stl源码剖析的第61页有上面的位运算，不过我不了解，求证明。
闭关修行中......
7 = 00000111b& is bitwise ANDso the x&7 has the only 3 low bits of x.
~7 = 11111000b
so the x&~7 will set x's 3 low bits to 0, and it is the multiple of 8.
(x+n-1)&(n-1)，把x增大稍大于x的n的倍数，前提是n必须为 2^mbecause only 2^m-1 has the type : high bits must be 1, low bits must be 0, not intersected.
(x+n-1)&(n-1)
2^m-1
subnet mask 子网掩码
7 = 00000111b
& is bitwise AND
so the x&7 has the only 3 low bits of x.
~7 = 11111000b
so the x&~7 will set x's 3 low bits to 0, and it is the multiple of 8.
(x+n-1)&(n-1)，把x增大稍大于x的n的倍数，前提是n必须为 2^mbecause only
2^m-1has the type : high bits must be 1, low bits must be 0, not intersected.subnet mask 子网掩码