位的安全和模操作 [英] Bit hacking and modulo operation

问题描述

阅读本：<一href=\"http://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith64BitsDiv\">http://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith64BitsDiv

我来到了那句话：

最后一步，其由2 ^ 10涉及模数除法 - 1，具有
  每组10位合并在一起（从位置0-9的效果，
  10-19，20-29，...）中的64位值。

The last step, which involves modulus division by 2^10 - 1, has the effect of merging together each set of 10 bits (from positions 0-9, 10-19, 20-29, ...) in the 64-bit value.

（它是关于在一个数字反转比特）...

(it is about reversing the bits in a number)...

所以我做了一些计算：

reverted = (input * 0x0202020202ULL & 0x010884422010ULL) % 1023;

b = 74          :                                 01001010
b 
 * 0x0202020202 :       1000000010000000100000001000000010
   = 9494949494 :01001010010010100100101001001010010010100
  & 10884422010 :10000100010000100010000100010000000010000 
    = 84000010  :         10000100000000000000000000010000
  % 1023        :                               1111111111
    = 82        :                                 01010010

现在，这是有点不清楚的部分是那里的大数目由1023（2 ^ 10 - 1）模的零件包，给我倒位...我没有找到一点关系有什么好的文档操作和取模运算（旁边的x％的2 ^ N == X'放大器;（2 ^ N - 1））），所以也许如果有人施放这一点光将是非常富有成效的。

Now, the only part which is somewhat unclear is the part where the big number modulo by 1023 (2^10 - 1) packs and gives me the inverted bits... I did not find any good doc about relationship between bit operations and the modulo operation (beside x % 2^n == x & (2^n - 1))) so maybe if someone would cast a light on this it would be very fruitful.

推荐答案

模运算不给你倒位本身，它只是一个分级操作。

The modulo operation does not give you the inverted bits per se, it is just a binning operation.

B * 0x0202020202 = 01001010 01001010 01001010 01001010 01001010 0

b * 0x0202020202 = 01001010 01001010 01001010 01001010 01001010 0

乘法操作具有卷积特性，这意味着它复制输入变量几次（5在这里，因为它是一个8位字）。

The multiplication operation has a convolution property, which means it replicate the input variable several times (5 here since it's a 8-bit word).

这就是黑客的最棘手的部分。你要记住，我们是在一个8位字工作： B = ABCDEFGH ，其中[A-H]是1或0

That's the most tricky part of the hack. You have to remember that we are working on a 8-bit word : b = abcdefgh, where [a-h] are either 1 or 0.

b  * 0x0202020202 = abcdefghabcdefghabcdefghabcdefghabcdefgha
    & 10884422010 = a0000f000b0000g000c0000h000d00000000e0000

最后一行：字分级

模有特殊的属性： 10≡1（MOD 9）所以 100≡10 * 10≡10 * 1（MOD 9）≡ 1（MOD 9）。

更一般地，对于一个基地 B ， B≡1（MOD b - 1）那么所有数字 A≡总和（a_k * b ^ K）≡总和（a_k）（MOD b - 1）

More generally, for a base b, b ≡ 1 (mod b - 1) so for all number a ≡ sum(a_k*b^k) ≡ sum (a_k) (mod b - 1).

在这个例子中，基地= 1024 （10位）等等。

In the example, base = 1024 (10 bits) so

b ≡ a0000f000b0000g000c0000h000d00000000e0000 
  ≡ a*base^4 + 0000f000b0*base^3 + 000g000c00*base^2 + 00h000d000*base +00000e0000 
  ≡ a + 0000f000b0 + 000g000c00 + 00h000d000 + 00000e0000 (mod b - 1)
  ≡  000000000a
   + 0000f000b0 
   + 000g000c00 
   + 00h000d000 
   + 00000e0000 (mod b - 1)
 ≡   00hgfedcba (mod b - 1) since there is no carry (no overlap)

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