二进制和八进制与十六进制之间是如何换算的?
我想知道它们换算的具体方法。
推荐回答(2个)
最简单的方法:
如二进制的“10110111011”:
换八进制时,从右到左,三位一组,不够补0,即成了:
010 110 111 011
然后每组中的3个数分别对应4、2、1的状态,然后将为状态为1的相加【备注】,如:
010 = 2
110 = 4+2 = 6
111 = 4+2+1 = 7
011 = 2+1 = 3
结果为:2673
而换十六进制时其实也类似,只要每组4位,分别对应8、4、2、1就行了,如分解为:
0101 1011 1011
运算为:
0101 = 4+1 = 5
1011 = 8+2+1 = 11(由于10为A,所以11即B)
1011 = 8+2+1 = 11(由于10为A,所以11即B)
结果为:5BB
注意,二进制数转化为八进制数、十六进制数时,对于分组,整数部分是从右向左分组,不够在左边补0,而小数部分则是从左向右分组,不够在右边补0,
如二进制小数:
0.1100111
转化为八进制数,应分组为:0.110 011 100 ,转化后为0.634
转化为十六进制数,应分组为:0.1100 1110,转化后为0.CD
备注:常见进制数之间的算关系表:
二 …… 八……十……十六(进制)
0000 ……0 ……0 ……0
0001 ……1…… 1…… 1
0010 ……2…… 2…… 2
0011 ……3…… 3…… 3
0100 ……4 ……4 ……4
0101 ……5 ……5 ……5
0110 ……6 ……6 ……6
0111 ……7 ……7 ……7
1000 ……10……8 ……8
1001 ……11……9 ……9
1010 ……12……10……A
1011 ……13……11……B
1100 ……14……12……C
1101 ……15……13……D
1110 ……16……14……E
1111 ……17……15……F
1 、转化为八进制的方法
比如10 110 101,从低位开始,每三位一组,最后一组如果不足三位,则前补零,将每组3位二进制数分别转换成相应的1位八进制即可,上面的数转化之后就是265
2、转化成十六进制
同理,只不过是每四位一组,也是从低位开始
如,101 1110 1000 1010,转化之后是5E8A
你那么聪明,一定学得会的,哈哈
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