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微機(jī)原理、匯編與接口技術(shù)第二版朱定華版課后答案(編輯修改稿)

2025-06-27 18:02 本頁(yè)面
　

【文章內(nèi)容簡(jiǎn)介】 的ASCIIBCD數(shù)變?yōu)锳SCII碼MOVOBUF1AHMOVOBUF2ALMOVOBUF3′′MOVDXOFFSETOBUFMOVAH9INT21Hretbeginendpcodeendsendbegin將8位二進(jìn)制數(shù)轉(zhuǎn)換為十進(jìn)制數(shù)也可以用除10取余法實(shí)現(xiàn)。轉(zhuǎn)換程序段如下MOVALBVARXORAHAH將8位二進(jìn)制數(shù)擴(kuò)展為16位無(wú)符號(hào)數(shù)擴(kuò)展MOVDL10DIVDLADDAH30H將第一個(gè)余數(shù)即個(gè)位變?yōu)锳SCII碼MOVOBUF2AHXORAHAH將AL中的二進(jìn)制數(shù)擴(kuò)展為16位DIVDLADDAX3030HAH為十位AL為百位MOVWORDPTROBUFAXMOVOBUF3′′還可以用左移二進(jìn)制數(shù)的方法將二進(jìn)制數(shù)轉(zhuǎn)換為十進(jìn)制數(shù)XORAXAX清0AXMOVCX8左移二進(jìn)制數(shù)8次AGAIN:SHLBVAR1ADCALALDAA19ADCAH0AH的值小于等于2不需校正LOOPAGAINADDAH30H百位轉(zhuǎn)換為ASCII碼MOVOBUFAHMOVAHAL保存十位和個(gè)位MOVCL4SHRALCLBCD數(shù)十位移至AL低4位ANDAHOFHBCD數(shù)個(gè)位在AH低4位ADDAX3030H個(gè)位和十位轉(zhuǎn)換為ASCII碼MOVWORDPTROBUF1AXMOVOBUF3′′以二進(jìn)制數(shù)11111101B轉(zhuǎn)換為BCD數(shù)0253H為例說明轉(zhuǎn)換的過程全過程如下圖所示。37stacksegmentstack′stack′dw32dup0stackends20datasegmentTABDW0DB′0′DW1DB′1′DW8DB′8′┇DW64000DB′64000′LFZDW0IBUFDB303DUBP0dataendscodesegmentbeginprocfarassumess:stackcs:codeds:datapushdssubaxaxpushaxmovaxdatamovdsaxMOVDXOFFSETIBUFMOVAH10INT21HMOVAH2MOVDL0AHINT21HMOVAXWORDPTRIBUF2個(gè)位送AH十位送ALXCHGAHAL個(gè)位與十位交換CMPIBUF12判別N是一位數(shù)還是二位數(shù)JECOM21MOVALAHXORAHAH若N為一位數(shù)則要清AHCOM:ANDAX0F0FH將AX中的ASCII碼變?yōu)锳SCIIBCD數(shù)AAD將AX中的ASCIIBCD數(shù)變?yōu)槎M(jìn)制數(shù)MOVCL3SHLAXCL將AX中的二進(jìn)制數(shù)乘以8MOVBXAXMOVAXTABBXMOVLFZAXMOVDXOFFSETTAB2DX指向立方值的ASCII碼ADDDXBX將DX下移8N指向N的ASCII碼MOVAH9INT21Hretbeginendpcodeendsendbegin38stacksegmentstack′stack′dw32dup0stackendsdatasegmentWADW0ABCDHBADB0dataendscodesegmentbeginprocfarassumess:stackcs:codeds:datapushdssubaxaxpushax22movaxdatamovdsaxMOVAH11號(hào)功能調(diào)用鍵入ii03INT21HMOVDXWATESTAL1判別鍵入數(shù)i是奇數(shù)還是偶數(shù)D0位為0則為偶數(shù)JZNRSMOVCL4i為奇數(shù)將aa1移至各自字節(jié)的低4位SHRDXCLNRS:TESTAL2判別鍵入數(shù)i的D1位是0還是1JZHNSLMOVDLDHi的D1位為1則是取a3或a2將其送DLHNSL:ANDDL0FH取低4位MOVBADLretbeginendpcodeendsendbegin本題還可以將鍵入值乘以4得到將ai移到WA的低4位的移位次數(shù)對(duì)WA執(zhí)行移位操作即可將ai移到WA的低4位。取WA的低4位送入BA中。即將分支程序設(shè)計(jì)成較簡(jiǎn)單的順序程序。用此方法的編制的程序段如下MOVAH1INT21HMOVCLAL將鍵入數(shù)ASCII碼送移位次數(shù)寄存器CLANDCL0FH將移位次數(shù)的ASCII碼變?yōu)槎M(jìn)制數(shù)SHLCL1將鍵入i值乘以4SHLCL1MOVAXWASHRAXCL右移4i次將ai移到最低4位ANDAL0FH取aiMOVBAAL2339stacksegmentstack′stack′dw32dup0stackendsdatasegmentW2DD8920B094Hdataendscodesegmentstartprocfarassumess:stackcs:codeds:datapushdssubaxaxpushaxmovaxdatamovdsaxTESTWORDPTRW228000H檢測(cè)原碼數(shù)的符號(hào)JZBAK正數(shù)的原碼與補(bǔ)碼相同不用再求ANDWORDPTRW227FFFH求負(fù)數(shù)的絕對(duì)值NOTWORDPTRW22雙字求補(bǔ)NOTWORDPTRW2ADDWORDPTRW21ADCWORTPTRW220BAK:retstartendpcodeendsendstart310stacksegmentstack′stack′dw32dup024stackendsdatasegmentXDW8100HYDW9420HKDB0dataendscodesegmentbeginprocfarassumess:stackcs:codeds:datapushdssubaxaxpushaxmovaxdatamovdsaxANDX0FFFFH根據(jù)X是否為0置ZFJZK0ANDY0FFFFH根據(jù)Y是否為0置ZFJZK0TESTX8000H測(cè)試X的符號(hào)JZK14Xgt0去K14TESTY800
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