Given an integer, how do I find the next largest power of two using bit-twiddling?

Question

If I have a integer number n, how can I find the next number k > n such that k = 2^i, with some i element of N by bitwise shifting or logic.

Example: If I have n = 123, how can I find k = 128, which is a power of two, and not 124 which is only divisible by two. This should be simple, but it eludes me.

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Related questions for some specific languages:

• C: Rounding up to next power of 2
• C++: Algorithm for finding the smallest power of two that's greater or equal to a given value (including C++20 std::bit_ceil())

Answer 1

For 32-bit integers, this is a simple and straightforward route:

unsigned int n;

n--;
n |= n >> 1;   // Divide by 2^k for consecutive doublings of k up to 32,
n |= n >> 2;   // and then or the results.
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;           // The result is a number of 1 bits equal to the number
               // of bits in the original number, plus 1. That's the
               // next highest power of 2.

Here's a more concrete example. Let's take the number 221, which is 11011101 in binary:

n--;           // 1101 1101 --> 1101 1100
n |= n >> 1;   // 1101 1100 | 0110 1110 = 1111 1110
n |= n >> 2;   // 1111 1110 | 0011 1111 = 1111 1111
n |= n >> 4;   // ...
n |= n >> 8;
n |= n >> 16;  // 1111 1111 | 1111 1111 = 1111 1111
n++;           // 1111 1111 --> 1 0000 0000

There's one bit in the ninth position, which represents 2^8, or 256, which is indeed the next largest power of 2. Each of the shifts overlaps all of the existing 1 bits in the number with some of the previously untouched zeroes, eventually producing a number of 1 bits equal to the number of bits in the original number. Adding one to that value produces a new power of 2.

Another example; we'll use 131, which is 10000011 in binary:

n--;           // 1000 0011 --> 1000 0010
n |= n >> 1;   // 1000 0010 | 0100 0001 = 1100 0011
n |= n >> 2;   // 1100 0011 | 0011 0000 = 1111 0011
n |= n >> 4;   // 1111 0011 | 0000 1111 = 1111 1111
n |= n >> 8;   // ... (At this point all bits are 1, so further bitwise-or
n |= n >> 16;  //      operations produce no effect.)
n++;           // 1111 1111 --> 1 0000 0000

And indeed, 256 is the next highest power of 2 from 131.

If the number of bits used to represent the integer is itself a power of 2, you can continue to extend this technique efficiently and indefinitely (for example, add a n >> 32 line for 64-bit integers).

Answer 2

There is actually a assembly solution for this (since the 80386 instruction set).

You can use the BSR (Bit Scan Reverse) instruction to scan for the most significant bit in your integer.

bsr scans the bits, starting at the most significant bit, in the doubleword operand or the second word. If the bits are all zero, ZF is cleared. Otherwise, ZF is set and the bit index of the first set bit found, while scanning in the reverse direction, is loaded into the destination register

(Extracted from: http://dlc.sun.com/pdf/802-1948/802-1948.pdf)

And than inc the result with 1.

so:

bsr ecx, eax  //eax = number
jz  @zero
mov eax, 2    // result set the second bit (instead of a inc ecx)
shl eax, ecx  // and move it ecx times to the left
ret           // result is in eax

@zero:
xor eax, eax
ret

In newer CPU's you can use the much faster lzcnt instruction (aka rep bsr). lzcnt does its job in a single cycle.

Answer 3

A more mathematical way, without loops:

public static int ByLogs(int n)
{
    double y = Math.Floor(Math.Log(n, 2));

    return (int)Math.Pow(2, y + 1);
}

Answer 4

Here's a logic answer:

function getK(int n)
{
  int k = 1;
  while (k < n)
    k *= 2;
  return k;
}

Answer 5

Here's John Feminella's answer implemented as a loop so it can handle Python's long integers:

def next_power_of_2(n):
    """
    Return next power of 2 greater than or equal to n
    """
    n -= 1 # greater than OR EQUAL TO n
    shift = 1
    while (n+1) & n: # n+1 is not a power of 2 yet
        n |= n >> shift
        shift <<= 1
    return n + 1

It also returns faster if n is already a power of 2.

For Python >2.7, this is simpler and faster for most N:

def next_power_of_2(n):
    """
    Return next power of 2 greater than or equal to n
    """
    return 2**(n-1).bit_length()

Answer 6

This answer is based on constexpr to prevent any computing at runtime when the function parameter is passed as const

Greater than / Greater than or equal to

The following snippets are for the next number k > n such that k = 2^i
(n=123 => k=128, n=128 => k=256) as specified by OP.

If you want the smallest power of 2 greater than OR equal to n then just replace __builtin_clzll(n) by __builtin_clzll(n-1) in the following snippets.

C++11 using GCC or Clang (64 bits)

#include <cstdint>  // uint64_t

constexpr uint64_t nextPowerOfTwo64 (uint64_t n)
{
    return 1ULL << (sizeof(uint64_t) * 8 - __builtin_clzll(n));
}

Enhancement using CHAR_BIT as proposed by martinec

#include <cstdint>

constexpr uint64_t nextPowerOfTwo64 (uint64_t n)
{
    return 1ULL << (sizeof(uint64_t) * CHAR_BIT - __builtin_clzll(n));
}

C++17 using GCC or Clang (from 8 to 128 bits)

#include <cstdint>

template <typename T>
constexpr T nextPowerOfTwo64 (T n)
{
   T clz = 0;
   if constexpr (sizeof(T) <= 32)
      clz = __builtin_clzl(n); // unsigned long
   else if (sizeof(T) <= 64)
      clz = __builtin_clzll(n); // unsigned long long
   else { // See https://stackoverflow.com/a/40528716
      uint64_t hi = n >> 64;
      uint64_t lo = (hi == 0) ? n : -1ULL;
      clz = _lzcnt_u64(hi) + _lzcnt_u64(lo);
   }
   return T{1} << (CHAR_BIT * sizeof(T) - clz);
}

Other compilers

If you use a compiler other than GCC or Clang, please visit the Wikipedia page listing the Count Leading Zeroes bitwise functions:

• Visual C++ 2005 => Replace __builtin_clzl() by _BitScanForward()
• Visual C++ 2008 => Replace __builtin_clzl() by __lzcnt()
• icc => Replace __builtin_clzl() by _bit_scan_forward
• GHC (Haskell) => Replace __builtin_clzl() by countLeadingZeros()

Contribution welcome

Please propose improvements within the comments. Also propose alternative for the compiler you use, or your programming language...

See also similar answers

• nulleight's answer
• ydroneaud's answer

Answer 7

Here's a wild one that has no loops, but uses an intermediate float.

//  compute k = nextpowerof2(n)

if (n > 1) 
{
  float f = (float) n;
  unsigned int const t = 1U << ((*(unsigned int *)&f >> 23) - 0x7f);
  k = t << (t < n);
}
else k = 1;

This, and many other bit-twiddling hacks, including the on submitted by John Feminella, can be found here.

Answer 8

If you use GCC, MinGW or Clang:

template <typename T>
T nextPow2(T in)
{
  return (in & (T)(in - 1)) ? (1U << (sizeof(T) * 8 - __builtin_clz(in))) : in;
}

If you use Microsoft Visual C++, use function _BitScanForward() to replace __builtin_clz().

Answer 9

assume x is not negative.

int pot = Integer.highestOneBit(x);
if (pot != x) {
    pot *= 2;
}

Answer 10

function Pow2Thing(int n)
{
    x = 1;
    while (n>0)
    {
        n/=2;
        x*=2;
    }
    return x;
}

Answer 11

Bit-twiddling, you say?

long int pow_2_ceil(long int t) {
    if (t == 0) return 1;
    if (t != (t & -t)) {
        do {
            t -= t & -t;
        } while (t != (t & -t));
        t <<= 1;
    }
    return t;
}

Each loop strips the least-significant 1-bit directly. N.B. This only works where signed numbers are encoded in two's complement.

Answer 12

What about something like this:

int pot = 1;
for (int i = 0; i < 31; i++, pot <<= 1)
    if (pot >= x)
        break;

Answer 13

You just need to find the most significant bit and shift it left once. Here's a Python implementation. I think x86 has an instruction to get the MSB, but here I'm implementing it all in straight Python. Once you have the MSB it's easy.

>>> def msb(n):
...     result = -1
...     index = 0
...     while n:
...         bit = 1 << index
...         if bit & n:
...             result = index
...             n &= ~bit
...         index += 1
...     return result
...
>>> def next_pow(n):
...     return 1 << (msb(n) + 1)
...
>>> next_pow(1)
2
>>> next_pow(2)
4
>>> next_pow(3)
4
>>> next_pow(4)
8
>>> next_pow(123)
128
>>> next_pow(222)
256
>>>

Answer 14

Forget this! It uses loop !

     unsigned int nextPowerOf2 ( unsigned int u)
     {
         unsigned int v = 0x80000000; // supposed 32-bit unsigned int

         if (u < v) {
            while (v > u) v = v >> 1;
         }
         return (v << 1);  // return 0 if number is too big
     }

Answer 15

private static int nextHighestPower(int number){
    if((number & number-1)==0){
        return number;
    }
    else{
        int count=0;
        while(number!=0){
            number=number>>1;
            count++;
        }
        return 1<<count;
    }
}

Answer 16

#define nextPowerOf2(x, n) (x + (n-1)) & ~(n-1)

or even

#define nextPowerOf2(x, n)  x + (x & (n-1))