为什么如果（N＆安培; -n）== N，然后按n为2的幂？ [英] Why if (n & -n) == n then n is a power of 2?

问题描述

线java.util.Random中源说的294

if ((n & -n) == n) // i.e., n is a power of 2
    // rest of the code

这是为什么？

推荐答案

的描述是不完全准确，因为（0安培; -0）== 0 0，但为不是两个电源。一个更好的办法可以说是

The description is not entirely accurate because (0 & -0) == 0 but 0 is not a power of two. A better way to say it is

（（N安培; -n）== n）的当n为二的幂，或两种，或零功率的负

((n & -n) == n) when n is a power of two, or the negative of a power of two, or zero.

如果n为二的幂，则n在二进制是一个1随后零。
-n以二的补是逆+ 1，因此位线高达从而

If n is a power of two, then n in binary is a single 1 followed by zeros. -n in two's complement is the inverse + 1 so the bits lines up thus

 n      0000100...000
-n      1111100...000
 n & -n 0000100...000

要明白为什么这项工作，可以考虑补的逆+ 1 -n ==〜N + 1

To see why this work, consider two's complement as inverse + 1, -n == ~n + 1

n          0000100...000
inverse n  1111011...111
                     + 1
two's comp 1111100...000

因为你通过携带一个一路增加一个时，为了获得二进制补码。

since you carry the one all the way through when adding one to get the two's complement.

如果N的以外的任何其他两个与匕首电源;那么结果将是缺少了一点，因为补不会有最高位设置，由于携带。

If n were anything other than a power of two† then the result would be missing a bit because the two's complement would not have the highest bit set due to that carry.

＆匕首​​; - 或零或二的幂的一个负...如在顶部说明

† - or zero or a negative of a power of two ... as explained at the top.

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