ORG0100HMAIN:MOVR2,#10HMOVR3,#20H MOV R6,#0FHMOVR7,#0FFHMOVDPL,#00HMOVDPH,R2LOP1:MOVXA,@DPTRMOV DPH,R3MOVX@DPTR,AINCDPTRMOVR3,DPHCLRCMOVA,R3SUBBA,#10H

4.15編程將外部RAM的1000H～1FFFH區(qū)域的數(shù)據(jù)送到2000H～2FFFH區(qū)域。

MOVR2,AMOVDPH,R2MOVA,R7JNZLOP2ORLA,R6JZENDPDECR6LOP2:DECR7SJMPLOP1ENDP:SJMP $ENDBLOCK1EQU30HBLOCK2EQU40HORG0000HLJMPSTARTSTART:MOVR0,#BLOCK1MOVR1,#BLOCK2MOVR7,#1LP:MOVR6,#5CLRCLP1:MOVA,@R0SUBBA,@R1MOV@R0,AINCR0INCR1DJNZR6,LP1DJNZR7,LPSJMP$

END4.16已知一內(nèi)部RAM以BLOCK1和BLOCK2為起始地址的存儲(chǔ)區(qū)中分別有5字節(jié)無(wú)符號(hào)被減數(shù)和減數(shù)（低位在前，高位在后。請(qǐng)編寫(xiě)減法子程序令它們相減，并把差放入以BLOCK1為起始地址的存儲(chǔ)單元。THREEEQU1CHORG0000HLJMPSTARTSTART:MOVONE,#00HMOVTWO,#00HMOVTHREE,#00HMOVR0,#20HMOVR7,1FHLOOP3:MOVA,@R0INCR0JZLOOP1JBACC.7,LOOP2INCONESJMPHALTLOOP1:INCTWOLOOP2:INCTHREEHALT:DJNZR7,LOOP3

4.17從內(nèi)部RAM20H單元開(kāi)始存有一組帶符號(hào)數(shù)，其個(gè)數(shù)已存放在1FH單元中。要求統(tǒng)計(jì)出大于0、等于0和小于0的數(shù)的數(shù)目，并把統(tǒng)計(jì)結(jié)果分別存放在ONE、TWO、THREE三個(gè)單元中。4.18設(shè)內(nèi)部RAM30H單元有兩個(gè)非零的BCD數(shù)，請(qǐng)編寫(xiě)求兩個(gè)BCD數(shù)的積，并將積以壓縮型BCD數(shù)形式送入31H單元的程序。ORG0000H

MOV A,30H ANL A,#0F0H SWAP A MOV B,A MOV A,30H ANL A,#0FH MUL A,B MOV B,#0AH DIV AB SWAPA ORL A,B MOV 31H,A SJMP $4.20編制絕對(duì)值函數(shù)程序。絕對(duì)值函數(shù)方程如下：假設(shè)X存于30H單元，Y存于40H單元。X及Y均為補(bǔ)碼數(shù).

XEQU30HYEQU40HORG0000HLJMPSTARTSTART:MOVA,XJZLOOPJNBACC.7,LOOPCPLAADDA,#01HLOOP:MOVY,ASJMP$

END4.21試編寫(xiě)統(tǒng)計(jì)數(shù)據(jù)區(qū)長(zhǎng)度的程序，設(shè)數(shù)據(jù)區(qū)從內(nèi)RAM30H開(kāi)始，該數(shù)據(jù)區(qū)以0結(jié)束，統(tǒng)計(jì)結(jié)果送人2FH中。且數(shù)據(jù)區(qū)結(jié)束字符，也計(jì)入統(tǒng)計(jì)數(shù)據(jù)區(qū)的長(zhǎng)度中去.

ORG0000HLJMPSTARTSTART:MOVR0,#30HMOVR7,#00HLOOP:MOVA,@R0INCR0INCR7CJNEA,#00H,LOOPMOV2FH,R7SJMP$4.23巳知R7中為2位十六進(jìn)制數(shù)，試編程將其轉(zhuǎn)換為ASCII碼，存人內(nèi)部RAM31H、32H中（低字節(jié)在前）。英文字母為大寫(xiě).

ORG0000HLJMPSTARTSTART:MOVSP,#60HMOVA,R7LCALLBCDASCIIMOV31H,AMOVA,R7SWAPALCALLBCDASCIIMOV32H,ABCDASCII:ANLA,#0FHCJNEA,#10H,LOOPLOOP:JCAD30HADDA,#07HAD30H:ADDA,#30HRET

ENDORG0000HLJMPSTARTSTART:MOVDPTR,#TABMOVA,R7ANLA,#0FHMOVCA,@A+DPTRMOV31H,AMOVA,R7ANLA,#0F0HSWAPAMOVCA,@A+DPTRMOV32H,ASJMP$TAB:DB30H,31H,32H,33H,34H,35H,36H,37H,38H,39H,41H,42H,43H,44H,45H,46H

END4.24設(shè)在MA和MB單元中有兩個(gè)補(bǔ)碼形式的8位二進(jìn)制帶符號(hào)數(shù)。請(qǐng)編寫(xiě)求兩數(shù)之和并把它放在SUML和SUMH單元(低8位在SUML單元)的子程序。提示：在兩個(gè)8位二進(jìn)制帶符號(hào)數(shù)相加時(shí)，其和很可能會(huì)超過(guò)8位數(shù)所能表示的范圍，從而需要采用16位數(shù)形式來(lái)表示。因此，在進(jìn)行加法時(shí)、可以預(yù)先把這兩個(gè)加數(shù)擴(kuò)張成16位二進(jìn)制補(bǔ)碼形式，然后對(duì)它完成雙字節(jié)相加。

ORG0000HLJMPSTARTSTART:MOVA,MAJBACC.7,LLAMOVR7,#00HSJMPLLLLLA:MOVR7,#0FFHLLL:MOVA,MBJBACC.7,LLBMOVR6,#00HSJMPLADDLLB:MOVR6,#0FFHLADD:MOVA,MAADDA,MBMOVSUML,AMOVA,R6ADDCA,R7MOVSUMH,ASJMP$

4.25設(shè)外部RAM1000H～10FFH數(shù)據(jù)區(qū)中的數(shù)均為無(wú)符號(hào)數(shù)。試編寫(xiě)程序，找出該區(qū)域中的數(shù)的最大值，并放人內(nèi)部RAM30H單元中。方法1:

ORG0000HLJMPSTARTSTART:MOVDPTR,#1000HMOVR7,#0FFHMOVXA,@DPTRMOVB,ALOOP:INCDPTRMOVXA,@DPTRCJNEA,B,LOOP1LOOP1:JNCLOOP2MOVA,BLOOP2:DJNZR7,LOOPMOV30H,ASJMP$

4.25設(shè)外部RAM1000H～10FFH數(shù)據(jù)區(qū)中的數(shù)均為無(wú)符號(hào)數(shù)。試編寫(xiě)程序，找出該區(qū)域中的數(shù)的最大值，并放人內(nèi)部RAM30H單元中。方法2:MOV OPTR,#1000H MOV R7,#00H ;計(jì)數(shù)器

MOV B,#00H ;放中間結(jié)果L1: MOVX A,@DPTR CJNE A,B,L2L2: JC L3 MOV B,A ;B中放大值L3: INC DPTR DJNZ R7,L1 MOV 30H,A SJMP $4.28已知a、b、c均為0～9的整數(shù)，試編程求解表達(dá)式的值（要求使用子程序）。

AA EQU 31H BB EQU 32H CC EQU 33H Y EQU 34H ORG 0100H MOV SP,#60H MOV A,AA CLR C SUBB A,BB ACALL CHUCI MOV R6,A;……………. MOV A,BB CLR C SUBB A,CC ACALL CHUCI ADD A,R6 MOV R6,A;……………….

MOV A,CC CLR C SUBB A,AA ACALL CHUCI ADD A,R6 MOV R6,A MOV Y,R6 SJMP $;’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’;入口:項(xiàng)內(nèi)差在A中.;出口:平方值在A中CHUCI: JNB ACC.7,SQR CPL A;差為負(fù)則先求其相反數(shù)

ADD A,#01HSQR: ADD A,#01H MOVC A,@A+PC RETSQRTAB:DB0,1,4,9,16,25,36,49,64,81AAEQU31HBBEQU32HCCEQU33HYEQU34HORG0000HLJMPSTARTSTART:MOVSP,#60HMOVA,AACLRCSUBBA,BBACALLCHUNCIMOVR6,AMOVA,BBCLRCSUBBA,CCACALLCHUNCIADDA,R6MOVR6,A