Round up to Power of

This headline should explain what we are going to do, Chinese means finding a 2^n number so that it is not less than the given number. For example: If you give a number 63, then I need to get a number of not less than 63, but this number needs to be 2 of the n-th square, so

• 63 corresponds to 64 (2^6)
• 64 corresponds to still 64 (2^6)
• 100 corresponds to 128 (2^7)

Here's the problem:

How to calculate this result quickly?

Perhaps the first thing that comes to our eyes is the calculation of log or some other non-bitwise operation algorithms, which are not explained again, and look at how the JDK and Android source packages are calculated.

HashMap is a common data result, the bottom is the combination of arrays and linked lists, in order to be able to make the key evenly distributed, reduce collisions, hashmap capacity is 2^n, the capacity is 2^n another benefit is to calculate hashcode% size can be used with the operation instead of ( Implementation away can be viewed on Google), when we need to construct a specified capacity (recorded as Sizea) HashMap, HashMap help us calculate the Sizeb,sizeb to meet 2^n not less than Sizea.
Specifically implemented in Android Java.util.Collections:

 /** * Returns the smallest power of the >= its argument, with several caveats: * If the argument is Negativ e but not Integer.min_value, the method returns * zero. If the argument is > 2^30 or equal to Integer.min_value, the method * returns Integer.min_value.     If The argument is zero, the method returns * zero. * * @hide * *    Public   static int rounduptopoweroftwo(int i) {i--;//If input is a power of both, shift its high-order bit right.        //"Smear" the High-order bit all the the-the-right.I |= i >>>1; I |= i >>>2; I |= i >>>4; I |= i >>>8; I |= i >>> -;returni +1; }

and the implementation in the JDK source code:

     private static int roundUpToPowerOf2(int number) {//Assert number >= 0: "Number must be non-negative";        returnNumber >= maximum_capacity? Maximum_capacity: (Number >1) ? Integer.highestonebit (Number-1) <<1) :1; }Public   static int highestonebit(int i) {//HD, figure 3-1I |= (i >>1); I |= (i >>2); I |= (i >>4); I |= (i >>8); I |= (i >> -);returnI-(i >>>1); }

Do you understand the principle? can go to the little fat Xuan View principle explanation

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Round up to Power of