微機(jī)原理編程題集合-wenkub
2023-04-09 01:55:51 本頁(yè)面
【正文】 MOV AX,DATA MOV DS,AX MOV AL,BUF CMP AL,0 JGE L1 MOV AL,1 JMP L3L1:JZ L2 MOV AL,1 JMP L3L2:MOV AL,0L3:MOV RESULT,AL MOV AH,4CH INT 21HCODE ENDS END START 2. N個(gè)數(shù)中求最大值、=10,已知原始數(shù)據(jù)存放在BUF開(kāi)始的內(nèi)存單元中，將結(jié)果存入MAX，MIN內(nèi)存單元中。Z=|XY|；DATA SEGMENTBUF1 DB 05HBUF2 DB 02HRESULT DB ?DATA ENDSCODE SEGMENT ASSUME CS:CODE,DS:DATASTART:MOV AX,DATA MOV DS,AX MOV AL,BUF1 MOV BL,BUF2 SUB AL,BL JNC GREAT NEG ALGREAT:MOV RESULT,ALEXIT: MOV AH,4CH INT 21H CODE ENDS END START ②.編寫(xiě)一段程序，已知BUF單元中有一無(wú)符號(hào)字節(jié)數(shù)據(jù)X，假設(shè)為8，根據(jù)以下函數(shù)關(guān)系編寫(xiě)程序求Y的值，結(jié)果存入RESULT 單元。Z=5X X10,X5 X=10。DATA SEGMENTBUF DB 3,5,7,8,5,3,9,7,13,1COUNT EQU $BUFMAX DB ?MIN DB ? DATA ENDSCODE SEGMENT ASSUME CS:CODE,DS:DATASTART: MOV AX,DATA MOV DS,AX MOV CX,COUNT1 MOV AL,BUF MOV MAX,AL MOV MIN,AL LEA SI,BUFAGAIN: MOV AL,[SI+1] CMP AL,MIN JGE S1 MOV MIN,ALS1: CMP AL,MAX JLE S2 MOV MAX,ALS2: INC SI LOOP AGAIN MOV AH,4CH INT 21HCODE ENDS END START 3. 編寫(xiě)程序求1+2+3+……+N100時(shí)最大的N值，將N值送NUM單元中，同時(shí)將1+2+3+……+N的和送到SUM單元中。結(jié)尾，試編寫(xiě)程序，統(tǒng)計(jì)字符串的長(zhǎng)度，并將結(jié)果存入COUNT單元。$39。DATA SEGMENTBUF1 DB 01H,01H,01H,01H,01H,01H,01H,01H,01H,01HBUF2 DB 03H,03H,03H,03H,03H,03H,03H,03H,03H,03HBUF3 DB 11 DUP(?) DATA ENDSCODE SEGMENT ASSUME CS:CODE,DS:DATASTART: MOV AX,DATA MOV DS,AX MOV CX,10 LEA SI,BUF1 LEA DI,BUF2 LEA BX,BUF3AGAIN: MOV AL,[SI] ADC AL,[DI] DAA MOV [BX],AL INC SI INC DI INC BX LOOP AGAIN ADC AH,0 MOV [BX],AH MOV AH,4CH INT 21HCODE ENDSEND START6. 8255編程題：(課后實(shí)驗(yàn)) 8255工作于方式0，此時(shí)PA、PB、PC均為可獨(dú)立輸入/輸出的并行口。設(shè)置為A口輸入,C口輸出 MOV AL,90H
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