微機(jī)原理習(xí)題參考答案(黃冰版)-文庫(kù)吧資料
2025-07-04 18:19本頁(yè)面
【正文】 87FFFHEprom 2Y1任意, 10, 001, 變化 *8800H~*8FFFFHEprom 3Y2任意, 10, 010, 變化*9000H~*97FFFH項(xiàng)目片選端地址線A19~A16,A15A14,A13~A11,A10,A 9~A0起始和結(jié)束地址Ram 1Y4,A10任意, 10, 100, 1, 變化*A400H~*A7FFHRam 2Y4,A10任意, 10, 100, 0, 變化*A000H~*A3FFHRam 3Y5,A10任意, 10, 101, 1, 變化*AC00H~*AFFFHRam 4Y5,A10任意, 10, 101, 0, 變化*A800H~*ABFFH第六章 I/O接口技術(shù)答：1）程序控制方式2）中斷控制方式3）直接存儲(chǔ)訪問(wèn)（DMA）方式區(qū)別：程序控制方式和中斷控制方式都是采用軟件形式，實(shí)現(xiàn)CPU與外設(shè)的數(shù)據(jù)傳送，都要占用CPU的寶貴時(shí)間。JZ DONEINC SILOOP LOP1DONE: MOV ADRBUF,AL MOV AH,4CHINT 21HCODE ENDS END STARTSTACK SEGMENT STACKDW 256 DUP(?)TOP LABEL WORDSTACK ENDSDATA SEGMENTBUFF DB 1,2,3,4,5,6,7,8SUM DW 0000HDATA ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA,SS:STACKSTART: MOV AX,DATAMOV DS,AXMOV AX,STACKMOV SS,AXMOV SP,OFFSET TOPMOV CH,0MOV CL,8LOP1: MOV SI,OFFSET BUFFMOV AX,OFFSET SUMPUSH AXMOV AH,0MOV AL, [SI]PUSH AXCALL FAR PTR FACTMOV DX,SUMADD SUM,DXINC SILOOP LOP1MOV AH,4CHINT 21HCODE ENDSCODES SEGMENT ASSUME CS:CODESFACT PROC FARPUSH BPMOV BP,SPPUSH BXPUSH AXMOV BX,[BP+8]MOV AX,[BP+6]CMP AX,0JE DONEPUSH BXDEC AXPUSH AXCALL FACTMOV BX,[BP+8]MOV AX,[BX]MUL WORD PTR [BP+6]JMP RETURNDONE: MOV AX,01HRETURN:MOV [BX],AX POP AX POP BXPOP BPRET 4FACT ENDPCODES ENDSEND START第五章 半導(dǎo)體存儲(chǔ)器 1) Intel 1024(1K1位) ： =1024（片） 2） Intel 2114（1K4位）：=256（片） 3） Intel 2128（2K8位）：=64（片） 4） Intel 2167（16K1位）：=16（片） 1）=128(片)2）1K=210 222。COUNT DB $STRBUFADRBUF DB ?DATA ENDSCODE SEGMENTASSUME CS:CODE, DS:DATASTART: MOV AX, DATAMOV DS, AXLEA SI, STRBUFMOV CH,0MOV CL, COUNTLOP1: MOV AL, [SI]CMP AL,39。JGE DONEJMP SHOWNEXT2:SUB AL,20HSHOW:MOV DL,ALMOV AH,2INT 21HLOOP LOP1DONE:MOV AH,4CHINT 21HCODE ENDSEND STARTDATA SEGMENTSTRBUF DB 39。JGE NEXT2CMP AL,39。JZ DONENEXT1 :CMP AL,39。HEXNUM變量中原來(lái)的內(nèi)容未知，程序段執(zhí)行后的內(nèi)容是字符A的十六進(jìn)制數(shù)0ADATA_SEG SEGMENT N=10DATA1 DB N DUP(?) DATA2 DB N DUP(?) ADR1 DW ? ADR2 DW ? DATA_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA_SEGSTART: MOV AX, DATA_SEGMOV DS, AXLEA SI, DATA1LEA DI,DATA2MOV CX, NLOP1: MOV AH,[SI]CMP AH,[DI] JNZ NOTEQUINC SIINC DILOOP LOP1MOV AH,0FFHSAHFJMP DONENOTEQU: MOV AH,0SAHFMOV ADR1,[SI]MOV ADR2,[DI]DONE:MOV AH,4CHINT 21HCODE ENDSEND STARTDATA_SEG SEGMENTCOUNT=100BUF DB COUNT NUP(?) MAX DB ?DATA_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA_SEGSTART: MOV AX, DATA_SEG MOV DS, AXMOV SI, OFFSET BUFMOV CX,COUNTLOP1: MOV AH,1INT 21HMOV [SI], ALINC SILOOP LOP1MOV SI,OFFSET BUFMOV CX,COUNTMOV AL, [SI]LOP2:CMP AL, [SI+1]JA NEXT2XCHG AL,[SI+1]INC SINEXT2:LOOP LOP2MOV MAX,ALMOV AH, 4CHINT 21HCODE ENDSEND STARTDATA_SEG SEGMENT BUF DB 10 DUP(?)STR1 DB ‘Do you want input number(y/n)?’,0DH,0AH,’$’STR2 DB ‘Please input the numbers’ ,0DH,0AH,’$’ MAX DB ? MIN DB ?DATA_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA_SEGSTART: MOV AX, DATA_SEGMOV DS, AXMOV DX,OFFSET STR1MOV AH,09HINT 21HMOV DX,OFFSET STR2MOV AH,09HINT 21HMOV SI, OFFSET BUFMOV CX,10LOP1:MOV AH,1INT 21HMOV [SI], ALINC SILOOP LOP1MOV SI,OFFSET BUFMOV CX,9MOV AL, [SI]LOP2:CMP AL, [SI+1]JA NEXT2XCHG AL,[SI+1]INC SINEXT2:LOOP LOP2MOV MAX,ALMOV SI,OFFSET BUFMOV CX,9MOV AL, [SI]LOP3: CMP AL, [SI+1]JB NEXT3XCHG AL,[SI+1]INC SINEXT3:LOOP LOP3MOV MIN,ALMOV AH,4CHINT 21HCODE ENDSEND STARTDATA SEGMENTBUFF DB 10 DUP(?)DATA ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATASTART: MOV AX,DATAMOV DS,AXMOV SI , OFFSET BUFFMOV CX,0AHLOP1: MOV AH,7INT 21HCMP AL,39。 COUNT DB $STR2NUM DB ?DATA_SEG ENDSCODE SEGMENT ASSUME CS: C
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