Next Power of 2

Next Power of 2

Write a function that, for a given no n, finds a number p which be greater than or equal to N and is a power of 2.

    IP 5    op 8         IP +    op         ip +    32     

There is plenty of solutions for this. Let us take the example of explain some of them.

Method 1 (Using Log of the number)

    1.  Calculate Position of Set bit in P (next power of 2):        pos =  ceil (LGN)  (ceiling of log n with base 2) 
    
     2.  Now calculate P:        p   
    

Example

    Let us try for +            pos = 5            p   = 32    

Use the LG () function to pull up the whole

Method 2 (by getting the position of only set bit in result)

    /* If n is a power of 2 then return n    *  /1 if (N &! n& (n-1)) then return n     2  Else keep right shifting n until it becomes zero and         count no of shifts        a. I Nitialize:count = 0        B. While n! = 0                n = n>>1                count = count + 1/* Now count have the     position of set bit in result */    3  retur N (1 << count)  

Example:

    Let us try for                 page count = 5                 p     = 32   

The position of the highest 1 in the binary of the number n

1#include <iostream>2#include <ctime>3 using namespacestd;4 5UnsignedintNEXTPOWEROF2 (intN) {6         /*First n in the below condition are for the case where n is 0*/7     if(n &&!) (N & (n1)))8         returnN;9     intCNT =0;Ten      while(n! =0){ OneN >>=1; Acnt++; -     } -     return(1<<CNT); the }  -  - intMain () { -cout << NextPowerOf2 ( -) <<Endl; +cout << NextPowerOf2 ( -) <<Endl; -     return 0; +}

Reference: http://www.geeksforgeeks.org/next-power-of-2/

Next Power of 2