__fls 源码分析

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__fls 用于找到一个long型数最后哪个bit为1，例如0x1010的话，__fls 就返回12，表示最高为是第12bit
源码分析如下：

static __always_inline unsigned long __fls(unsigned long word)
{
   int num = BITS_PER_LONG - 1;

#if BITS_PER_LONG == 64
   if (!(word & (~0ul << 32))) {
       num -= 32;
       word <<= 32;
   }
#endif
   if (!(word & (~0ul << (BITS_PER_LONG-16)))) {
       num -= 16;
       word <<= 16;
   }
   if (!(word & (~0ul << (BITS_PER_LONG-8)))) {
       num -= 8;
       word <<= 8;
   }
   if (!(word & (~0ul << (BITS_PER_LONG-4)))) {
       num -= 4;
       word <<= 4;
   }
   if (!(word & (~0ul << (BITS_PER_LONG-2)))) {
       num -= 2;
       word <<= 2;
   }
   if (!(word & (~0ul << (BITS_PER_LONG-1))))
       num -= 1;
   return num;
}
假定BITS_PER_LONG ==32
则~0ul ==0xFFFFFFFF
第一步：!(word & (~0ul << (BITS_PER_LONG-16))) =！（0x1010 & 0xFFFF0000）=1，则num -=16=15.word = 0x1010 0000
第二步：!(word & (~0ul << (BITS_PER_LONG-8))) =！（0x1010 & 0xFF000000）=0，则num=15.word = 0x1010 0000
第三步：!(word & (~0ul << (BITS_PER_LONG-4))) =！（0x1010 & 0xc0000000）=0，则num=15.word = 0xf0000000
第三步：!(word & (~0ul << (BITS_PER_LONG-2))) =！（0x1010 & 0xc0000000）=1，则num=13.word = 0x40400000
第四步：!(word & (~0ul << (BITS_PER_LONG-2))) =！（0x1010 & 0x80000000）=1，则num=12.word = 0x40400000
所以最终返回12

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