPOPCX

RET

SKIPLINESENDP

7.設(shè)有10個學生的成績分別是76,69,81,90,73,88,99,63,100和80分.試編制一個子程序統(tǒng)計60-69,70-79,80-89,90-99和100分的人數(shù)并分別存放到S6,S7,S8,S9和S10單元中.DSEGSEGMENT

NUMDW76,69,84,90,73,88,99,63,100,80NDW10S6DW0S7DW0S8DW0S9DW0S10DW0DSEGENDSCODESEGMENTMAINPROCFAR

ASSUMECS:CODE,DS:DSEGSTART:PUSHDSSUBAX,AXPUSHAX

MOVAX,DSEGMOVDS,AXCALLSUB1RET

MAINENDP

SUB1PROCNEARPUSHAXPUSHBXPUSHCXPUSHSIMOVSI,0MOVCX,NNEXT:

MOVAX,NUM[SI]MOVBX,10DIVBLMOVBL,ALCBW

SUBBX,6SALBX,1INCS6[BX]ADDSI,2

LOOPNEXTPOPSIPOPCXPOPBXPOPAXRET

SUB1ENDPCODEENDSENDSTART(解法二)

datasgsegment

gradedb76,69,84,90,73,88,99,63,100,80s6db0s7db0s8db0s9db0s10db0

mess6db'60~69:$'mess7db'70~79:$'mess8db'80~89:$'mess9db'90~99:$'mess10db'100:$'datasgendscodesgsegmentmainprocfar

assumecs:codesg,ds:datasgstart:

pushdssubax,axpushax

movax,datasgmovds,axcallsub1

leadx,mess6calldispstrmovdl,s6

calldispscorecallcrlf

leadx,mess7calldispstrmovdl,s7

calldispscore

callcrlf

leadx,mess8calldispstrmovdl,s8

calldispscorecallcrlf

leadx,mess9calldispstrmovdl,s9

calldispscorecallcrlf

leadx,mess10calldispstrmovdl,s10

calldispscorecallcrlfretmainendp

sub1procnearmovcx,10movsi,0

loop1:moval,grade[si]cmpal,60jlnext5cmpal,70jgenext1incs6

jmpshortnext5next1:cmpal,80jgenext2incs7

jmpshortnext5next2:cmpal,90jgenext3incs8

jmpshortnext5next3:cmpal,100jgnext5jenext4incs9

jmpshortnext5next4:incs10next5:incsi

looploop1retsub1endpdispstrprocnearmovah,9int21hdispstrendpdispscoreprocnearadddl,30hmovah,2int21hdispscoreendpcrlfprocnearmovdl,0dhmovah,2int21hmovdl,0ahmovah,2int21hretcrlfendpcodesgendsendstart

8.編寫一個有主程序和子程序結(jié)構(gòu)的程序模塊.子程序的參數(shù)是一個N字節(jié)數(shù)組的首地址TABLE,數(shù)N及字符CHAR.要求在N字節(jié)數(shù)組中查找字符CHAR,并記錄該字符的出現(xiàn)次數(shù).;主程序則要求從鍵盤接收一串字符以建立字節(jié)數(shù)組TABLE,并逐個顯示從鍵盤輸入的每個字符CHAR以及它在TABLE數(shù)組中出現(xiàn)的次數(shù).(為簡化起見,假設(shè)出現(xiàn)次數(shù)COUNTENDP

DISPLAYPROCNEAR;DISPLAY子程序CALLCRLF;顯示回車和換行MOVDL,CHARMOVAH,2INT21H

MOVDL,20HMOVAH,2INT21HMOVAL,BLANDAL,0FHADDAL,30HCMPAL,3AHJLPRINTADDAL,7PRINT:MOVDL,ALINT21HCALLCRLFRET

DISPLAYENDP

CRLFPROCNEAR;CRLF子程序MOVDL,0DHMOVAH,2INT21H

MOVDL,0AHMOVAH,2INT21HRET

CRLFENDPCODEENDSENDSTART

9.編寫一個子程序嵌套結(jié)構(gòu)的程序模塊,分別從鍵盤輸入姓名及8個字符的電話號碼,并以一定的格式顯示出來.主程序TELIST:

(1)顯示提醒符INPUTNAME:;

(2)調(diào)用子程序INPUT_NAME輸入姓名:

(3)顯示提醒符INPUTATELEPHONENUMBER:;(4)調(diào)用子程序INPHONE輸入電話號碼;

(5)調(diào)用子程序PRINTLINE顯示姓名及電話號碼;子程序INPUT_NAME:

(1)調(diào)用鍵盤輸入子程序GETCHAR,把輸入的姓名存放在INBUF緩沖區(qū)中;(2)把INBUF中的姓名移入輸出行OUTNAME;

子程序INPHONE:

(1)調(diào)用鍵盤輸入子程序GETCHAR,把輸入的8位電話號碼存放在INBUF緩沖區(qū)中;(2)把INBUF中的號碼移入輸出行OUTPHONE.子程序PRINTLINE:

顯示姓名及電話號碼,格式為:NAMETEL

************

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

;編寫一個子程序嵌套結(jié)構(gòu)的程序模塊，分別從鍵盤輸入姓名及8個字符的電話號碼，并以一定的格式顯示出來(解法一)

datasegment

No_of_namedb20No_of_phonedb8inbufdb20dup(?)

outnamedb20dup(?),'$'outphonedb8dup(?),'$'

message1db'pleaseinputname:','$'

message2db'pleaseinputatelephonenumber:','$'message3db'NAME',16dup(20h),'TEL.',13,10,'$'

errormessage1db'youshouldinput8numbers!',13,10,'$'errormessage2db'youinputawrongnumber!',13,10,'$'flagdb?dataendsprogsegmentmainprocfar

assumecs:prog,ds:datastart:

pushdssubax,axpushax

movax,datamovds,axmovflag,0leadx,message1movah,09int21hcallinput_namecallcrlf

leadx,message2movah,09int21hcallinphonecallcrlfcmpflag,1jeexitcallprintlineexit:

retmainendp

input_nameprocnearpushaxpushdxpushcx

movcx,0

movcl,No_of_namecallgetchar

movax,0

moval,No_of_namesubax,cx

movcx,axmovsi,0next1:

moval,inbuf[si]movoutname[si],alincsi

loopnext1popcxpopdxpopaxret

input_nameendp

inphoneprocnearpushaxpushdx

movcx,0

movcl,No_of_phonecallgetcharcmpcx,0jnzerror1

movcl,No_of_phonemovsi,0

next2:

moval,inbuf[si]

cmpal,30hjlerror2cmpal,39hjaerror2

movoutphone[si],alincsi

loopnext2jmpexit2error1:

callcrlf

leadx,errormessage1movah,09int21h

movflag,1jmpexit2error2:

callcrlf

leadx,errormessage2movah,09int21hmovflag,1jmpexit2exit2:

popdxpopaxretinphoneendpgetcharprocnearpushaxpushdxmovdi,0rotate:

movah,01int21hcmpal,0dhjeexit1movinbuf[di],alincdilooprotateexit1:

popdxpopaxretgetcharendpcrlfprocnearpushaxpushdxmovdl,0dhmovah,2int21hmovdl,0ahmovah,2int21h

popdxpopaxret

crlfendp

printlineprocnearpushaxpushdxleadx,message3movah,09int21h

leadx,outnamemovah,09int21h

leadx,outphonemovah,09int21h

popdxpopaxretprintlineendpprogendsendmain

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

;從鍵盤輸入姓名及8個字符的電話號碼，并以一定的格式顯示出來datareasegment

inbuflabelbyte;nameparameterlist:maxnlendb9;max.length

namelendb?;no.charenterednameflddb9dup(?)crlfdb13,10,'$'

messg1db'INPUTNAME:',13,10,'$'

messg2db'INPUTATELEPHONENUMBER:',13,10,'$'messg3db'NAMETEL',13,10,'$'OUTNAMEdb21dup(?),'$'OUTPHONEdb8dup(?),'$'datareaendsprognamsegment

assumecs:prognam,ds:datareastart:pushdssubax,axpushax

movax,datarea

movds,axmoves,axTELISTprocfarmovah,09leadx,messg1int21h

callINPUT_NAMEmovah,09leadx,messg2int21hcallINPHONEcallPRINTLINEretTELISTendp

INPUT_NAMEprocnearcallGETCHARmovbh,0

movbl,namelenmovcx,21subcx,bx

b20:movnamefld[bx],20hincbxloopb20

leasi,namefldleadi,OUTNAMEmovcx,9cld

repmovsbretINPUT_NAMEendp

INPHONEprocnearcallGETCHARmovbh,0

movbl,namelenmovcx,9subcx,bx

b30:movnamefld[bx],20hincbxloopb30

leasi,namefldleadi,OUTPHONEmovcx,8cld

repmovsb

retINPHONEendp

PRINTLINEprocnearmovah,09hleadx,messg3int21hmovah,09

leadx,OUTNAMEint21hmovah,09

leadx,OUTPHONEint21hretPRINTLINEendp

GETCHARprocnearmovah,0ahleadx,inbufint21hmovah,09leadx,crlfint21hretGETCHARendpprognamendsendstart

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

10.編寫子程序嵌套結(jié)構(gòu)的程序,把整數(shù)分別用二進制和八進制形式顯示出來.主程序BANDO:把整數(shù)字變量VAL1存入堆棧,并調(diào)用子程序PAIRS;子程序PAIRS:從堆棧中取出VAL1;調(diào)用二進制顯示程序OUTBIN顯示出與其等效的二進制數(shù);輸出8個空格;調(diào)用八進制顯示程序OUTOCT顯示出與其等效的八進制數(shù);調(diào)用輸出回車及換行符的子程序..modelsmall.stack100h.data

val1dw0ffffh.code

mainprocfarstart:pushdssubax,axpushax

movax,@datamovds,axpushval1

callpairsret

mainendp

pairsprocnearpushbpmovbp,s