Approach 1:

we can use the knowledge of binary numbers.A well known property of numbers that are a power of 2 is that in binary representation they are represented by a single set bit , followed by some zeros.

So if we are given a number , if we can determine , the leftmost bit that is set ( let that be in position x ) , then a 1 followed by x zeros is the required number .

For ex :

Number = 103 ( binary equivalent : 1100111 )

Now counting from right to left , we find out that 7th bit is the last one that is set .

So the number 1 followed by 7 zeros ... i.e. 10000000 = 128 is the required solution .

Code:

int  nextPower_2 ( unsigned  int  x )

{

int  counter  =  0 ;

 while ( x > 0 )

{

  count ++ ;

  x  =  x  >>  1 ;

 }

return  (1 << counter) ;

}

Approach 2:

Another much easier approach exists to solve the above problem:

1. Start with some variable and initialize it with value 1.

2. Compare the variable with given number.

3. If variable is smaller than the number, do left shift.

4. Do the above step, until the variable becomes larger than the given number.

Code:

int  nextPower_2 ( unsigned  int  x )

{

int  value  =  1 ;

while  ( value  <=  x)

{

value  =  value  <<  1 ;

}

return  value ;

}

Approach 3:

Use of the functions implemented in Math library.

1. Find log2X , this would give us the power P , such that 2 ^ P = X.

2. Use the ceil function to get the next number.

3. Then, perform 2 ^ next number and that's the required result.

Code:

  int   nextPower_2 ( unsigned  int  x)

{

double nextnum = ceil  ( log2  x) ;

double result = pow  (2.0  ,  nextnum) ;

return  int ( result ) ;

}