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128，64，2，32，()
A．16
B．8
C．1
D．1/16
参考答案
更多 “ 128，64，2，32，()A．16B．8C．1D．1/16 ” 相关考题
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以下语句的输出结果是( ) printf("%d，%d，%u\n"，016，16，-16)；A．16，16，65420B．14，16，6542C．16，16，-16D．14，16，-16
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What are two rules for compacting IPv6 addresses?()
A. Every 16 - bit segment segment that consists of all zeroes can be represented with a single colon.B. The trailing zeroes in any 16 - bit segment do not have to be written.C. The leading zeroes in any 16 - bit segment do not have to be written.D. Any single, continuous string of one or more 16 - bit segments that consists of all zeroes can be represented with a double colon.E. The maximum number of times a double colon can replace a 16 - bit segment that consists of all zeroes is two.F. Two zeroes in the middle of any 16 -bit segment do not have to be written.
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WhataretworulesforcompactingIPv6addresses?()
A.Themaximumnumberoftimesadoublecoloncanreplacea16-bitsegmentthatconsistsofallzeroesistwoB.Theleadingzeroesinany16-bitsegmentdonothavetobewritten.C.Every16-bitsegmentthatconsistsofallzeroescanberepresentedwithasinglecolonD.Thetrailingzeroesinany16-bitsegmentdonothavetobewritten.E.Anysingle，continuousstringofoneormore16-bitsegmentsthatconsistsofallzeroescanberepresentedwithadoublecolonF.Twozeroesinthemiddleofany16-bitsegmentdonothavetobewritten
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填空： （1) （1234)10＝（)2＝（)16 （2) （34.6875)10＝（)2＝（)16 （3) （271.33)10＝（)2＝（)16 （4) （101011001001)2＝（)10＝（)16 （5) （1AB.E)16＝（)10＝（)2 （6) （10101010.0111)2＝（)10＝（)16
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（1011100011101.11）2转换成16进制数（）16A．（171D.C）16B．（17113.12）16C．（871D.3）16D．（171D.3）16
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