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

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

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

对于IEEE浮点，你可以这样做。

int next_power_of_two(float a_F){
    int f = *(int*)&a_F;
    int b = f << 9 != 0; // If we're a power of two this is 0, otherwise this is 1

    f >>= 23; // remove factional part of floating point number
    f -= 127; // subtract 127 (the bias) from the exponent

    // adds one to the exponent if were not a power of two, 
    // then raises our new exponent to the power of two again.
    return (1 << (f + b)); 
}

如果你需要一个整数的解决方案，并且你能够使用内联汇编，BSR会在x86上给你一个整数的log2。它计算有多少位是正确的，这正好等于这个数字的log2。其他处理器(通常)有类似的指令，比如CLZ，根据你的编译器，可能有一个内在的可用指令来为你做这项工作。

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)); }

next = pow(2, ceil(log(x)/log(2)));

这是通过找到你想要2乘以x的数字来实现的(取这个数字的对数，然后除以想要的底数的对数，详见维基百科)。然后把它四舍五入，得到最接近的整数幂。

这是一个比其他地方链接的按位方法更通用的方法(即更慢!)，但很好地了解数学，不是吗?