怎么办＆QUOT;＆的indexOf QUOT;和＆QUOT; lastIndexOf＆QUOT;与位标志的整数？ （获取发现指数的*功率*） [英] How to do "indexOf" and "lastIndexOf" on an integer with bit flags? (Get the *power* of the found index)

问题描述

这是对我的<一个跟进href=\"http://stackoverflow.com/questions/23259194/why-is-1-plus-31-zeros-not-a-valid-integer-if-bit-flags-are-31-length-plus-one\">$p$pvious在，在那里我清理了一些大的误解位标志的问题。

This is a follow up on my previous question on bit flags, where I cleared up some big misconceptions.

我需要创建这些功能以发现含有零个或多个标志单个位标志在一个int：

I need to create these functions to find a single bit-flag in an int containing zero or more flags:

BitBinaryUtil
   int twoExponentOfHighestOneBit(int all_flags)

   int twoExponentOfHighestOneBitAtMost(int all_flags,
         int twoExponentOfTwo_max0Thr30Incl)

   int twoExponentOfLowestOneBit(int all_flags)

   int twoExponentOfLowestOneBitAtLeast(int all_flags,
         int twoExponentOfTwo_min0Thr30Incl)

这是大致相同的 String.indexOf 和 lastIndexOf ，只是它们返回的指数的两个发现位。例如（共31位）

Which are roughly similar to String.indexOf and lastIndexOf, except they return the exponent of two of the found-bit. For example (all 31 bits)

twoExponentOfHighestOneBit:
   1000000000000000000000000000000  -->  30 (2^30=1073741824)
   0000000000000000000000000000001  -->  0  (2^0=1)
   0000000000000100000000000000001  -->  18
   0000000000000000000000000000000  -->  -1

twoExponentOfLowestOneBit:
   1000000000000000000000000000000  -->  30
   0000000000000000000000000000001  -->  0
   0000000000000100000000000000001  -->  0

我怎样才能做到这一点？

How can I do this?

推荐答案

Integer类有类似的 Integer.highestOneBit 。
对于迭代和位操作类BitSet中提供了很好的功能了。
但整数就行了。

The class Integer has many bit functions like Integer.highestOneBit. For iteration and bit manipulation the class BitSet offers nice features too. But Integer will do.

是这样的：

int n = Integer.numberOfTrailingZeros(Integer.highestOneBit(x)) + 1;

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