`(x+7)&~7`，把x增大为稍大于x的8的倍数。这怎么理解？

Question

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。

Answer 1

7 = 00000111b
& is bitwise AND
so the x&7 has the only 3 low bits of x.

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~7 = 11111000b

so the x&~7 will set x's 3 low bits to 0, and it is the multiple of 8.

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(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^m
because only 2^m-1 has the type : high bits must be 1, low bits must be 0, not intersected.

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