[x]双补=x1＞x≥0２２+x0＞x≥–1（mod4）[x]双补=0，x2n＞x≥02n+２+x0＞x≥2n（mod2n+２）[A]补=０0.1010[A]补=１1.0001[B]补=０0.1001[A+B]补=０1.0011[B]补=１1.0101[A+B]补=１0.0110UsingDoublesignsTwosComplementinoverflowdetectionMethod2:双符号位法溢出的检测方法00,×××××11,×××××10,×××××01,×××××whentwosignbitsoftheresultaresamethereisnooverflowsumisnegativesumispositivewhentwosignbitsoftheresultaredifferentthereisoverflowNegativeoverflowPositiveoverflowMSBisthesignoftheresult溢出的检测方法CarryInCarryInCarryOutCarryOutALU31ALU30…result31(符号位)result30(符号位)CarryIn29CarryIn31CarryIn30a31a30b31b30OverflowMethod2:双符号位法UsingDoublesignsTwosComplementinoverflowdetectionwhentwosignbitsoftheresultaredifferentthereisoverflowwhentwosignbitsoftheresultaresamethereisnooverflowOverflow=1，溢出；0，没有溢出注：最高两位是符号位OverflowDetectionLogic(溢出检测逻辑)A-B=A+(–B)=A+B+1formtwocomplementbyinvertandaddoneSoweonlyneedadditionandcomplementcircuitsTaketwoscomplimentofsubtrahendandaddtominuend-i.e.a-b=a+(-b)HowtoperformSubtractionontheALUweconstructedbefore?abCarryInCarryOutresultabCarryOutCarryIn＋Operation210OperationresultCarryInCarryOutALU01BinvertBinvertHowtoperformSubtractionontheALU

weconstructedbefore?WhenBinvert=1inputsoffulladderareaandb已有全加器加入输入为Ba0CarryIn=1result0Operationresult1CarryInALU0CarryOutCarryInCarryInCarryInCarryOutCarryOutCarryOutALU1ALU2ALU31b0result2result31a1a2a31b1b2b31……Binvert=1CarryIn=1Binvert=1Subtractionon32-bitALUForSubtractionA-B=A+(-B)=A+B+1a0CarryInresult0Operationresult1CarryInALU0CarryOutCarryInCarryInCarryInCarryOutCarryOutCarryOutALU1ALU2AL