微機(jī)原理習(xí)題4答案

1、習(xí)題4：請編寫完整匯編程序：1. 內(nèi)存中以BUF單元開始存放8個16位二進(jìn)制數(shù)，試編程將8個數(shù)倒序后存放于BUF開始的單元，試編程；（提示：采用堆棧實(shí)現(xiàn)）DATA SEGMENTORG 0000HBUF DW 1111H,2222H,3333H,4444H,5555H,6666H,7777H,8888HCOUNT EQU ($-BUF)/2DATA ENDSSTACK1 SEGMENT STACK DW 256 DUP(0)STACK1 ENDSCODE SEGMENT ASSUME CS:CODE,SS:STACK1,DS:DATASTART: MOV AX,DATA MOV DS,AX M

2、OV AX,STACK1 MOV SS,AX LEA SI,BUF MOV CX,COUNTLOP1: MOVAX,SI PUSH AX INC SI INC SI LOOP LOP1 LEA SI,BUF MOV CX,COUNTLOP2: POP AXMOVSI,AX INC SI INC SI LOOP LOP2CODE ENDS END START2. 將8個16位無符號數(shù)相加，結(jié)果保存在32位無符號數(shù)SUM中；DATA SEGMENTBUF DW 1111H,2222H,3333H,4444H,5555H,6666H,7777H,8888HCOUNT EQU ($-BUF)/2SUM

3、 DD 0DATA ENDSSTACK1 SEGMENT STACK DW 100 DUP(0)STACK1 ENDSCODE SEGMENT ASSUME CS:CODE,DS:DATA,SS:STACK1START: MOV AX,DATA MOV DS,AX LEA BX,BUF MOV CX,COUNT MOV DX,0000HLOP: MOV AX,BX ADD WORD PTR SUM,AX ADC WORD PTR SUM+2,DX INC BX INC BX LOOP LOP MOV AX,4C00H INT 21HCODE ENDS END START3. 以十進(jìn)制形式在計(jì)算

4、機(jī)屏幕上顯示內(nèi)存中的一個8位有符號數(shù)，例如：若內(nèi)存單元中存放的數(shù)據(jù)為7FH，則在屏幕上顯示+127，若內(nèi)存單元存放的數(shù)據(jù)為0FFH，則應(yīng)在屏幕上顯示-1；DATA SEGMENT VARDB0FFH STR1 DB 'THE RESULT IS: $'DATA ENDSSS_SEG SEGMENT STACKDW100 DUP(0)SS_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA, SS:SS_SEGSTART:MOVAX, DATAMOVDS, AXLEADX,STR1MOVAH,09HINT21HMOVDL, '+

5、9;CMPVAR, 0 JGE NEXTMOVDL, '-'NEGVARNEXT:MOVAH,02HINT21HMOVAL, VARMOVBL, 10MOV CL,0LOP1: AND AH,0 DIV BL PUSH AX INC CL CMP AL,0 JNZ LOP1LOP2: POPAX MOV DL,AH ADD DL,30H MOV AH,2 INT 21H LOOP LOP2 MOV AX, 4C00HINT21HCODE ENDSEND START4. 從鍵盤輸入一個4位十進(jìn)制數(shù)，然后以16進(jìn)制形式顯示在屏幕上，試編程；例如：輸入1 0 2 4 在屏幕上應(yīng)該顯示

6、0400HDATA SEGMENT STR1 DB 'INPUT DATA: $'BUFDB20DB4DB4 DUP(?) STR2 DB 0AH,0DH,'THE RESULT IS: ','$'DATA ENDSSS_SEG SEGMENT STACKDB100 DUP(0)SS_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA, SS:SS_SEGSTART:MOVAX, DATAMOVDS, AXLEADX,STR1MOVAH,09HINT21HMOVAH,0AHLEADX,BUFINT21HMOV

7、CX,03HLEASI,BUF+2ANDBX,0HMOVDL,0AHLOP1:MOVAL,SISUBAL,30HPUSHCXLOP2:MULDL LOOP LOP2POPCXADDBX,AXINCSILOOP LOP1ANDCH,00HMOVCL,SISUBCL,30HADDBX,CXLEADX,STR2MOVAH,09HINT21HMOVAX,BX AND CH,00HMOVCL,04HMOVDH,04HMOVDL,00HAAA1:ANDAX,000FHPUSHAXDECDHINCDLSHRBX,CLMOVAX,BXCMPDH,0JAAAA1 MOV CL,DLBBB:POPDX CMPDL

8、,09HJBNEXTADDDL,07H NEXT: ADD DL,30H MOV AH,2 INT 21HLOOPBBBMOVDL,'H'MOVAH,02HINT21H MOVAX,4C00HINT21HCODEENDSENDSTART5. 數(shù)據(jù)段中存放有一個無符號字?jǐn)?shù)據(jù)VAR，將其轉(zhuǎn)換成非壓縮格式的BCD碼，存于BUF開始的單元中（高位在前）；例如：若VAR為0800H，則轉(zhuǎn)換后(BUF) = 20H(BUF+1) = 48HDATA SEGMENT PARAVARDW0800HBUFDB2 DUP(0)DATA ENDSSS_SEG SEGMENT STACKDW100 D

9、UP(0)SS_SEG ENDSCODE SEGMENT PARAASSUME CS:CODE, DS:DATA, SS:SS_SEGSTART:MOVAX, DATAMOVDS, AXMOVCX, 16MAIN1:SHLVAR, 1MOVBX, 4PUSHCXMOVCX, 5MAIN2:MOVAL, BUFBXADCAL, ALAAAMOVBUFBX, ALDECBXLOOPMAIN2POPCXLOOPMAIN1EXIT:MOVAX, 4C00HINT21HCODE ENDSEND START6. 內(nèi)存中以str1和str2開始分別存放了兩個字符串，結(jié)束符為NULL（ASCII碼為0），將s

10、tr2連接到str1后，形成1個字符串，并將連接后的字符串str1輸出到屏幕上；DATA SEGMENT STR1DB'GOOD MORNING,',00HSTRDB50 DUP(0)STR2DB'Mr WANG!',0AH,0DH,00HDATA ENDSSS_SEG SEGMENT STACKDW100 DUP(0)SS_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA, SS:SS_SEGSTART: MOV AX,DATAMOVDS,AX LEA SI,STR1MOVAL,SICMPAL,00HJEJP1JP2:I

11、NCSIMOVAL,SICMPAL,00HJAJP2JP1:MOVCX,01HLEABX,STR2MOVAH,BXMOVBX,AHCMPAH,00HJEJP3MOVSI,AHJP4:INCSIINCBXINCCXMOVAH,BXMOVSI,AHCMPAH,00HJAJP4JP3: INC SI MOV SI,BYTE PTR '$'LEADX,STR1MOVAH,09HINT21HMOVAX,4C00HINT21HCODEENDSENDSTART7. 統(tǒng)計(jì)10個有符號字節(jié)數(shù)中，大于0、小于0、等于0的個數(shù)，分別存放在NUM1、NUM2、NUM3三個變量中，并找出最大值、最小值分

12、別存放到MAX、MIN變量中，再求10個數(shù)的和，將結(jié)果存放到16位有符號數(shù)SUM中。DATA SEGMENT NUMDB0F0H,03H,0B4H,0AH,0AAH,00H,80H,7FH,99H,21HCOUNTEQU($-NUM)ORG0010HNUM1DB0NUM2DB0NUM3DB0MINDB0MAXDB0SUMDW0DATA ENDSSS_SEG SEGMENT STACKDW100 DUP(0)SS_SEG ENDSCODE SEGMENT ASSUME CS:CODE, DS:DATA, SS:SS_SEGSTART: MOV AX,DATAMOVDS,AXMOVSS,AXLEA

13、SI,NUMMOVCX,COUNTMOVBX,0000HMOVDX,0000HLOP:MOVAL,SIINCSICMPAL,0JGDAJLXIAOJEDENGDA:INCBHJMPAAA1XIAO:INCBLJMPAAA1DENG:INCDHAAA1:LOOPLOPLEASI,NUM1MOVSI,BHLEASI,NUM2MOVSI,BLLEASI,NUM3MOVSI,DHMOVCX,COUNT-1MAIN1: LEA BX,NUMPUSHCXMAIN2: MOV AL,BX INC BX CMP AL,BX JLE NEXT XCHG AL,BX MOV BX-1,ALNEXT:LOOPMAI

14、N2POPCXLOOPMAIN1LEASI,MIN MOV AL,NUM MOV SI,ALLEASI,MAX MOV BL,NUM+9 MOV SI,BL MOV SI,OFFSETNUM MOVCX,COUNT MOVAX,0LOP2: AND BX,0MOVBL,SIADDAX,BXINC SI LOOP LOP2LEASI,SUMMOVSI,AXMOVAX,4C00HINT21HCODEENDSENDSTART8 若程序的數(shù)據(jù)段定義如下，寫出各指令語句獨(dú)立執(zhí)行后的結(jié)果。DSEG SEGMENTDATA1 DB 10H, 20H, 30HDATA2 DW 10 DUP (?)String

15、 DB 123DSEG ENDS(1) MOV AL, DATA1 10H->AL(2) MOV BX, offset DATA2DATA2代表的首地址賦給BX(3) LEA SI, StringString代表的首地址賦給BXADD DI,SISI+DI->DI9 假設(shè)數(shù)據(jù)項(xiàng)定義如下：DATA1 DB HELLO!GOOD MORNINGDATA2 DB 20 DUP(?)用串操作指令編寫程序段，使其分別完成一下功能。（1） 從左到右將DATA1中的字符串傳送到DATA2中；LEASI,DATA1LEADI,DATA2MOVCX,20CLDREPMOVSB（2） 傳送完后，比較D

16、ATA1和DATA2中的內(nèi)容是否相同；LEASI,DATA1LEADI,DATA2MOVCX,20CLDREPECMPSB（3） 把DATA1中的第3個字節(jié)和第四個字節(jié)裝入AX;LEASI,DATA1ADDSI,2LODSW（4） 將AX的內(nèi)容存入DATA2+5開始的字節(jié)單元中；LEADI,DATA2ADDDI,5STOSW10 執(zhí)行下列指令后，AX寄存器中的內(nèi)容是多少？TABLE DW 10,20,30,40,50ENTRY DW 3MOV BX, Offset TABLEADD BX, ENTRYMOV AX,BXAX=1400H存儲器11 圖示以下數(shù)據(jù)段在存儲器中的存放形式；10H34H07H09H42H00H42H00H48H45H4CH4CH4FH21HCDHABH00H00HDATA SEGMENT DATA1DB 10H, 34H, 07H, 09H DATA2DW 2 DUP(42H) DATA3 DB HELLO!DATA4 EQU 12DATA5 DD ABCDHDATA ENDS12 閱讀下邊的程序段，試說明它實(shí)