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lua底层对log2求幂
思想
1 将256以内的数的log2结果存到数组,这部分可直接得到结果
2 反向思维,进行一次左移:1左移n位,就是2的n次方,(假设n=3,效果是左移之后,1的右边都是0,这些0任意修改后的值s,log2(s)的结果不变)故,s变换到一定程度之后才会引起n的变换,所以: l += 8; x >>= 8; 成立。
int luaO_log2 (unsigned int x) {
static const lu_byte log_2[256] = {
0,1,2,2,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,
6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,
7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,
7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8
};
int l = -1;
while (x >= 256) { l += 8; x >>= 8; }
return l + log_2[x];
}
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