许多处理器架构都支持log以2为底或非常类似的操作——计数前导零。许多编译器都有针对它的内在特性。参见https://en.wikipedia.org/wiki/Find_first_set

from math import ceil, log2
pot_ceil = lambda N: 0x1 << ceil(log2(N))

测试:

for i in range(10):
    print(i, pot_ceil(i))

输出:

1 1
2 2
3 4
4 4
5 8
6 8
7 8
8 8
9 16
10 16

import sys


def is_power2(x):
    return x > 0 and ((x & (x - 1)) == 0)


def find_nearest_power2(x):
    if x <= 0:
        raise ValueError("invalid input")
    if is_power2(x):
        return x
    else:
        bits = get_bits(x)
        upper = 1 << (bits)
        lower = 1 << (bits - 1)
        mid = (upper + lower) // 2
        if (x - mid) > 0:
            return upper
        else:
            return lower


def get_bits(x):
    """return number of bits in binary representation"""
    if x < 0:
        raise ValueError("invalid input: input should be positive integer")
    count = 0
    while (x != 0):
        try:
            x = x >> 1
        except TypeError as error:
            print(error, "input should be of type integer")
            sys.exit(1)
        count += 1
    return count

g++编译器提供了一个内置函数__builtin_clz，用于计算前导零:

所以我们可以这样做:

int nextPowerOfTwo(unsigned int x) {
  return 1 << sizeof(x)*8 - __builtin_clz(x);
}

int main () {
  std::cout << nextPowerOfTwo(7)  << std::endl;
  std::cout << nextPowerOfTwo(31) << std::endl;
  std::cout << nextPowerOfTwo(33) << std::endl;
  std::cout << nextPowerOfTwo(8)  << std::endl;
  std::cout << nextPowerOfTwo(91) << std::endl;
  
  return 0;
}

结果:

8
32
64
16
128

但请注意，对于x == 0， __builtin_clz return是未定义的。

如果您想要单行模板。在这里

int nxt_po2(int n) { return 1 + (n|=(n|=(n|=(n|=(n|=(n-=1)>>1)>>2)>>4)>>8)>>16); }

or

int nxt_po2(int n) { return 1 + (n|=(n|=(n|=(n|=(n|=(n-=1)>>(1<<0))>>(1<<1))>>(1<<2))>>(1<<3))>>(1<<4)); }

unsigned long upper_power_of_two(unsigned long v)
{
    v--;
    v |= v >> 1;
    v |= v >> 2;
    v |= v >> 4;
    v |= v >> 8;
    v |= v >> 16;
    v++;
    return v;

}