我认为这也是可行的:

int power = 1;
while(power < x)
    power*=2;

答案就是力量。

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;

}

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

你可能会发现以下的澄清有助于达到你的目的:

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;
    }
}

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;
}