单片机第四章作业讲解1.doc

第四章作业讲解4－5求8个数的平均值设计思路：先用8次循环求8个数的和（和用两个字节表示），再将该和右移三位以实现除8运算。方法一：ORG1000HMOVR0,#30HMOVR7,#08HMOVR2,#00HMOVR3,#00HLOOP1:MOVA,R3ADDA,@R0MOVR3,AMOVA,R2ADDCA,#00HMOVR2,AINCR0DJNZR7,LOOP1;右移3位相当于除8MOVR7,#03HLOOP2:CLRCMOVA,R2RRCAMOVR2,AMOVA,R3RRCAMOVR3,ADJNZR7,LOOP2MOV38H,R3SJMP$END方法二：ORG0030HSTART:MOVR7,#7MOVR3,#0MOVA,30HMOVR0,#31HLOOP:ADDA,@R0JNCNEXTINCR3NEXT:INCR0DJNZR7,LOOPMOVR2,A;右移3位相当于除8MOVR7,#03HLOOP2:CLRCMOVA,R2RRCAMOVR2,AMOVA,R3RRCAMOVR3,ADJNZR7,LOOP2MOV38H,R3SJMP$END4-6把外部RAM的数据移入内部RAM方法一：ORG1000HMOVDPTR,#2000HMOVR7,#50HMOVR0,#30HLOOP:MOVXA,@DPTRMOV@R0,AINCR0INCDPLDJNZR7,LOOPSJMP$END方法二：MOVSP,#2FHMOVDPTR,#2000HMOVR7,#50HNEXT:MOVXA,@DPTRPUSHAINCDPL;或者INCDPTRDJNZR7,NEXTSJMP$END4－7求内部RAM数据块最小数ORG1000HBLOCKDATA30HLENDATA10HMINIDATA20HMOVR0,#BLOCKMOVR7,LENDECR7MOVA,@R0LOOOP:CLRCMOVB,ASUBBA,@R0JCNOP1MOVA,@R0SJMPNOP2NOP1:MOVA,BNOP2:INCR0DJNZR7,LOOPMOVMINI,ASJMP$END4-13SIGN函数ORG1000HVARDATA30HFUNCDATA31HMOVA,VARJNBACC.7,GOEMOVFUNC,#0FFHSJMPFINISHGOE:JNZGRMOVFUNC,#0SJMPFINISHGR:MOVFUNC,#1FINISH:SJMP$END注意：使用CJNE指令，会影响Cy位4-1616分支跳转JMP16:MOVA,R3RLAADDA,R3MOVDPTR,#PRGTBLJMP@A+DPTRPRGTBL:LJMPROOT00LJMPROOT01LJMPROOT02……………………………LJMPROOT15注意：LJMPAddr16和AJMPAddr11的区别4-18查平方表SQR:MOVA,R3;R3是入口参数，也是出口参数ANLA，#15ADDA,#02HMOVCA,@A+PCMOVR3,ARETSQRTAB:DB00,01,04,09,16DB25,36,49,64,81DB100,121,144,169,196测试芯片功能方法一：ORG1000HA1BITP1.0B1BITP1.1C1BITP1.2D1BITP1.3E1BITP1.4F1BITP1.5G1BITP1.6H1BITP1.7Y1BITP3.0TEST:MOVC,A1ANLC,B1ANLC,C1ANLC,D1MOVF0,CMOVC,E1ANLC,F1ANLC,G1ANLC,H1ORLC,F0CPLCMOVY1,CSJMP$END方法二：ORG1000HA1BITP1.0B1BITP1.1C1BITP1.2D1BITP1.3E1BITP1.4F1BITP1.5G1BITP1.6H1BITP1.7Y1BITP3.0TEST:MOVC,A1ANLC,B1ANLC,C1ANLC,D1JCLOOPMOVC,E1ANLC,F1ANLC,G1ANLC,H1JCLOWSETBY1SJMPNEXTLOW:CLRY1NEXT:SJMP$END