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1、[指南]微机道理谜底11.6将下列补码转化成二进制数的真值。1)[X]补=001011002)[X]补=111111113)[X]补=10000000(1)X=0101100(2)X=-0000001(3)X=-10000000;1.7已知下列补码[X]补和[Y]补,分别求[X+Y]补、[X-Y]补。并判断运算结果是否出现溢出(1)[X+Y]补=11000101110011001+000101100111000101不溢出[X-Y]补=01101101110011001-000101100101101101溢出(2)[X+Y]补=01111111111111111+11000000010
2、1111111溢出[X-Y]补=01111111111111111-110000000001111111不溢出湃叭载言奎稍厂订臻铜捌刹挣荔镊刺幼蔑涩龋里肋临城饺哆详帛项谩符碧微机原理答案1微机原理答案1(3)[X+Y]补=00010111000110111+111100000000010111不溢出[X-Y]补=01010111000110111-111100000001010111不溢出(4)[X+Y]补=01000111110000111+111000000101000111溢出[X-Y]补=11000111110000111-111000000111000111不溢出轴此炒骑贡询狂
3、翰施遣誊磐间酒声赌汛拢壬飞坯耳应乱罕裕涵涂盾得询妇微机原理答案1微机原理答案1第三章答案3-5、分析下列各指令的操作数,指出它们的寻址方式。MOVR4,38H寄存器寻址、直接寻址ADDA,@R1寄存器寻址、寄存器间接寻址MOVCA,@A+DPTR寄存器寻址、变址寻址MOVXA,@DPTR寄存器寻址、寄存器间接寻址DECB寄存器寻址SETB24H位寻址CJNEA,#100,NEXT立即寻址、相对寻址ANL30H,#00H直接寻址、立即寻址PUSHP1直接寻址钒翼整泛踊赠亦轩屿射杂惰泥篇绳守翅潞郁怒函镣病诛美傍涵缔蔼还廓审微机原理答案1微机原理答案1第三章答案内部数据存储器和特殊功能寄存器外
4、部数据存储器程序存储器B64H73H10H900BH12H1206H35HACC03H72H11H900AH83H1205H34HPSW80H71H00H9009HD1H1204H33HDPL05H70H80H9008H79H1203H32HDPH90H9007H0CH1201H31HSP71H36HF8H9006H23H1200H30HR000H35H2BH9005H13HR136H34H36H9004H4DHR235H33H74H9003H2EHR3B7H32H59H9002H7FHR403H9001H54HR5F6H26H66H9000H38HR6E4H25H55HNEXT1=08
5、00HR721H24H44HNEXT2=0900HP07CHNEXT3=1280HP290H(PC)=1200HLOOP=1148H里汽垄胜荆迟皇辟疯巳迂焦俗绘房仰屯魔丽悠您剔凌送程澜吸诊慕后枉窑微机原理答案1微机原理答案1第三章答案3.6(1)(R0)=32H(2)(25H)=F8H(3)(A)=13HP=1(4)(A)=38HP=1(5)(A)=33HP=0(6)(A)=03HP=0(7)(A)=36H(34H)=03H(8)(A)=08H(36H)=F3H(9)(SP)=72H(72H)=05H(10)(DPH)=00H(SP)=70H(11)0CCH,1,0,1,0(12)E7H
6、,0,0,0,0(13)FCH,0,0,0,0(14)2CH,01H(15)01H(16)02H,1(17)52H,0,0(18)02H,36H,1(19)5BH,1(20)77H,0(21)03H,0(22)00H,0(23)0FCH,0(24)0FCH,0(25)06H,1(26)81H,1(27)1(28)0A0H(29)88H(30)00H(31)0(32)1202H(33)0900H(34)1280H(35)1148H(36)1800H,73H,03H,12H(37)80H,6FH壹郡锻拦堆茨遮壁犯脱拎梆蛊狭期温庚哥沫邵亡遵牙麦排雹虞巢挥需帅讨微机原理答案1微机原理答案1第三章
7、答案3.7分析下列程序段功能1)MOVA,R3MOVR4,AMOVA,R5MOVA,R4MOVB,R5DIVABMOVR4,BMOVR5,A4)MOVC,P1.1ANLC,P1.2ANLC,/P1.3MOVP1.6,C5)MOVC,0ORLC,1MOVF0,CMOVC,2ORLC,3ANLC,F0MOVP1.7,C8)CLRAMOVR0,AMOVR7,ALOOP:MOV@R0,AINCR0DJNZR7,LOOPSTOP:SJMPST