c++ 14 clp2的constexpr版本

#include <iostream>
#include <type_traits>

// Closest least power of 2 minus 1. Returns 0 if n = 0.
template <typename UInt, std::enable_if_t<std::is_unsigned<UInt>::value,int> = 0>
  constexpr UInt clp2m1(UInt n, unsigned i = 1) noexcept
    { return i < sizeof(UInt) * 8 ? clp2m1(UInt(n | (n >> i)),i << 1) : n; }

/// Closest least power of 2 minus 1. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value && std::is_signed<Int>::value,int> = 0>
  constexpr auto clp2m1(Int n) noexcept
    { return clp2m1(std::make_unsigned_t<Int>(n <= 0 ? 0 : n)); }

/// Closest least power of 2. Returns 2^N: 2^(N-1) < n <= 2^N. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
  constexpr auto clp2(Int n) noexcept
    { return clp2m1(std::make_unsigned_t<Int>(n-1)) + 1; }

/// Next power of 2. Returns 2^N: 2^(N-1) <= n < 2^N. Returns 1 if n = 0. Returns 0 if n < 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
  constexpr auto np2(Int n) noexcept
    { return clp2m1(std::make_unsigned_t<Int>(n)) + 1; }

template <typename T>
  void test(T v) { std::cout << clp2(v) << std::endl; }

int main()
{
    test(-5);                          // 0
    test(0);                           // 0
    test(8);                           // 8
    test(31);                          // 32
    test(33);                          // 64
    test(789);                         // 1024
    test(char(260));                   // 4
    test(unsigned(-1) - 1);            // 0
    test<long long>(unsigned(-1) - 1); // 4294967296

    return 0;
}

如果你需要OpenGL相关的东西:

/* Compute the nearest power of 2 number that is 
 * less than or equal to the value passed in. 
 */
static GLuint 
nearestPower( GLuint value )
{
    int i = 1;

    if (value == 0) return -1;      /* Error! */
    for (;;) {
         if (value == 1) return i;
         else if (value == 3) return i*4;
         value >>= 1; i *= 2;
    }
}

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));
}

假设你有一个好的编译器&它可以做bit twiddling在这一点上我以上，但无论如何这是工作!!

    // http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogObvious
    #define SH1(v)  ((v-1) | ((v-1) >> 1))            // accidently came up w/ this...
    #define SH2(v)  ((v) | ((v) >> 2))
    #define SH4(v)  ((v) | ((v) >> 4))
    #define SH8(v)  ((v) | ((v) >> 8))
    #define SH16(v) ((v) | ((v) >> 16))
    #define OP(v) (SH16(SH8(SH4(SH2(SH1(v))))))         

    #define CB0(v)   ((v) - (((v) >> 1) & 0x55555555))
    #define CB1(v)   (((v) & 0x33333333) + (((v) >> 2) & 0x33333333))
    #define CB2(v)   ((((v) + ((v) >> 4) & 0xF0F0F0F) * 0x1010101) >> 24)
    #define CBSET(v) (CB2(CB1(CB0((v)))))
    #define FLOG2(v) (CBSET(OP(v)))

测试代码如下:

#include <iostream>

using namespace std;

// http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogObvious
#define SH1(v)  ((v-1) | ((v-1) >> 1))  // accidently guess this...
#define SH2(v)  ((v) | ((v) >> 2))
#define SH4(v)  ((v) | ((v) >> 4))
#define SH8(v)  ((v) | ((v) >> 8))
#define SH16(v) ((v) | ((v) >> 16))
#define OP(v) (SH16(SH8(SH4(SH2(SH1(v))))))         

#define CB0(v)   ((v) - (((v) >> 1) & 0x55555555))
#define CB1(v)   (((v) & 0x33333333) + (((v) >> 2) & 0x33333333))
#define CB2(v)   ((((v) + ((v) >> 4) & 0xF0F0F0F) * 0x1010101) >> 24)
#define CBSET(v) (CB2(CB1(CB0((v)))))
#define FLOG2(v) (CBSET(OP(v))) 

#define SZ4         FLOG2(4)
#define SZ6         FLOG2(6)
#define SZ7         FLOG2(7)
#define SZ8         FLOG2(8) 
#define SZ9         FLOG2(9)
#define SZ16        FLOG2(16)
#define SZ17        FLOG2(17)
#define SZ127       FLOG2(127)
#define SZ1023      FLOG2(1023)
#define SZ1024      FLOG2(1024)
#define SZ2_17      FLOG2((1ul << 17))  // 
#define SZ_LOG2     FLOG2(SZ)

#define DBG_PRINT(x) do { std::printf("Line:%-4d" "  %10s = %-10d\n", __LINE__, #x, x); } while(0);

uint32_t arrTble[FLOG2(63)];

int main(){
    int8_t n;

    DBG_PRINT(SZ4);    
    DBG_PRINT(SZ6);    
    DBG_PRINT(SZ7);    
    DBG_PRINT(SZ8);    
    DBG_PRINT(SZ9); 
    DBG_PRINT(SZ16);
    DBG_PRINT(SZ17);
    DBG_PRINT(SZ127);
    DBG_PRINT(SZ1023);
    DBG_PRINT(SZ1024);
    DBG_PRINT(SZ2_17);

    return(0);
}

输出:

Line:39           SZ4 = 2
Line:40           SZ6 = 3
Line:41           SZ7 = 3
Line:42           SZ8 = 3
Line:43           SZ9 = 4
Line:44          SZ16 = 4
Line:45          SZ17 = 5
Line:46         SZ127 = 7
Line:47        SZ1023 = 10
Line:48        SZ1024 = 10
Line:49        SZ2_16 = 17

这是我用c语言的解决方案，希望能有所帮助!

int next_power_of_two(int n) {
    int i = 0;
    for (--n; n > 0; n >>= 1) {
        i++;
    }
    return 1 << i;
}

c++ 14 clp2的constexpr版本

#include <iostream>
#include <type_traits>

// Closest least power of 2 minus 1. Returns 0 if n = 0.
template <typename UInt, std::enable_if_t<std::is_unsigned<UInt>::value,int> = 0>
  constexpr UInt clp2m1(UInt n, unsigned i = 1) noexcept
    { return i < sizeof(UInt) * 8 ? clp2m1(UInt(n | (n >> i)),i << 1) : n; }

/// Closest least power of 2 minus 1. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value && std::is_signed<Int>::value,int> = 0>
  constexpr auto clp2m1(Int n) noexcept
    { return clp2m1(std::make_unsigned_t<Int>(n <= 0 ? 0 : n)); }

/// Closest least power of 2. Returns 2^N: 2^(N-1) < n <= 2^N. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
  constexpr auto clp2(Int n) noexcept
    { return clp2m1(std::make_unsigned_t<Int>(n-1)) + 1; }

/// Next power of 2. Returns 2^N: 2^(N-1) <= n < 2^N. Returns 1 if n = 0. Returns 0 if n < 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
  constexpr auto np2(Int n) noexcept
    { return clp2m1(std::make_unsigned_t<Int>(n)) + 1; }

template <typename T>
  void test(T v) { std::cout << clp2(v) << std::endl; }

int main()
{
    test(-5);                          // 0
    test(0);                           // 0
    test(8);                           // 8
    test(31);                          // 32
    test(33);                          // 64
    test(789);                         // 1024
    test(char(260));                   // 4
    test(unsigned(-1) - 1);            // 0
    test<long long>(unsigned(-1) - 1); // 4294967296

    return 0;
}