这是我用来让它成为一个常数表达式的，如果输入是一个常数表达式的话。

#define uptopow2_0(v) ((v) - 1)
#define uptopow2_1(v) (uptopow2_0(v) | uptopow2_0(v) >> 1)
#define uptopow2_2(v) (uptopow2_1(v) | uptopow2_1(v) >> 2)
#define uptopow2_3(v) (uptopow2_2(v) | uptopow2_2(v) >> 4)
#define uptopow2_4(v) (uptopow2_3(v) | uptopow2_3(v) >> 8)
#define uptopow2_5(v) (uptopow2_4(v) | uptopow2_4(v) >> 16)

#define uptopow2(v) (uptopow2_5(v) + 1)  /* this is the one programmer uses */

例如，这样的表达式:

uptopow2(sizeof (struct foo))

会很好地化简为常数。

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

试图为这个问题找到一个“终极”解决方案。下面的代码

针对的是C语言(不是c++)， 使用编译器内置生成有效的代码(CLZ或BSR指令)，如果编译器支持任何， 是便携式的(标准C和没有汇编)，除了内置，和 处理所有未定义的行为。

如果你用c++编写，你可以适当地调整代码。注意，c++ 20引入了std::bit_ceil，它做了完全相同的事情，只是在某些条件下行为可能是未定义的。

#include <limits.h>

#ifdef _MSC_VER
# if _MSC_VER >= 1400
/* _BitScanReverse is introduced in Visual C++ 2005 and requires
   <intrin.h> (also introduced in Visual C++ 2005). */
#include <intrin.h>
#pragma intrinsic(_BitScanReverse)
#pragma intrinsic(_BitScanReverse64)
#  define HAVE_BITSCANREVERSE 1
# endif
#endif

/* Macro indicating that the compiler supports __builtin_clz().
   The name HAVE_BUILTIN_CLZ seems to be the most common, but in some
   projects HAVE__BUILTIN_CLZ is used instead. */
#ifdef __has_builtin
# if __has_builtin(__builtin_clz)
#  define HAVE_BUILTIN_CLZ 1
# endif
#elif defined(__GNUC__)
# if (__GNUC__ > 3)
#  define HAVE_BUILTIN_CLZ 1
# elif defined(__GNUC_MINOR__)
#  if (__GNUC__ == 3 && __GNUC_MINOR__ >= 4)
#   define HAVE_BUILTIN_CLZ 1
#  endif
# endif
#endif

/**
 * Returns the smallest power of two that is not smaller than x.
 */
unsigned long int next_power_of_2_long(unsigned long int x)
{
    if (x <= 1) {
        return 1;
    }
    x--;

#ifdef HAVE_BITSCANREVERSE
    if (x > (ULONG_MAX >> 1)) {
        return 0;
    } else {
        unsigned long int index;
        (void) _BitScanReverse(&index, x);
        return (1UL << (index + 1));
    }
#elif defined(HAVE_BUILTIN_CLZ)
    if (x > (ULONG_MAX >> 1)) {
        return 0;
    }
    return (1UL << (sizeof(x) * CHAR_BIT - __builtin_clzl(x)));
#else
    /* Solution from "Bit Twiddling Hacks"
       <http://www.graphics.stanford.edu/~seander/bithacks.html#RoundUpPowerOf2>
       but converted to a loop for smaller code size.
       ("gcc -O3" will unroll this.) */
    {
        unsigned int shift;
        for (shift = 1; shift < sizeof(x) * CHAR_BIT; shift <<= 1) {
            x |= (x >> shift);
        }
    }
    return (x + 1);
#endif
}

unsigned int next_power_of_2(unsigned int x)
{
    if (x <= 1) {
        return 1;
    }
    x--;

#ifdef HAVE_BITSCANREVERSE
    if (x > (UINT_MAX >> 1)) {
        return 0;
    } else {
        unsigned long int index;
        (void) _BitScanReverse(&index, x);
        return (1U << (index + 1));
    }
#elif defined(HAVE_BUILTIN_CLZ)
    if (x > (UINT_MAX >> 1)) {
        return 0;
    }
    return (1U << (sizeof(x) * CHAR_BIT - __builtin_clz(x)));
#else
    {
        unsigned int shift;
        for (shift = 1; shift < sizeof(x) * CHAR_BIT; shift <<= 1) {
            x |= (x >> shift);
        }
    }
    return (x + 1);
#endif
}

unsigned long long next_power_of_2_long_long(unsigned long long x)
{
    if (x <= 1) {
        return 1;
    }
    x--;

#if (defined(HAVE_BITSCANREVERSE) && \
    ULLONG_MAX == 18446744073709551615ULL)
    if (x > (ULLONG_MAX >> 1)) {
        return 0;
    } else {
        /* assert(sizeof(__int64) == sizeof(long long)); */
        unsigned long int index;
        (void) _BitScanReverse64(&index, x);
        return (1ULL << (index + 1));
    }
#elif defined(HAVE_BUILTIN_CLZ)
    if (x > (ULLONG_MAX >> 1)) {
        return 0;
    }
    return (1ULL << (sizeof(x) * CHAR_BIT - __builtin_clzll(x)));
#else
    {
        unsigned int shift;
        for (shift = 1; shift < sizeof(x) * CHAR_BIT; shift <<= 1) {
            x |= (x >> shift);
        }
    }
    return (x + 1);
#endif
}

如果您正在使用GCC，您可能想要看看Lockless Inc.优化next_pow2()函数。本页描述了一种使用内置函数builtin_clz()(计数前导零)的方法，然后直接使用x86 (ia32)汇编指令bsr(位扫描反向)，就像它在另一个答案的游戏开发站点链接中所描述的那样。此代码可能比前面的回答中描述的更快。

顺便说一下，如果你不打算使用汇编指令和64位数据类型，你可以使用这个

/**
 * return the smallest power of two value
 * greater than x
 *
 * Input range:  [2..2147483648]
 * Output range: [2..2147483648]
 *
 */
__attribute__ ((const))
static inline uint32_t p2(uint32_t x)
{
#if 0
    assert(x > 1);
    assert(x <= ((UINT32_MAX/2) + 1));
#endif

    return 1 << (32 - __builtin_clz (x - 1));
}

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